fall 2012 math 8230 (vector bundles) lecture notes

FALL 2012 MATH 8230 (VECTOR BUNDLES) LECTURE NOTES

1. DEFINITIONS: VECTOR BUNDLES AND STRUCTURE GROUPS

A vector bundle over a topological space M (or “with base space M”) is, essentially, familyof vector spaces continuously parametrized by M . (I’m using the letter M to denote the basespace of the vector bundle as a concession to the fact that in most of the applications we’ll beinterested in the base space will be a smooth manifold; however for basic definitions and resultsit there is no need to restrict to this case.) One way of making this precise is as follows:

Definition 1.1. A (real, rank-k) vector bundle over a topological space M is a continuous mapπ: E→ M where E is a topological space such that, for all m ∈ M :

(i) the “fiber” Em := π−1(m) is equipped with the structure of a vector space over R.

(ii) There is an neighborhood U ⊂ M of m and a “local trivialization” Φ : π−1(U)→ U ×Rk

which, for each x ∈ U , maps the fiber Ex to x ×Rk by a linear isomorphism.

Remark 1.2. In practice one says things like, “Let E be a vector bundle over M” instead of “Letπ: E→ M” be a vector bundle over M . But one should keep in mind that, even so, the “vectorbundle E” refers not just to the topological space but also to the projection map and the localtrivializations.

Example 1.3. There is one very obvious example (for any M and any k): Simply take E = M×Rk

and let π be the projection. For item (ii) in Definition 1.1 one can just set U = M (for any

m ∈ M), so that π−1(M) = M ×Rk, and let Φ : π−1(M)→ M ×Rk be the identity map.Appropriately, this bundle is called the “trivial bundle of rank k over M .”

Definition 1.4. • Let πE : E → M and πF : F → M be two vector bundles over M . Anisomorphism from E to F is a homeomorphism Ψ : E → F such that for each m ∈M Ψ restricts to Em as a linear map (and hence a linear isomorphism, since Ψ is ahomeomorphism) from Em to Fm.• The vector bundleπ: E→ M is called trivial it is isomorphic to the trivial rank-k bundle

over M for some k.

Exercise 1.5. Suppose that πE : E→ M and πF : F → M are two vector bundles and Ψ : E→ F

is a continuous map that maps each fiber Em to Fm by a linear isomorphism. Prove that Ψ is an

isomorphism. (The nontrivial part of this is showing that Ψ−1 is continuous. Hint: By restrictingto appropriate local trivializations, you can reduce to the case where Ψ is a continuous map of

the trivial bundle U × Rk to itself which restricts to each x × Rk as a linear automorphism,

and you should be able to show then that the fact that Ψ is continuous implies that Ψ−1 is alsocontinuous.)

At first glance one might not be sure whether there exist any nontrivial bundles. A firstexample of one is given by the following.

Example 1.6. Let E denote the topological space which can be expressed as a quotient in eitherof the following two equivalent ways:

E =[0, 1]×R

(1, t)∼ (0,−t)=

R×R

(s+ 1, t)∼ (s,−t).

1

2 FALL 2012 MATH 8230 (VECTOR BUNDLES) LECTURE NOTES

You should recognize this topological space as an (open, i.e. with no boundary) Möbius strip.

Where we identify the circle S1 with R/Z, there is a continuous map π: E → S1 defined by

π([s, t]) = [s]. It is not hard to see that this is a rank-1 vector bundle over S1. Indeed, using

the second expression for E above, for any [s0] ∈ S1 (where s0 ∈ R), we have π−1([s0]) =

[s0, t]|t ∈ R, which is naturally identified with R and so inherits the vector space structure ofR. To construct a local trivialization around s0, let U = [s]|s0 − 1/2 < s < s0 + 1/2, and then

define Φ : π−1(U) → U × R by Φ([(s, t)]) = ([s], t) for s0 − 1/2 < s < s0 + 1/2. This is easilyseen to be a homeomorphism (it is well-defined and injective because s is taken from an openinterval of length only one) which satisfies the requirements of a local trivialization.

There are various ways of seeing that this “Möbius bundle” is not trivial. For one thing, if itwere trivial then the open Möbius strip would be homeomorphic to the open cylinder, which itis not, though showing this is not exactly straightforward (but perhaps already familiar to you).There is also a somewhat easier argument (which is best phrased in terms of the languageof the upcoming Definition 1.10) in which one starts from a system of local trivializations forthe Möbius bundle and derives a contradiction from the way that a hypothetical isomorphismto the trivial bundle would have to act with respect to the transition functions for the localtrivializations.

Example 1.7. As one learns in an introductory smooth manifolds course, for any smooth man-ifold M there is a naturally associated vector bundle with rank equal to dim M , namely thetangent bundle π: T M → M , whose fiber at a point m ∈ M is the tangent space TmM . (Thus,loosely speaking, TmM consists of all possible velocity vectors of curves passing through m. Fora review of this see any introductory text on smooth manifolds, e.g. [Lee, Chapter 3], or [U1,Section 3.1]).

Exercise 1.8. Show that TS1→ S1 is trivial, but that the Möbius bundle over S1 is not.

Remark 1.9. In fact, every rank-1 vector bundle over S1 is either trivial or isomorphic to theMöbius bundle, as we should be able to prove within the next few weeks.

In general, a smooth manifold M such that T M is trivial is called parallelizable. The example

of S1 should not be considered representative, as most manifolds are not parallelizable—the

only spheres with this property are S0, S1, S3, and S7, as was proven by Kervaire and Bott-Milnor in the late 1950s.

I’d now like to give an equivalent definition of a vector bundle, which will have some usefulgeneralizations when we wish to speak of vector bundles carrying some additional structure.

Definition 1.10. A (real, rank-k) vector bundle over a topological space M is a continuous mapπ: E→ M where E is a topological space such that there is an open cover Uαα∈A of M and a

collection of “local trivializations” Φα : π−1(Uα)→ Uα ×Rk such that

(i) Each Φα is a homeomorphism, which for every x ∈ Uα maps π−1(x) to x ×Rk.

(ii) For each α,β ∈ A, the map Φβ Φ−1α : (Uα ∩ Uβ )×R

k→ (Uα ∩ Uβ)×Rk has the form

Φβ Φ−1α (x , v) = (x , gαβ (x)v)

where for every x ∈ Uα ∩ Uβ , gαβ(x): Rk→ Rk is a linear map.

This definition is clearly closely related to Definition 1.1; in particular it should be clear that avector bundle in the sense of Definition 1.1 gives a vector bundle in the sense of Definition 1.10(just take the open cover to consist of all the sets U in Definition 1.1 obtained for various valuesof m ∈ M). Conversely, given a vector bundle in the sense of Definition 1.10 and m ∈ M , one canchoose α so that x ∈ Uα and then Φα is an obvious candidate for the local trivialization around m

FALL 2012 MATH 8230 (VECTOR BUNDLES) LECTURE NOTES 3

required in Definition 1.1(ii). But perhaps at this point one notices a difference between the twodefinitions, namely that in Definition 1.1 I assumed that the fibers Em were all vector spaces andthat the local trivializations were fiberwise linear, whereas I didn’t say anything about a vectorspace structure on the fibers in Definition 1.10. This discrepancy is resolved by the followingproposition, whose proof generalizes in important ways and so should be paid close attention.

Proposition 1.11. Given a vector bundle in the sense of Definition 1.10, for each m ∈ M there is

a unique vector space structure on the fiber Em = π−1(m) such that for every α ∈ A with m ∈ Uα

the local trivialization restricts as a linear isomorphism Φα|Em: Em→ m ×R

k.

Proof. For any v, w ∈ Em and c ∈ R, our job is to define v + w, cv ∈ Em in such a way thatthe vector space axioms are satisfied and such that each of the Φα restrict to Em as a linear

isomorphism to Rk.If we fix an α such that m ∈ Uα, there is one and only one way of trying to do this: If v ∈ Em

we have Φα(v) = (m, vα) for some unique vα ∈ Rk, and if Φα is to be a linear isomorphism we

must define addition +α and scalar multiplication ·α by

v +α w = Φ−1α (m, vα + wα) c ·α v = Φ−1

α (m, cvα).

These are well-defined algebraic operations on Em which inherit the vector space axioms from

m×Rk because we have defined the operations by forcing the bijection Φα|Em: Em→ m×R

k

to intertwine them with the standard operations on m ×Rk.What remains to be checked is that these operations are independent of the choice of α with

m ∈ Uα; here we use a fact that we haven’t used before, namely that the “transition functions”gαβ(m) from Definition 1.10(ii) are linear. So choose α,β ∈ A such that m ∈ Uα∩Uβ ; we are to

show that for any v, w ∈ Em and c ∈ R we have v +α w = v +β w and c ·α v = c ·β v.

Note first that, where as before vα ∈ Rk is defined by the property that Φα(v) = (m, vα) and

likewise for vβ , we have

(m, vβ) = Φβ (v) = (Φβ Φ−1α )Φα(v)

= (Φβ Φ−1α )(x , vα) = (x , gαβ (m)vα).

Thus vβ = gαβ (m)vα, and likewise wβ = gαβ (m)wα.

Therefore,

v +β w = Φ−1β (m, vβ + wβ ) = Φ

−1β (m, gαβ(m)vα + gαβ(m)wα)

= Φ−1β (m, gαβ(m)(vα + wα)) = Φ

−1β

Φβ Φ

−1α (m, vα + wα)

= Φ−1

α (m, vα + wα)

= v +α w,

where in the third equality we have used the fact that gαβ(m) is linear. In exactly the same way

one shows that c ·α v = c ·β v. Thus the operations +α and ·α are in fact independent of the local

trivialization Φα used to define them, and so give well-defined vector space operations on Em

satisfying the required properties.

This result gets at an important idea which I want to introduce now—we’ll come back to itin more detail later. Note first of all that one could imagine a version of Definition 1.10 whichis exactly the same except that in the final line one eliminates the requirement that the gαβ (m)

are linear and just requires them to be homeomorphisms from Rk to itself. (For what it’s worth,

the name for the object defined in this way is a “fiber bundle with fiber Rk.”) Proposition 1.11

shows that by requiring the transition functions of a fiber bundle with fiber Rk to be not justarbitrary homeomorphisms but rather linear isomorphisms (in other words, to belong to the


group of those homeomorphisms which preserve the vector space structure of Rk), we obtain anatural vector space structure on the fibers.

Working now within the class of vector bundles (so that the transition functions gαβ are

automatically linear), we could restrict the transition maps further, and ask that they preserve

some additional structure on Rk.

Definition 1.12. Let G ≤ GL(k;R) be any subgroup. A vector bundle π: E→ M is said to have

structure group G if the transition functions gαβ(m) from Definition 1.10(ii) all belong to the

group G. More generally, we say that the structure group of E reduces to G if E is isomorphic toa vector bundle which has structure group G.

Exercise 1.13. Suppose that π: E → M has structure group e (where e is the k × k identitymatrix). Prove that E is trivial.

Many familiar subgroups G of GL(k;R) can be characterized as consisting of linear maps

which preserve some linear-algebraic structure SG on Rk. Consistently with the last few para-graphs, there is a general principle that if E has structure group G, then the fibers of E can be

naturally regarded as carrying the structure SG . More specifically, exactly as in the proof ofProposition 1.11 one can use the local trivializations Φα to impose the structure SG on thefibers, and the fact that the transition functions preserve SG implies that the imposed structureis independent of which local trivialization is used.

Example 1.14. Let G = GL+(k;R) := A ∈ GL(k;R)|detA > 0. GL+(k;R) can be regarded

as the group of linear maps which preserve the standard orientation of Rk.1 Thus if E hasstructure group GL+(k;R) then the fibers of E can naturally be regarded as oriented vectorspaces, in which case we say that the bundle is oriented. (Note that, of course, regardless ofthe structure group, the fibers of E can all perfectly well be given orientations—the issue hereis whether orientations can be assigned to the fibers in a consistent, continuous way.) One canshow that the structure group of the Möbius bundle does not reduce to GL+(k;R)

Example 1.15. Let G = O(k) := A ∈ GL(k;R)|(∀v, w ∈ Rk) ((Av) · (Aw) = v ·w) be the groupof orthogonal matrices. As the above expression indicates, O(k) can be described as the group

of matrices which preserves the standard dot product · on Rk, and so the general principle saysthat a bundle with structure group O(k) can be given a continuous family of inner products(such a continuous family of inner products is sometimes called a Euclidean or Riemannianor orthogonal structure on E). In fact, for any vector bundle, at least over a reasonable basespace M (e.g. any manifold—where as usual manifolds are required to be Hausdorff and secondcountable and hence paracompact—or any cell complex), the structure group reduces to O(k).On one level can be seen as a consequence of the fact that GL(k;R) deformation retracts toO(k). Alternately one can use a partition of unity to first construct a continuous family of innerproducts on the fibers, and then construct new local trivializations with orthogonal transitionmaps by modifying the original local trivializations so that they send the inner product on thefibers to the standard inner product.

Example 1.16. Suppose that k = 2n. Then R2n can be identified with Cn = Rn × (iRn). We

can think of “multiplication by i” in Cn as a linear automorphism of R2n given in terms of

1There are various ways of saying what an orientation on a vector space V is; one way is to say that an orientation

is equivalent to specifying a distinction between “positive” and “negative” ordered bases for V—in the standard orien-

tation for R3 the bases (e1, e2, e3) and (e2, e3, e1) are positive while (e2, e1, e3) is negative, since the right-hand rule is

satisfied by the first two but violated by the third. A more sophisticated way of saying this is that an orientation of a

k-dimensional vector space V is equivalent to specifying a positive direction in the one-dimensional vector space ΛkV ,

or (as is convenient for working with differential forms) in ΛkV ∗.


the standard basis by the block matrix J0 =

0 −11 0

. Now if A ∈ GL(2n;R) is any linear

isomorphism of R2n, if we now think of R2n = Cn as a vector space over C we can ask whetheror not A continues to be a (C)-linear map of C. It is easy to see that this holds if and only if A

commutes with multiplication by i. Thus inside GL(2n;R) we have the subgroup

GL(n;C) = A∈ GL(2n;R)|AJ0 = J0A.

If E has structure group GL(n;C), then the fibers of E can naturally be seen as complex vectorspaces, giving E what is called a “complex structure.” The construction is just as before—we justneed to say how to “multiply by i” in the fibers Em, and we do this by directly transporting thecorresponding operation via the local trivializations; this is independent of which trivializationwe use precisely because the transition maps are complex linear. It is rather common to findthis sort of structure—the tangent bundles to complex manifolds (i.e. manifolds defined justlike smooth manifolds but with the transition maps between coordinate charts required to beholomorphic maps between open subsets in Cn) and to symplectic manifolds all have structuregroup which reduces to GL(n;C).

2. TOWARD THE CLASSIFICATION OF VECTOR BUNDLES OVER A SPACE

Before too long, will introduce characteristic classes as a method for studying vector bun-dles. For the moment, we will set our sights toward a more ambitious goal, namely a completeclassification of the (finite-type, at least) vector bundles over a given (paracompact) space, upto isomorphism. As occurs in most areas of mathematics, the full classification problem turnsout to be somewhat unmanageable, but very satisfactory partial results can be obtained. To getat the methods underlying these partial results, we will first precisely reformulate the classifi-cation problem into a problem about homotopy classes of maps from the given space to certainauxiliary spaces (Grassmannians). Of course, although fully classifying homotopy classes ofmaps is typically a hard problem (for instance homotopy classes of maps between spheres arestill not completely understood in spite of 80 years of intensive effort), the methods of algebraictopology provide a great deal of insight—in our case they will give us characteristic classes.

For this to work we need to impose at least a modest hypothesis on the base spaces of ourfiber bundles. The following assumption is in force throughout the rest of this section:

Assumption 2.1. The topological space M is Hausdorff and paracompact, or equivalently, M is

Hausdorff and for every open cover Uαα∈A of X there is a continuous partition of unity subordinate

to Uα (i.e., continuous functions χα : M → [0, 1] such that supp(χα) ⊂ Uα, such that each point

in m has a neighborhood on which all but finitely many χα are identically zero, and such that∑αχα(m) = 1 for all m).

Note that paracompact Hausdorff spaces include all manifolds (provided that, as is customary,the definition of a manifold includes a second-countability assumption), and all CW complexes.

2.1. Pre-classifying maps. Out of a rank-k vector bundle π: E → M we will eventually con-struct something called a classifying map from M to a Grassmannian (to be defined later).Before that, we introduce the following notion.

Definition 2.2. Let π: E → M be a rank-k vector bundle. A pre-classifying map for E is a con-

tinuous map F : E→ RN (for some natural number N), such that for each fiber Em = π−1(m)

of E the restriction F |Em: Em→ R

N is linear and injective for all m ∈ M .


Of course, if E = M × Rk were the trivial bundle, then letting F : E → Rk be the projec-tion onto the second factor would give a pre-classifying map. Conversely, if there exists a pre-

classifying map F : E→ Rk (so that the parameters k and N in Definition 2.2 are the same) then

(π, F): E → M ×Rk would give an isomorphism between E and the trivial bundle. However,by allowing N to be larger than k we can accommodate nontrivial bundles into Definition 2.2:

Example 2.3. Let E→ S1 be the Möbius bundle, so we may identify E =[0,1]×R

(1,t)∼(0,−t). Now define

F : E→ R2

[s, t] 7→ (cos(πs)t, sin(πs)t)

(Note that F([1, t]) = (−t, 0) = F([0,−t]), so this map is well-defined on E). Then F is clearly

a pre-classifying map. Note that the image F(Es) of the fiber over s ∈ S1 is the line of slope

tan(πs)—in particular the image of the fiber rotates by 180 degrees as s varies through S1.We will see later that this specific rotational behavior of the image of the fiber under the pre-classifying map can be used to distinguish the Möbius bundle from the trivial bundle.

For the simplest-to-state-and-prove results we will need to restrict to a certain class of vectorbundle:

Definition 2.4. A vector bundle π: E → M is said to be of finite type if there is a collection of

local trivializations Φα : π−1(Uα)→ Uα×Rk as in Definition 1.10, such that Uα is a finite open

cover of M .

As should be obvious, if M is compact then any vector bundle over M is of finite type. Even inthe noncompact case, in my experience it is unusual to run across non-finite-type vector bundles,though of course such can be constructed. By working with maps to R∞ instead of to RN it isusually possible to do without finite-type assumptions, as is detailed in [MS, Chapter 5], but inany case it seems best to understand the finite-type case first.

Proposition 2.5. If π: E→ M is a finite-type vector bundle over a paracompact Hausdorff space

M then for some N there exists a classifying map F : E→ RN .

Proof. By the finite-type assumption there are open subsets U1, . . . , UT with M = U1 ∪ · · · ∪ UT

and local trivializations Φi : π−1(Ui) → Ui × R

k. We can write Φi(e) = (π(e),φi(e)) for some

φi : π−1(Ui)→ R

k which restricts to each fiber Em for m ∈ Ui as a linear isomorphism.Since M is paracompact there is a partition of unity ξi

Ti=1

subordinate to the open cover

Ui. Since for each i χi is identically zero on an open subset containing M \ Ui , for each i the

function e 7→ χi(π(e))φi(e) (initially defined as a map from π−1(Ui) to Rk) extends by zero to

give a continuous function E → Rk. With this continuous extension understood, let N = kT

and define F : E→ RN by

F(e) =χ1(π(e))φ1(e), . . . ,χT (π(e))φT (e)

.

The continuity of π and the χi and φi obviously implies that F is continuous. For m ∈ M , F |Emis

given by e 7→ (χ1(m)φ1(e), . . . ,χT (m)φT (e)), which is linear since for all i either χi(m) = 0 orφi |Em

is linear. If i is such that χi(m) 6= 0 (and of course i exists since∑χi = 1) then if e ∈ Em

with F(e) = 0 we would have φi(e) = 0 and hence e = 0 since φi |Em: Em → R

k is a linear

isomorphism. Thus F |Emis linear and injective for all m, and so F is a pre-classifying map.

While pre-classifying maps F : E → RN exist for all finite-type vector bundles E over M

(where as always we assume M is paracompact Hausdorff), they are clearly not unique: for


example if A ∈ GL(N ;R) we could obviously replace F by A F . Also, the number N is notuniquely determined. In particular, if N < N ′ we have an obvious “stabilization” map

IN ,N ′ : RN → RN ′

~x 7→ (~x , ~0),

and if F : E→ RN is a pre-classifying map for E then clearly so is IN ,N ′ F : E→ RN ′ .

Definition 2.6. • If F0, F1 : E→ RN are two pre-classifying maps, we say that F0 and F1 areisotopic if they are homotopic through pre-classifying maps, i.e. if there is a continuousmap F : [0, 1] × E → RN such that for all t the map e 7→ F(t, e) is a pre-classifyingmap for E, and such that for i = 0, 1 the given pre-classifying maps Fi are given byFi(m) = F(i, m).• If F0 : E → RN0 and F1 : E → RN1 are two pre-classifying maps, we say that F0 and

F1 are stably isotopic if there is N ′ ≥ maxN0, N1 such that the pre-classifying mapsIN0,N ′ F0 and IN1,N ′ F1 are isotopic.

It should be clear that both isotopy and stable isotopy are equivalence relations.

Exercise 2.7. Give, with proof, an example of a pair of pre-classifying maps F0, F1 : E→ RN forthe same vector bundle E (and with the same value of N) which are not isotopic.

Theorem 2.8. Let F0 : E→ RN0 and F1 : E→ RN1 be two pre-classifying maps for the same vector

bundle. Then F0 and F1 are stably isotopic.

Proof. Without loss of generality let us assume that N0 ≥ N1. Then let N ′ = N0+N1. First of all

note that F1 is stably isotopic to the map F ′1: E → RN ′ defined by F ′

1(e) = (~0N0

, F1(e)) (where

~0N0denotes the 0-vector in RN0 ; we will use similar notation below). Indeed,

(t, e) 7→(1− t)F1(e), ~0N0−N1

, t F1(e)

gives an isotopy from IN1,N ′ F1 to F ′1. But the map F ′ : [0, 1] × E → RN ′ defined by F ′(e) =

((1 − t)F0(e), t F1(e)) gives an isotopy from IN0,N ′ F0 to F ′1. Thus (by the transitivity of the

isotopy relation) IN0,N ′ F0 is isotopic to IN1,N ′ F1, and so F0 and F1 are stably isotopic.

Summing up, any finite-type vector bundle over M has pre-classifying maps, and any twosuch maps (for a given vector bundle) are stably isotopic. This begins to suggest an approachto distinguishing one vector bundle from another: we should construct pre-classifying mapsfor both of them and then (somehow) show that these are not stably isotopic. (It might be

instructive to try to do this for the Möbius and trivial bundles over S1.) To do this in generalwill require us to introduce Grassmannians and classifying maps.

2.2. The Grassmannian and the tautological vector bundle. As was already sort of suggestedin Example 2.3, the real information in a pre-classifying map F : E→ RN lies in the way that thek-dimensional subspaces F(Em) vary with m. Accordingly, we introduce the following auxiliaryset:

Definition 2.9. The Grassmannian of k-planes in RN is the set

Grk(RN ) = V ≤ RN |dim V = k.


Then any pre-classifying map F : E → RN , where E is a rank-k vector bundle over M , givesrise to a function

f : M → Grk(RN )

m 7→ F(Em)

This map f is an example of what we will later call a classifying map, though our formal defi-nition will be a bit different.

The plan is to study the map f using the methods of algebraic topology; one obvious prereq-uisite to doing this is making the Grassmannian Grk(R

N ) into a topological space rather thanjust a set. There are various equivalent definitions for a topology on Grk(R

N ) in the literature;for definiteness we will use the following. First introduce the Stiefel manifold

Vk(RN ) =

(~v1, . . . , ~vk) ∈ (R

N )k ~v1, . . . , ~vk is a linearly independent set

.

Of course Vk(RN ) has a natural topology, as it is an open subset of (RN )k. Moreover there

is an obvious map p : Vk(RN ) → Grk(R

N ) defined by setting p(~v1, . . . , ~vk) = span~v1, . . . , ~vk.We define the topology on Grk(R

N ) by requiring p to be a quotient map, i.e. by saying that

U ⊂ Grk(RN ) is open if and only if p−1(U) is open.

Exercise 2.10. In the literature it is somewhat more common to see the Stiefel manifold definedas the subset of Vk(R

N ) given by

Vk(RN ) =

(~v1, . . . , ~vk) ∈ (R

N )k ~vi · ~v j = δi j

(i.e. as the set of orthonormal k-tuples of vectors in RN ). Prove that we would obtain thesame topology on Grk(R

N ) as in the previous paragraph if we instead defined the topology byrequiring p|Vk(R

N ) to be a quotient map.

Note that since Vk(RN ) is compact (it is a closed subset of (SN−1)k), Exercise 2.10 shows that

Grk(RN ) is compact. One can show that Grk(R

N ) is a smooth manifold (indeed this is a popularexample in differential topology textbooks, see e.g. [Lee, Chapter 1]), but at least for now wewon’t need this fact.

Exercise 2.11. Let F : E → RN be a pre-classifying map for a rank-k vector bundle π: E → M ,and define f : M → Grk(R

N ) by f (m) = F(Em). Prove that f is continuous. (It’s probablyslightly easier to do this using our initial definition for the topology on Grk(R

N ) rather than theone from Exercise 2.10.)

There is a very important vector bundle over Grk(RN ): the tautological bundle2 γk(RN ). We

defineγk(RN ) =

(V, ~v) ∈ Grk(R

N )×RN |~x ∈ V

,

with the bundle projection π: γk(RN ) → Grk(RN ) defined by projection onto the first factor.

Thus for V ∈ Grk(RN ), (i.e. for V a k-dimensional subspace of RN ), π−1(V ) = V × V—the

fiber over V is (basically) the subspace V itself.

Proposition 2.12. π: γk(RN )→ Grk(RN ) is a rank-k vector bundle.

Proof. The topology on γk(RN ) is the subspace topology coming from the product topology

on Grk(RN ) × RN (where we topologize Grk(R

N ) as before). So the projection π: γk(RN ) →Grk(R

N ) is continuous, and its fibers have obvious vector space structures. It remains to con-struct local trivializations.

2In [MS] and some other references this is called the “canonical bundle.” I prefer to avoid this name in order to

avoid conflict with some terminology from algebraic geometry.


Endow RN with its standard inner product, and let V0 ∈ Grk(RN ); we will construct a local

trivialization around V0. The open set U over which the trivialization is defined will be

U = V ∈ Grk(RN )|V ∩ V⊥

0= 0.

We should first check that U is open—by the definition of the quotient topology this is equivalent

to the statement that p−1(U) is open where p : Vk(RN ) → Grk(R

N ) is the projection. Now if

~v⊥1

, . . . , ~v⊥N−k is an orthonormal basis for V⊥

0, then an element ( ~w1, . . . , ~wk) belongs to p−1(U)

if and only if ~v⊥1

, . . . , ~v⊥N−k

, ~w1, . . . , ~wk is a basis for RN , i.e. if and only if the N × N matrix

whose columns are ~v⊥1

, . . . , ~v⊥N−k

, ~w1, . . . , ~wk has nonzero determinant. This is clearly an open

condition on ( ~w1, . . . , ~wk), proving that p−1(U) ⊂ Vk(RN ) is open, and hence that U ⊂ Grk(R

N )

is open.

We now construct a local trvialization for γk(RN ) over U . Choose an orthonormal basis~v1, . . . , ~vk for V0. Then where Π : RN → V0 is the orthogonal projection, Π is given by the

formula Π(~v) =∑k

i=1(~v · ~vi)~vi , and since kerΠ = V⊥

0, for any V ∈ U the restriction Π|V : V → V0

is an isomorphism. Define

Φ : π−1(U)→ U ×Rk

by

Φ(V, ~v) = (V, ~v · ~v1, . . . , ~v · ~vk).

This is clearly continuous and fiberwise linear, and for each V ∈ U restricts to π−1(V) as alinear isomorphism since Π|V is a linear isomorphism. To complete the proof we need to show

that Φ−1 : U ×Rk→ π−1(U) is continuous. Now Φ−1(V, c1, . . . , ck) =V,∑k

i=1ci(Π|V )

−1(~vi), so

the desired continuity will follow if we show that, for each i, the map φi : U → RN defined by

φi(V ) = (Π|V )−1(vi) is continuous.

Now where as before p : Vk(RN ) → Grk(R

N ) is the quotient projection, let φi = φi p.It follows from the definition of the quotient topology that φi is continuous if and only if

φi : p−1(U)→ RN is continuous. Denote a general element of p−1(U) by W = ( ~w1, . . . , ~wk). LetA(W ) denote the k×k matrix representingΠ|p(W ) : p(W )→ V0 in terms of the basis ~w1, . . . , ~wkfor W and our fixed orthonormal basis ~v1, . . . , ~vk for V0. Then A(W )i j = ~vi · ~w j , and so the map

W 7→ A(W ) is continuous. Moreover A(W ) is invertible for all W ∈ p−1(U), so since inversionis a continuous operation on invertible matrices it follows that W 7→ A(W )−1 is continuous as a

function of W . We then have, where ~v⊥1

, . . . , ~v⊥N−k is the orthonormal basis for V⊥

0from earlier,

φi(W ) = (Π|p(W ))−1(vi) =

k∑

j=1

(A(W )−1) ji ~w j

= ~vi +

k∑

j=1

(A(W )−1) ji

N−k∑

m=1

( ~w j · ~v⊥m)~v⊥

m,

where in the last line we have split up∑k

j=1(A(W )−1) ji ~w j into its components with respect to the

orthogonal direct sum RN = V0⊕V⊥0

. This formula for φi(W )makes clear that φi is continuous

as a function of W , and hence that Φ−1 is continuous, completing the proof.

2.3. Pullbacks and the main classification result. Here we describe a general constructionwhich in particular will serve to link the constructions from the previous two sections. Letπ: V → Y be a vector bundle, and let f : X → Y be a continuous map. We then define, first asa toplogical space,

f ∗V =(x , v) ∈ X × V |π(v) = f (x)


where of course the topology is given by the subspace topology induced by the product topologyon X × V . Using πV and πX to denote the projections from (subspaces of) X × V to V and X ,we have a commutative diagram

f ∗V

πX

πV // V

π

X

f // Y

,

where all maps involved are continuous.

The fiber (πX )−1(x) of f ∗V over a point x ∈ X is just x×Vf (x); thus the fibers of f ∗V → X

inherit vector space structures from those of π: V → Y . f ∗V → X also inherits local trivializa-

tions from V → Y : if W ⊂ Y is an open set and if Φ : π−1(W )→W ×Rk is a local trivialization,

say given byΦ(v) = (π(v),φ(v)), then we obtain a local trivialization for f ∗V overπ−1X( f −1(W ))

by the formula Ψ((x , v)) = (x ,φ(v)). Thus f ∗V is a vector bundle. Since if Wαα∈A is an open

cover of Y then f −1(Wα)α∈A is a cover of X , clearly f ∗V will be of finite type if V is of finitetype. Moreover the transition maps gαβ for a collection of local trivializations for V pull back

in an obvious way to transition maps for a collection of local trivializations for π∗V , in view ofwhich if V has structure group G ≤ GL(k;R) then so does f ∗V . In particular if V is trivial thenso is f ∗V .

As a special case, if A ⊂ Y is some subset (viewed as a topological space with the subspacetopology), then where i : A→ Y is the inclusion for any vector bundle π: V → Y we can formthe bundle i∗V → A. This is the same as (or, strictly speaking, isomorphic to by a rather obviousisomorphism) the “restriction of V to A,” sometimes written V |A and defined simply by setting

V |A = π−1(A) and having the bundle projection be the restriction π|π−1(A) : π

−1(A)→ A.

In particular one could have A be a single point. Of course every vector bundle over a pointis trivial, so i∗V would in this case be trivial, though of course the original vector bundle V → Y

typically would not be. This shows that the implications two paragraphs above (e.g. that if V istrivial then so is f ∗V ) usually cannot be reversed—in general, the bundle f ∗V is “more trivial”than V .

Proposition 2.13. Given a pair of continuous maps Xf // Y

g // Z and a vector bundle

π: V → Z there is an isomorphism of vector bundles

f ∗(g∗V )∼= (g f )∗V.

Proof. We have

(g f )∗V = (x , v) ∈ X × V |g( f (x)) = π(V )

while

f ∗(g∗V ) = (w, x) ∈ (g∗V )× X | f (x) = πY (w)

=

x , (y, v)∈ X × (Y × V )|g(y) = π(v), f (x) = y

=

x , ( f (x), v)|g( f (x)) = π(v)

.

The map (x , v) 7→ (x , ( f (x), v)) is then obviously a bundle isomorphism from (g f )∗V tof ∗(g∗V ).

Remark 2.14. In categorical language, Proposition 2.13 shows that there is a contravariantfunctor Vectk from the category of topological spaces to the category of sets which,for any topological space X , has Vectk(X ) equal to the set of isomorphism classes of rank-kvector bundles. To any continuous function f : X → Y (i.e. any morphism in ) this functor


assigns the function f ∗ : Vectk(Y )→ Vectk(X ) defined by [E] 7→ [ f ∗E] where we denote theisomorphism class of a vector bundle E by [E].

We now return to the topic of pre-classifying maps. Let π: E→ X be a rank-k vector bundle.Recall that to any pre-classifying map F : E→ RN we may associate the map f : M → Grk(R

N )

defined by f (m) = F(Em); for the moment we will call this map f the “induced map” of F . ByExercise 2.11, f is continuous.

Proposition 2.15. If F : E → RN is a pre-classifying map with induced map f : M → Grk(RN )

then E is isomorphic to f ∗γk(RN ). Conversely, if f : M → Grk(RN ) is a continuous map such that

E is isomorphic to f ∗γk(RN ), then there is a pre-classifying map F : E → RN whose induced map

is f .

Proof. For a continuous map f : M → Grk(RN ), we have

f ∗γk(RN ) =(m, (V, ~v)) ∈ M × (Grk(R

N )×RN )|~v ∈ V, f (m) = V

.

In particular the fiber of f ∗γk(RN ) over m ∈ M consists of all points of the form (m, ( f (m), ~v))where ~v ∈ f (m).

Suppose that f is the induced map of some pre-classifying map F : E→ RN , whereπ: E→ M

is a rank-k vector bundle. Define G : E→ f ∗γk(RN ) by

G(e) =π(e), ( f (π(e)), F(e))

.

Evidently G is continuous, and since by definition F maps the fiber Em by a linear isomorphism to

f (m) for each m ∈ M , we see that G restricts to each Em as a linear isomorphism to

f ∗γk(RN )

m.

So by Exercise 1.5, G is an isomorphism.

Conversely, suppose that G : E → f ∗γk(RN ) is an isomorphism of bundles over M . In view

of the formula for f ∗γk(RN ) above, G necessarily takes the form

G(e) = (π(e), ( f (π(e)), F(e)))

for some continuous map F : E → RN . For each m ∈ M , G restricts to Em as the map e 7→

(m, ( f (m)), F(e)), and the fact that this is an isomorphism to

f ∗γk(RN )

mamounts to the

statement that F maps Em isomorphically to the k-dimensional subspace f (m) ≤ RN . But thisprecisely means that F : E→ RN is a pre-classifying map for E.

This motivates the following important definition:

Definition 2.16. Letπ: E→ M be a rank-k vector bundle. A classifying map for E is a continuous

map f : M → Grk(RN ) (for some N) such that E is isomorphic to f ∗γk(RN ).

Corollary 2.17. If E is a finite-type rank-k vector bundle over a paracompact Hausdorff space M

then there exists a classifying map for E.

Proof. There is a pre-classifying map for E by Proposition 2.5, and the induced map of thispre-classifying map is a classifying map by Proposition 2.15.

Remark 2.18. It also follows from Proposition 2.15 that, conversely, if E admits a (pre-)classifying

map then E must be finite-type, since γk(RN ) is finite-type and pullbacks of finite-type vectorbundles are finite-type.

We now address the uniqueness of classifying maps; the situation quite closely correspondsto that for pre-classifying maps as described in Theorem 2.8. Corresponding to the stabilization


maps IN ,N ′ : RN → RN ′ for N ≤ N ′ discussed earlier (recall that these are simply given by

~x 7→ (~x , ~0)), we have a map

iN ,N ′ : Grk(RN )→ Grk(R

N ′)

V 7→ V × ~0.

It is trivial to check that iN ,N ′ is continuous. Moreover, if F : E → RN is a pre-classifying map

for E which induces the map f : M → Grk(RN ), it follows directly from the definitions that the

pre-classifying map IN ,N ′ F : E→ RN ′ induces the map iN ,N ′ F : M → Grk(RN ′).

Proposition 2.19. For N ′ ≥ N there is an isomorphism of vector bundles γk(RN )∼= i∗N ,N ′γk(RN ′).

Proof. It’s not hard to check this directly from the definition of the pullback. Alternately, we can

observe that the map F : γk(RN ) → RN ′ defined by F(V, ~v) = IN ,N ′(~v) is a pre-classifying map

for γk(RN ) which induces the map iN ,N ′ : Grk(RN )→ Grk(R

N ′), and so the proposition followsfrom Proposition 2.15.

Analogously to the relation of stable isotopy for pre-classifying maps defined before Theorem2.8, we make the following definition.

Definition 2.20. Two maps f0 : M → Grk(RN0) and f1 : Grk(R

N1) are stably homotopic if there isN ′ ≥maxN0, N1 such that iN0,N ′ f0 is homotopic to iN1,N ′ f1.

Theorem 2.21. Let M be a paracompact Hausdorff space. Then there is a well-defined, surjective

map

Φ :

§Isomorphism classes of finite-type,

rank-k vector bundles over M

ª→

§Stable homotopy classes of

maps f : M → Grk(RN )

ª

given by, where [E] denotes the isomorphism class of a vector bundle E, setting Φ([E]) equal to the

stable homotopy class of any classifying map for E.

Proof. Any finite-type rank-k vector bundle admits a classifying map by Corollary 2.15. If f0, f1

are two classifying maps for E, then by the second half of Proposition 2.15 there exist pre-classifying maps F0 : E → RN0 and F1 : E → RN1 inducing f0 and f1, respectively. By Theorem

2.8 there is N ′ ≥maxN0, N1 and an isotopy F ′ : [0, 1]× E→ RN ′ from IN0,N ′ F0 to IN1,N ′ F1.

Writing F ′t= F ′(t, ·) (so by the definition of isotopy each F ′

tis a pre-classifying map), for t ∈

[0, 1] let f ′t: M → Grk(R

N ′) be the map induced by F ′t, and define f : [0, 1]× M → Grk(R

N ).(It is straightforward to see that f is continuous, by the same method as in Exercise 2.11).Then f ′(0, ·) = iN0,N ′ f0 and f ′(1, ·) = iN1,N ′ f1. Thus iN0,N ′ f0 and iN1,N ′ f1 are homotopic.

This proves that the stable homotopy class of a classifying map for a finite-type vector bundledepends only on the vector bundle. Moreover, it is obvious from Definition 2.16 that isomorphicvector bundles have the same (sets of) classifying maps. This completes the proof that Φ is well-defined.

If f : M → Grk(RN ) is any continuous map, [ f ∗γk(RN )] is sent by Φ to the stable homotopy

class of f . Thus Φ is surjective.

This suffices to justify the following strategy for distinguishing vector bundles from eachother: construct classifying maps for each of them, and prove that the classifying maps are notstably homotopic. Here is a simple example:

Example 2.22. Let E be the Möbius bundle over S1, and let E′ = S1×R be the trivial rank-1 vectorbundle. Note that since we are dealing here with rank-1 bundles the relevant Grassmannians


Grk(RN ) are already familiar to you from algebraic topology: they are the real projective spaces

RPN−1.The trivial bundle E′ = S1 ×R admits the projection to the second factor as a pre-classifying

map. The induced classifying map is then the constant map to RP0 (which stabilizes under the

maps i1,N to constant maps to RPN−1).

Meanwhile in Example 2.3 we constructed a pre-classifying map F : E → R2 for the Möbiusbundle, given by the formula F([s, t]) = (cos(πs)t, sin(πs)t) (for 0 ≤ s ≤ 1). The induced

classifying map is then the map f : S1 → RP1 defined by f ([s]) = [cos(πs) : sin(πs)] where

we identify S1 with R/Z and use standard homogeneous coordinates on RP1. So to prove thenontriviality of the Möbius bundle it suffices to show that f is not stably homotopic to the

constant map, i.e. that for all N ≥ 2 the map i2,N f : S1 → RPN−1 given by i2,N f ([s]) =

[cos(πs) : sin(πs) : 0 : · · · : 0] is not homotopic to a constant.In fact, as you likely learned in algebraic topology, the above map i2,N f represents a gen-

erator for the fundamental group of π1(RPN−1). Viewing RPN−1 as the quotient of SN−1 by the

antipodal map, so that we have a two-to-one cover q : SN−1→ RPN−1, if i2,N f were homotopic

to a constant then it would lift to a closed curve in SN−1, whereas in fact it lifts to the non-closedpath s 7→ (cos(πs), sin(πs), 0, . . . , 0).

Thus the classifying map f for the Möbius bundle is not stably homotopic to a constant,demonstrating that Theorem 2.21 can be used to distinguish the Möbius bundle from the trivialbundle.

In fact, at least in principle, Theorem 2.21 could be used to distinguish any two nontrivialbundles, as the following shows.

Theorem 2.23. The map Φ from Theorem 2.21 is injective. Thus any finite-type rank-k vector

bundle over a paracompact Hausdorff space M has a classifying map f : M → Grk(RN ) (and so

the bundle is isomorphic to f ∗γk(RN )), and two such bundles are isomorphic if and only if their

classifying maps are stably homotopic.

Proof. This quickly follows from the following fundamental result, whose proof will be delayedto Section 2.4.

Theorem 2.24. Let X be a paracompact Hausdorff space,π: E→ Y a vector bundle, and h0, h1 : X →Y two homotopic continuous maps. Then h∗

0E is isomorphic to h∗

1V .

To prove Theorem 2.23 assuming Theorem 2.24, we must show that if E0, E1 → M arevector bundles with stably homotopic classifying maps, then E0 and E1 are isomorphic. Letf0 : M → RN0 and f1 : M → RN1 be classifying maps for E0 and E1 respectively, and chooseN ′ ≥maxN0, N1 so that iN0,N ′ f0 and iN1,N ′ f1 are homotopic. Then using Propositions 2.19

and 2.15 and Theorem 2.24 we obtain the following chain of isomorphisms:

E0∼= f ∗

0γk(RN0)∼= f ∗

0i∗N0,N ′

γk(RN ′)∼= (iN0,N ′ f0)∗γk(RN ′)∼= (iN1,N ′ f1)

∗γk(RN ′)

∼= f ∗1

i∗N1,N ′

γk(RN ′)∼= f ∗1γk(RN1)∼= E1.

2.4. Homotopy invariance of pullbacks. In this section we will complete the reformulationof the classification problem for finite-type vector bundles over a paracompact Hausdorff spaceby proving Theorem 2.24, which asserts that if h0, h1 : X → Y are homotopic maps where X isparacompact and Hausdorff, then for any vector bundle E over Y the pullbacks h∗

0E and h∗

1E are

isomorphic vector bundles over X .


Let H : [0, 1]×X → Y be a homotopy from h0 to h1, and for t ∈ [0, 1] define it : X → [0, 1]×X

by it(x) = (t, x). In particular h0 = H i0 and h1 = H i1. So we have a vector bundleH∗E→ [0, 1]× X and, in view of Proposition 2.13, the theorem amounts to the statement thati∗0(H∗E) is isomorphic to i∗

1(H∗E). (In effect, by replacing E with H∗E we have reduced to the

case where Y = [0, 1]× X and h0, h1 are the inclusions i0, i1.)First we prove:

Lemma 2.25. For any x ∈ X there is a neighborhood Ux such that the restriction (H∗E)|[0,1]×Ux

is trivial.

(As usual, the restriction (H∗E)|[0,1]×Uxmeans the pullback of H∗E by the inclusion of [0, 1]×

Ux into [0, 1]× X—in other words if πH∗E : H∗E→ [0, 1]× X is the bundle projection, the total

space of (H∗E)|[0,1]×Uxis just π−1

H∗E([0, 1] × Ux) and the bundle projection is the restriction of

πH∗E to this total space.)

Proof. By the definition of a vector bundle and of the product topology, for any s ∈ [0, 1] thereis an interval Is around s (with Is relatively open in [0, 1]), a neighborhood U s of x in X , and a

local trivialization Φs : (H∗E)|Is×U s → (Is×U s)×Rk. The relatively open intervals Is cover [0, 1],

so by the compactness of [0, 1] there are s1, . . . , sm such that [0, 1] = Is1∪ · · ·∪ Ism

. Without loss

of generality we may assume that none of the Isiis contained in another Is j

, and that the si are

ordered so that the left endpoints of the Isiare in increasing order. So necessarily 0 ∈ Is1

and

1 ∈ Ism, and Isi

∩ Isi+1= ∅ for each i. If for i = 1, . . . , m − 1 we choose t i ∈ Isi

∩ Isi+1, and set

t0 = 0 and tm = 1, then for each i = 1, . . . , m we will have [t i−1, t i] ∈ Isi, with the intervals

[t i−1, t i] covering [0, 1].Let U = U s1 ∩ · · · ∩ U sm , so U is an open neighborhood of x in X . For i = 1, . . . , m, by

restricting the local trivialization Φsifrom earlier, we obtain a trivialization (still denoted Φsi

) of

(H∗E)|[t i−1,t i]×U . For each i = 1, . . . , m−1, we have a mapΨi : t i×U → Rk→ t i×U×Rk given

by Ψ = ΦsiΦ−1

si+1. Ψi is a homeomorphism which is compatible with the projection to t i × U

and with the vector space structure on the fibers, so it has the form Ψi(t i , z, v) = (t i , z,ψi(z)v)

for some continuous map ψi : U → GL(k;R).We can now define a trivialization of H∗E over [0, 1]×U . Let us write the local trivializations

Φsi: (H∗E)|[t i−1,t i]×U → ([t i−1, t i]×U)×Rk asΦsi

(e) = (πH∗E(e),φi(e))whereφi : (H∗E)|[t i−1,t i]×U →

Rk restricts to each fiber as a linear isomorphism.

Then define Φ : (H∗E)|[0,1]×U → [0, 1]×U×Rk by declaring its restriction to (H∗E)(t,z) where

z ∈ U and t ∈ [t i−1, t i] to be given by e 7→ (πH∗E(e),ψ1(z) · · ·ψi−1(z)φi(e)) (for i = 1 thisshould be interpreted as just e 7→ (πH∗E(e),φ1(e))). While this appears to be multiply definedat the values t j (as there is a prescription both for i = j and i = j + 1), the definition of the ψi

ensures that both prescriptions give the same value, in view of which it easily follows that Φ isa trivialization of H∗E over [0, 1]× U .

The following would be obvious under rather modest hypotheses on X , e.g. if we assumedthat X was Lindelöf, but is needed for the general paracompact Hausdorff case.

Lemma 2.26. Let Uαα∈A be an open cover of a paracompact Hausdorff space X . Then there is

a countable open cover Vn∞n=1

of X such that each Vn is a disjoint union of sets each of which is

contained in some Uα.

Proof. The lemma is obvious if A is finite (or indeed even if A is countable) so we assume that A

is infinite. Let χαα∈A be a partition of unity subordinate to Uα (and recall that partitions ofunity are by definition locally finite). For each nonempty finite subset S ⊂ A let

WS = x ∈ X |(α ∈ S, β /∈ S⇒ χα(x)> χβ(x)).


By the local finiteness condition, for each x ∈ X there is a neighborhood U of X and a finitesubset T ⊂ A such that φγ|U = 0 unless γ ∈ U . Without loss of generality (by adding extraneous

elements of A to T) we may assume that T \ S 6= ∅. In this case WS ∩ U = ∅ unless S ⊂ T , andif S ⊂ T then

WS ∩ U =⋂

α∈S,β∈T\S

(χα −χβ)−1((0,∞))

which is a finite intersection of open sets and so is open. So since X is covered by such opensets U it follows that WS is open for all finite subsets S.

Also, if α ∈ S then WS ⊂ Uα since supp(χα) ⊂ Uα. The WS cover X as S varies through finitesubsets of T : for instance if x ∈ X we will have x ∈ WS if S is the set of indices α such thatχα(x)> 0.

Furthermore, if #(S) = #(S′), then WS ∩WS′ = ∅, since if #(S) = #(S′) then there existα ∈ S \ S′ and β ∈ S′ \ S, and if x ∈ WS then χα(x) > χβ(x), while if x ∈ WS′ the reverseinequality holds.

So for any positive integer n let Vn = ∪S:#(S)=nWS . Then each Vn is indeed a disjoint union of

sets each of which is contained in some Uα, and so Vn∞n=1

is the desired cover.

Combining the above two lemmas immediately gives:

Corollary 2.27. Let H : [0, 1]×X → Y be a continuous map where X is a paracompact Hausdorff

space, and let E → Y be a vector bundle. Then there is a countable open cover Vn∞n=1

of X such

that each restriction (H∗E)|[0,1]×Vnis trivial.

Proof. Lemma 2.25 gives an open cover Uxx∈X of X such that each (H∗E)|[0,1]×Uxis trivial.

Associate to the cover Ux the countable cover Vn∞n=1

as in Lemma 2.26. So for each n, Vn isa disjoint union Vn = ⊔β∈Bn

Wβ so that for all β ∈ Bn there is x ∈ X with Wβ ⊂ Ux . So we have

trivializations of (H∗E)|[0,1]×Wβfor each β ∈ Bn, and taking the union of these trivializations

gives a trivialization for (H∗E)|[0,1]×Vn.

Let us now finish the proof of Theorem 2.27. Let Vn∞n=1

be an open cover as in Corollary

2.27. Since X is paracompact and Hausdorff there is a partition of unity χn∞n=1

subordinate to

Vn∞n=1

. We will inductively use the following proposition:

Proposition 2.28. For a continuous function f : X → [0, 1] define i f : X → [0, 1]×X by i f (x) =

( f (x), x). Suppose that j ∈ Z+ and that f and f + χ j are both continuous functions from X to

[0, 1]. Then there is an isomorphism of vector bundles i∗f(H∗E)∼= i∗

f +χ j(H∗E) which is the identity

on all fibers over points x where ψ j(x) = 0.

Proof. Let Wj = X \ supp(χ j). The restrictions of the two bundles to Wj are then identical, since

i f (x) = i f +χ j(x) for all x ∈Wj , and so over Wj we will just use the identity as our bundle isomor-

phism. So since Vj ∪Wj = X it suffices to construct an isomorphism i∗f(H∗E)|Vj

→ i∗f +χ j(H∗E)|Vj

which is the identity over Vj ∩Wj .

But this is straightforward given that h∗E is trivial over [0, 1]× Vj . Let Ψ : (H∗E)|[0,1]×Vj→

[0, 1]×Vj×Rk be a trivialization, say given byΨ(e) = (πH∗E(e),ψ(e)), and defineΦ : i∗

f(H∗E)|Vj

→

i∗f +χ j(H∗E)|Vj

by, for x ∈ Vj and e ∈ (H∗E)( f (x),x),

Φ(x , e) =x ,Ψ−1( f (x) +χ j(x), x ,ψ(e))

.

This is easily seen to have the desired properties. (Intuitively, the trivialization Ψ identifies all

fibers over [0, 1] × Vj with Rk, and we are using this identification to identify the fiber over

( f (x), x) with the fiber over ( f (x) +χ j(x), x).)


The above proposition gives us, for each m ∈ Z+, an isomorphism

Ψm : i∗∑m−1

j=1 χ j

(H∗E)→ i∗∑m

j=1 χ j

(H∗E).

By the local finiteness of the partition of unity χ j, X is covered by open sets UM such that∑M

j=1χ j(x) = 1 for each x ∈ UM . In particular we have, for all N ≥ M ,

i∗∑N

j=1 χ j

(H∗E)

|UM=i∗1(H∗E)

|UM

.

We may then define the “infinite composition” Ψ : i∗0(H∗E) → i∗

1(H∗E) by setting Ψ|π−1

i∗0

H∗E(UM )

equal to ΨN · · · Ψ1 for any N ≥ M . This is a globally defined map which restricts over eachUM to the finite composition ΨM · · · Ψ1, which is a bundle isomorphism.

Therefore Ψ : i∗0(H∗E)→ i∗

1(H∗E) is a bundle isomorphism, proving that h∗

0E ∼= i∗

0(H∗E) and

h∗1E ∼= i∗

1(H∗E) are isomorphic.

Remark 2.29. There is a somewhat more intuitive proof of the homotopy invariance theoremin the case that X and Y are smooth manifolds, E → Y is a smooth vector bundle (i.e., E is asmooth manifold and the local trivializations are diffeomorphisms), and H : [0, 1]× X → Y isa smooth map (so that H∗E → [0, 1]× X is also a smooth vector bundle). In this case, as we’llprobably discuss more later, one can use a partition of unity to construct a connection on the

vector bundle H∗E, which is to say a splitting of the tangent bundle T (H∗E) = T v ⊕ T h wherethe “vertical bundle” T v is the kernel of the derivative of the projection map to [0, 1]×X and the

“horizontal bundle” T h is complementary to T v and is appropriately compatible with the vectorbundle structure. There will then be a unique vector field V on the smooth manifold H∗E which

lies in T h and projects down to [0, 1]× X as the vector field ∂t in the [0, 1] direction, and anisomorphism from h∗

0E to h∗

1E can be obtained by taking the time-one flow of the vector field V .

Remark 2.30. Directly from Theorem 2.24 (and without appealing to any classification results)one can infer that any vector bundle E→ M where M is paracompact, Hausdorff and contractible

is trivial. Indeed, if f : M → pt is a homotopy equivalence with homotopy inverse pt, theng f is homotopic to the identity and so by Theorem 2.24 E is isomorphic to (g f )∗E ∼= f ∗(g∗E).But g∗E is a vector bundle over a single point and so is obviously trivial, and so f ∗(g∗E) is trivial.

More generally, if X and Y are two paracompact spaces, a similar argument shows that ahomotopy equivalence f : X → Y induces a bijection f ∗ : Vectk(Y )→ Vectk(X ) of isomorphismclasses of vector bundles. This is consistent with the classification theorem 2.23, as f induces abijection [Y, Grk(R

N )]→ [X , Grk(RN )], by precomposition. (Here we use the standard notation

[A, B] for the set of homotopy classes of maps from A to B.)

2.5. Variants and generalizations of the main classification theorem.

2.5.1. The infinite Grassmannian Grk. One can in fact obtain a version of Theorem 2.23 whichis at least superficially simpler in that it avoids any reference to stabilization, by working withR∞ rather than RN for (varying) large N . Here R∞ is the direct sum of infinitely many copiesof R, i.e., it is the space of sequences x i

∞i=1

where each x i ∈ R and all but finitely many x i

are equal to zero. So for any integer N we have inclusions IN ,∞ : RN → R∞ defined as before

by ~x 7→ (~x , ~0) (though now the zero-vector is infinite-dimensional), and R∞ is the union of theimages of the IN ,∞. The topology on R∞ is the “direct limit” topology, defined by saying that

U ⊂ R∞ is open if and only if I−1N ,∞(U) is open for all N—thus this is the finest topology on

R∞ consistent with the IN ,∞ all being continuous, and IN ,∞ embeds RN homeomorphically as asubspace of R∞.


Similarly, we define

Grk = V ≤ R∞|dim V = k,

so that there are maps iN ,∞ : Grk(RN ) → Grk defined by V 7→ V × ~0. Since any basis for a

finite-dimensional subspace of R∞ has all of its elements in some IN ,∞(RN ), Grk is the union of

the iN ,∞(Grk(RN )), and we define a topology on Grk by saying that U ⊂ Grk is open if and only

if i−1N ,∞(U) is open for all N .

As with the finite-dimensional Grassmannians there is a tautological vector bundle γk →Grk, whose fiber over the subspace V ≤ R∞ is naturally identified with that subspace. A pre-classifying map F : E → RN for a vector bundle E → M with induced map f : M → Grk(R

N )

gives rise (with obvious extensions of the definitions) to the pre-classifying map IN ,∞ F : E→

R∞ with induced map iN ,∞ f . Clearly if the pre-classifying maps F0 : E→ RN0 and F1 : E→ RN1

are stably isotopic then IN0,∞ F0 and IN1,∞ F1 are isotopic and the induced maps iN0,∞ f0 and

iN1,∞ f1 are homotopic. (Note that we no longer have to say “stably” here.)

Define α: R∞→ R∞ and β : R∞→ R∞ by setting the jth element of α(x i∞i=1) equal to x i if

j = 2i+1 and to 0 if j is even, and by setting the jth element of β(x i∞i=1) equal to x i if j = 2i

and to 0 if j is odd (so α(x1, x2, x3, . . .) = (0, 0, x1, 0, x2, 0, x3, 0, . . .) and β(x1, x2, x3, . . .) =(0, x1, 0, x2, 0, x3, . . .)). It is easy to see that α and β are both continuous maps. Moreover,if F0, F1 : E → R∞ are pre-classifying maps, then there are “straight-line isotopies” from F0 toα F0, from α F0 to β F1, and from β F1 to F1. (Here by a straight-line isotopy from a pre-classifying map G0 to a pre-classifying map G1 we mean one given by G(t, e) = (1− t)G0(e) +

tG1(e). This is indeed an isotopy provided that it is injective on each fiber, which holds if G1(e)

is never a negative multiple of G0(e) for nonzero e—this is easily seen to hold for each of thethree pairs (G0, G1) = (F0,αF0), (αF0,β F1), or (β F1, F1) listed above since in each case forall nonzero e the set of indices j such that G0(e) j 6= 0 is different from the corresponding set for

G1(e)). This shows that any two pre-classifying maps F0, F1 : E→ R∞ are isotopic (this slightlygeneralizes the previous paragraph, which established this result for pre-classifying maps of theform IN ,∞ F).

Just as in the case (Proposition 2.15) of pre-classifying maps to RN for finite N , if F : E→ R∞

is a pre-classifying map with induced map f : M → Grk, then E is isomorphic to f ∗γk. We havejust shown that any two pre-classifying maps to R∞ are isotopic, and so their associated inducedmaps are homotopic. Conversely, given two homotopic maps f0, f1 : M → Grk, by Theorem 2.24

the bundles f ∗0γk and f ∗

1γk are isomorphic.

Moreover, one can show that for any rank-k vector bundle E → M , not necessarily of finitetype, where M is paracompact and Hausdorff, there exists a pre-classifying map F : E → R∞

(this is [MS, Theorem 5.6]; it can be deduced from Lemma 2.26 using the same method as inour earlier construction of pre-classifying maps in the finite-type case—details are left to thereader). Consequently one obtains:

Theorem 2.31. For any paracompact Hausdorff space M, there is a one-to-one correspondence be-

tween isomorphism classes of rank-k vector bundles over M and homotopy classes of maps f : M →Grk, according to which (the homotopy class of) f : M → Grk corresponds to (the isomorphism

class of) f ∗γk.

Accordingly Grk is described as a “classifying space for rank-k vector bundles” (or, alterna-tively, a classifying space for the group O(k), as all for all such bundles the structure groupreduces to O(k)).

2.5.2. Classifying bundles with additional structure. The discussion in this section can be mod-ified in a mostly-straightforward way to accommodate bundles with certain extra structures,


provided that we replace the Grassmannian Grk(RN ) with another space depending on the

structure.The most important case of this is that of complex vector bundles π: E → M , i.e. rank-2n

(real) vector bundles with structure group GL(n;C) ≤ GL(2n;R). (See Example 1.16.) Given

a collection of local trivializations Φα : π−1(Uα) → Uα × Cn for E whose transition functions

gαβ(m) belong to GL(n;C), one can endow each fiber Em of E with the structure of a complex

vector space by demanding that the restriction of eachΦα to Em be a complex-linear isomorphismto m×Cn. (Equivalently, one could define a rank-n complex vector bundle as a vector bundlein which the fibers carry complex vector space structures such that the local trivializations arecomplex linear.)

If π: E→ M is a (finite-type) complex vector bundle of rank n (and hence in particular a real

vector bundle of rank 2n) and if we form the pre-classifying map F : M → R2nT = CnT exactlyas in the proof of Proposition 2.5 we note that F restricts to each fiber Em as a complex-linearmap. So we can define a complex pre-classifying map to be a continuous map F : E→ CN whichrestricts to each fiber as a complex-linear injection. As in the real case we have stabilization

maps ICN ,N ′

: CN → CN ′ , and a replication of the proof of Theorem 2.8 shows that two complex

pre-classifying maps for the same complex vector bundle are stably isotopic (with the isotopypassing only through complex pre-classifying maps).

Now define the complex Grassmannian Grn(CN ) to be the set of n-dimensional complex vec-

tor subspaces of CN . Where Vn(CN ) is the space of tuples (~v1, . . . , ~vn) ∈ (C

N )n which are lin-

early independent over C, we have a projection p : Vn(CN ) → Grn(C

N ) taking (~v1, . . . , ~vn) tospan~v1, . . . , ~vn, and so we topologize Grn(C

N ) by using the quotient topology associated top. As in the real case, a pre-classifying map F : M → CN induces a classifying map f : M →Grn(C

N ) defined by f (m) = F(Em), and a stable isotopy of pre-classifying maps induces a stablehomotopy of induced maps f . Moreover, there is a (complex) tautological bundle γn(CN ) →Grn(C

N ) (with fiber over V ∈ Grn(CN ) naturally identified with the complex vector space V ),

and if the pre-classifying map F : E→ CN induces the map f : M → Grn(CN ) then E is isomor-

phic as a complex vector bundle to f ∗γn(CN ), while conversely just as in Proposition 2.15 iff : M → Grn(C

N ) then one can construct a complex pre-classifying map for the complex vector

bundle3 f ∗γn(CN ) which induces the map f . Moreover the proof of Theorem 2.24 general-izes straightforwardly to show that the pullbacks of a complex vector bundle by two homotopicmaps are isomorphic as complex vector bundles. Consequently we obtain the following directgeneralization of Theorem 2.23.

Theorem 2.32. Any complex finite-type rank-n vector bundle E over a paracompact Hausdorff

space M has a classifying map f : M → Grn(CN ), in which case E is isomorphic as a complex

vector bundle to f ∗γn(CN ). Moreover, two such bundles are isomorphic as complex vector bundles

if and only if their classifying maps are stably homotopic.

While this result exactly parallels Theorem 2.23, note that the spaces involved are somewhat

different: for instance the homology groups of Gr1(RN ) = RPN−1 are, as you know from al-

gebraic topology, rather different from the homology groups of Gr1(CN ) = CPN−1. Just as in

Section 2.5.1, if one instead uses the Grassmannian GrCn

of complex n-dimensional subspacesof C∞ one may obtain that isomorphism classes of rank-n complex vector bundles over a para-compact Hausdorff space M are in one-to-one correspondence with homotopy classes of mapsfrom M to GrC

n.

3It is complex because, as noted in the discussion of pullbacks, the pullback of a bundle with structure group G has

structure group G


Aside from complex vector bundles, another case of some note is that of oriented vector bun-dles π: E → M , in which the structure group is the group GL+(k;R) of matrices with positivedeterminant. In this case the targets for classifying maps may be taken to be the “oriented

Grassmannians”ÝGrk(RN ), whose elements are pairs (V,o) where V is a k-dimensional subspace

of RN and o is an orientation for V . The projection p : Vk(RN ) → Grk(R

N ) factors throughÝGrk(R

N ), with (~v1, . . . , ~vk) mapping to the pair consisting of the vector space span~v1, . . . , ~vktogether with the orientation obtained by declaring (~v1, . . . , ~vk) to be a positively oriented basis;

in particular this latter projection allows us to topologize ÝGrk(RN ) with the quotient topology.

If E is an oriented vector bundle, with the fiber over m having orientation om, then a pre-

classifying map F : E → RN gives rise to a map M → ÝGrk(RN ) which sends m to the pair

(F(Em), F∗om), where we push forward orientations in the obvious way (a positively-orientedbasis for the pushforward is the image under F of a positively-oriented basis for Em). Fromhere, an exact replication of the arguments from earlier allows one to set up a one-to-one cor-respondence between isomorphism classes of finite-type oriented bundles and stable homotopy

classes of maps to ÝGrk(RN ).

Moreover, just as in the unoriented case, one can form the oriented Grassmannian ÝGrk oforiented k-planes in R∞; it is topologized by the quotient topology induced by the projection

p : Vk(R∞) →ÝGrk which sends a k-tuple of linearly independent vectors to its span together

with the orientation induced by declaring the k-tuple to be a positive basis. We have a two-to-

one cover q : ÝGrk → Grk given by q(V,o) = V , and the pullback γk = q∗γk of the tautological

bundle is naturally oriented. (Its fiber over a point (V,o) ofÝGrk is V , and the orientation on thisfiber is just o.) The general classification theorem for oriented vector bundles is then:

Theorem 2.33. For any paracompact Hausdorff space M, there is a one-to-one correspondence

between isomorphism classes of rank-k oriented vector bundles over M and homotopy classes of

maps f : M →ÝGrk, according to which (the homotopy class of) f : M →ÝGrk corresponds to (the

isomorphism class of) f ∗γk.

A general vector bundle π: E → M is called orientable if there exists an oriented vector

bundle π′ : E′→ M such that E is vector-bundle-isomorphic to E′. In this case, if f ′ : M →ÝGrk

has the property that E′ is isomorphic as an oriented bundle to ( f ′)∗γk, then where q : ÝGrk →

Grk is the two-to-one cover mentioned above q f ′ : M → Grk is a classifying map for E′ andhence is homotopic to any classifying map f : M → Grk for E. Conversely, if a classifying map

f : M → Grk for E is homotopic to a map f ′ such that there exists f ′ : M →ÝGrk with f ′ = q f ′,

then E is isomorphic to ( f ′)∗γk = ( f ′)∗γk and so is orientable. This shows that a vector bundleis orientable if and only if it has a classifying map which is homotopic to a map f ′ : M → Grk

which lifts to a map f ′ : M →ÝGrk.

Exercise 2.34. a) When k = 1, prove that the space ÝGr1 is homeomorphic to the “infinite-dimensional sphere”

S∞ = x i∞i=1∈ R∞|

∑x2

i= 1.

b) Prove that S∞ is contractible.c) Prove that any orientable rank-1 vector bundle over a paracompact Hausdorff space is

trivial.

For the following exercise we first recall some linear algebra: if V is a finite-dimensional vec-tor space over R and p is a positive integer we denote by ΛpV ∗ the space of p-linear alternating


forms ω: V p→ R, i.e. maps satisfying, for v1, . . . , vp, w ∈ V and c ∈ R,

ω(v1, . . . , vl−1, cvl + w, vl+1, . . . , vp) = cω(v1, . . . , vl , . . . , vp) +ω(v1, . . . , w, . . . , vp) (1≤ l ≤ p)

andω(v1, . . . , vi , . . . , v j , . . . , vp) = −ω(v1, . . . , v j , . . . , vi , . . . , vp) (1≤ i < j <≤ p)

It is a standard fact that ΛpV ∗ is a vector space of dimension

dim V

p

, and in particular of

dimension one if p = dim V . Moreover a generator δ for Λdim V V ∗ is obtained by choosing abasis (e1, . . . , edim V ) for V and setting δ(v1, . . . , vdim V ) equal to the determinant of the squarematrix whose jth column is consists of the coefficients of v j in terms of the basis (ei).

A linear map A: V → W between two finite-dimensional vector spaces induces a (linear)pullback map A∗ : ΛpW ∗→ ΛpV ∗ by the formula

(A∗ω)(v1, . . . , vp) =ω(Av1, . . . , Avp).

In the case that V =W and p = dim V , so that A∗ is a linear endomorphism of a one-dimensionalvector space, A∗ is given by multiplication by det A.

Exercise 2.35. Let π: E → M be a rank-k vector bundle, with local trivializations Φα : E|Uα →

Uα ×Rk. For m ∈ Uα denote the restriction of Φ−1

α to m ×Rk by ψα(m), so for each m ∈ Uα,

ψα(m) is a linear isomorphism from m ×Rk to Em.(a) Define

ΛkE∗ =

⋃m∈M

m ×ΛkE∗m

.

Prove that there is a vector bundle structure on ΛkE∗ having local trivializations over Uα givenby (m,ω) 7→ (m,ψα(m)

∗ω). (Note that in order to do this problem you in particular have to put

a topology on ΛkE∗ and prove that various maps are continuous with respect to this topology.)If the transition functions for the original bundle are given by gαβ : Uα ∩ Uβ → GL(k;R), what

are the transition functions for ΛkE∗?(b) If M is paracompact and Hausdorff prove that a rank-k vector bundle over M is orientable

if and only if ΛkE∗ is trivial.

3. INTRODUCTION TO COHOMOLOGY

3.1. Some recollections about homology. In MATH 8200 you learned about the singular ho-

mology groups of a space X ; to set up notation and pave the way for some generalizations letus recall the construction. For k ∈ Z≥0 one has the k-simplex

∆k = (t1, . . . , tk) ∈ R

k|t i ≥ 0,∑

t i ≤ 1

=

(k∑

i=0

t i vki

k∑

i=0

t i = 1

).

where vk0

is the zero-vector and vki

is the ith standard basis vector in Rk for 1 ≤ i ≤ k. Forj = 0, . . . , k+ 1 there are face maps

φ j : ∆k→∆k+1

defined by

φ j

k∑

i=0

t i vki

!=

j−1∑

i=0

t i vk+1i+

k∑

i= j

t i vk+1i+1

which embed ∆k as part of the boundary of ∆k+1.


The singular chain complex of X is defined by letting Sk(X ) be the free abelian group (i.e.

free Z-module) generated by all continuous maps σ : ∆k→ X ; in other words

Sk(X ) =

(N∑

i=1

niσi

ni ∈ Z,σi ∈ C(∆k, X )

).

The singular boundary operator is the unique homomorphism ∂ : Sk+1(X )→ Sk(X ) whose val-

ues on the generators σ for Sk+1(X ) (i.e. on continuous maps σ : ∆k+1 → X ) is given by theformula

∂ σ =

k+1∑

j=0

(−1) j(σ φ j).

One then shows that ∂ ∂ = 0, and the kth (singular) homology of X (with coefficients in Z) isdefined by

Hk(X ;Z) =ker(∂ : Sk(X )→ Sk−1(X ))

Im(∂ : Sk+1(X )→ Sk(X )).

For a subspace A ⊂ X , we can view Sk(A) as a subgroup of Sk(X ), and form the relative

singular chain group Sk(X , A) :=Sk(X )

Sk(A). The fact that ∂ maps Sk+1(A) to Sk(A) implies that ∂

descends to a map ∂ : Sk+1(X , A)→ Sk(X , A), and we have the relative homology

Hk(X , A;Z) =ker(∂ : Sk(X , A)→ Sk−1(X , A))

Im(∂ : Sk+1(X , A)→ Sk(X , A)).

For any abelian group R one can form the relative singular homology of (X , A) with coefficients

in R as follows. Since R is a Z-module, we can form Sk(X , A; R) = Sk(X , A) ⊗ R, and then theboundary operator ∂ : Sk+1(X , A) → Sk(X , A) naturally induces a map ∂ ⊗ 1: Sk+1(X , A; R) →Sk(X , A;R)which again squares to zero, and so we obtain the relative homology with coefficientsin R

Hk(X , A;R) =ker(∂ : Sk(X , A; R)→ Sk−1(X , A; R))

Im(∂ : Sk+1(X , A; R)→ Sk(X , A; R)).

As a special case we have the absolute homology with coefficients in R, Hk(X ; R) := Hk(X ,∅; R).In the case that R is a commutative ring with unity, Sk(X ; R) and Sk(X , A; R) are naturally

R-modules, and the boundary operators on the respective complexes are R-module homomor-phisms. Hence in this case the homology with coefficients in R is also an R-module.

The basic results and computational tools about singular homology from 8200, such as theMayer-Vietoris sequence, the long exact sequence of a pair, invariance under homotopy equiv-alence, and excision, all extend to homology with coefficients in R in a straightforward wayprovided that one works with the same coefficients throughout. There exists a general formula(called the universal coefficient theorem for homology—we won’t need or precisely formulatethis, though it can be found in any algebraic topology book and we will discuss the correspond-ing result for cohomology in more detail) which relates the homology with coefficients in R

to the ordinary (Z-coeffcient) homology. In particular the Z-coefficient homology determinesthe homology with coefficients in any abelian group. (If R is a field of characteristic zero thenone has the simple formula Hk(X , A;R) ∼= Hk(X , A;Z) ⊗ R, but in general the relationship ismore complicated and involves Hn−1(X , A;Z), similarly to the situation that we’ll soon see forcohomology.)

As you learned in 8200, a useful way of computing the singular homology of a space X (atleast with coefficients in Z) involves first showing that X is homeomorphic (or even just homo-topy equivalent) to a cell complex (a.k.a. CW complex). Without fully rehashing the definition,


recall that a cell complex X consists of the images of various maps (“k-cells”) fα : Dk→ X where

Dk is the closed unit disk inRk (with k varying), with X equal to the disjoint union of the embed-

ded images of the interiors of Dk under the fα, while fα(∂ Dk) is contained in a union of cells

of dimension less than k. One can then form a chain complex (C cel l∗ (X ),∂

cel l), with C cel lk(X )

equal to the free abelian group having one generator for each k-cell, and the cellular boundary

operator ∂ cel l : C cel lk+1(X )→ C cel l

k(X ) is computed by the prescription on [H, p. 140].

An important theorem in algebraic topology asserts that H∗(X ;Z) is isomorphic to the homol-

ogy of the cellular chain complex (C cel l∗ (X ),∂

cel l). This is very useful, since many spaces havecell decompositions which aren’t too complicated (in particular any compact smooth manifoldis homeomorphic to a cell complex with only finitely many cells, so the cellular chain complex isfinitely-generated) and have homology which can be effectively computed, whereas the singu-lar chain complex itself is usually unmanageably large (typically uncountably-generated). Thistheorem extends to the case of homology with coefficients: for any abelian group R we can form

the chain complex (C cel l∗ (X )⊗R,∂ cel l⊗1), and the homology of this chain complex is isomorphic

to H∗(X ;R).

Example 3.1. Let us recall the standard cell decomposition for RPn, which we view for the

purposes of this example as the quotient of the sphere Sn =(x0, . . . , xn) ∈ R

n+1∑n

i=0x2

i= 1

by the equivalence relation which identifies ~x with −~x .

The 0-dimensional unit disk D0 is just a one-point space 0 (and ∂ D0 = ∅ and int(D0) =

D0), and we define a 0-cell f0 : D0 → RPn by f0(0) = [(1, 0, . . . , 0)]. For 1 ≤ k ≤ n define

fk : Dk→ RPn by

fk(x0, . . . , xk−1) =

x0, . . . , xk−1,

√√√√1−

k−1∑

i=0

x2i, 0, . . . , 0

An element [(y0, . . . , yn)] ∈ RPn lies in the image of fk(int(Dk)) for a unique value of k,namely the largest k such that yk 6= 0. Moreover for each k, fk|int(Dk) is a homeomorphism to

its image, while fk|∂ Dk maps ∂ Dk in two-to-one fashion to ∪k−1j=0

f j(int(D j)). From this one can

see that f0, . . . , fn give a cell decomposition for the space RPn.

Thus the cellular chain complex is, as an abelian group, given by C cel lk(X ) = Z⟨ fk⟩ (i.e. the

free Z-module with generator fk) for 0≤ k ≤ n and C cel lk(X ) = 0 otherwise. As is shown in [H,

Example 2.42], the cellular boundary operator is given by

∂ cel l fk =

§2 fk−1 1≤ k ≤ n and k is even0 otherwise

So ker∂ cel l is generated by f0, f1, f3, f5, . . . , fn if n is odd and by f0, f1, f3, f5, . . . , fn−1 if n is even,

while Im(∂ cel l) is generated by 2 f1, 2 f3, . . . , 2 fn−2 is n is odd and 2 f1, 2 f3, . . . , 2 fn−1 if n is even.So we obtain

Hk(RPn;Z)∼=

Z/2Z if k is odd and 1≤ k < n

Z if k = 0 or if (k = n and n is odd)0 otherwise

Things become rather simpler if we instead work with coefficients in the ring R= Z/2Z. Then

C cel lk(RPn)⊗Z/2Z is the free Z/2Z-module generated by fk for 0≤ k ≤ n and 0 otherwise, and

the boundary operator ∂ cel l ⊗ 1 over Z/2Z is just 0 since in Z/2Z one has 2 = 0. Of course,


for a chain complex with zero boundary operator the homology is just isomorphic to the chaincomplex, so we have

Hk(RPn;Z/2Z)∼=

§Z/2Z 0≤ k ≤ n

0 otherwise

In particular for even numbers k from 2 to n, RPn has trivial kth homology over Z but non-trivial kth homology over Z/2Z. If one knows the universal coefficient theorem for homologyone could use it to compute H∗(RPn;Z/2Z) from H∗(RPn;Z), but it seems easier to directly usethe coefficient-extended cellular boundary operator.

Exercise 3.2. For every integer m≥ 3, compute H∗(RPn;Z/mZ). Also, compute H∗(RPn;Q).

3.2. Cohomology and the universal coefficient theorem. The singular cohomology of a spaceX with coefficients in Z will be the (co)homology of the dual complex of the singular chaincomplex of X . Presently we will work out some of the algebra relating to this.

Definition 3.3. Let R be a commutative ring with unity.

(i) A chain complex of R-modules (C•,∂ ) is a sequence Ckk∈Z where each Ck is an R-module, together with R-module homomorphisms ∂k : Ck→ Ck−1 such that ∂k−1 ∂k =

0. For k ∈ Z, the kth homology of (C•,∂ ) is Hk(C•) =ker∂k

Im∂k+1

.

(ii) A cochain complex of R-modules (C•,δ) is a sequence C kk∈Z where each C k is an R-

module, together with R-module homomorphisms δk : C k→ C k+1 such that δk+1δk =

0. We denote C∗ = ⊕k∈ZCk. For k ∈ Z, the kth cohomology of (C•,δ) is Hk(C•) =

kerδk

Imδk−1

.

(iii) Let (C•,∂ ) be a chain complex of R-modules and let S be another R-module. The dual

cochain complex of (C•,∂ ) with coefficients in S, denoted (C•S,δ), is given by setting

C kS= HomR(Ck, S)4 and defining δk : C k

S→ C k+1

Sby, for each α ∈ C k

Sand c ∈ Ck+1,

(δkα)(c) = α(∂k+1c).

So in case S = R, the (co)boundary operator δ on C•R

is just the transpose of the boundaryoperator ∂ on C•.

Exercise 3.4. Where R= Z, compute, Hk(C•S) in the cases where C• is the cellular chain complex

for RPn from Example 3.1 and S = Z/2Z, Q, and Z.

Shortly, we will prove the universal coefficient theorem for cohomology, which will show inparticular that (provided at least that R is a PID, as is pretty much always true in applications—indeed for the most part for our purposes R will be either Z or a field) and the Ck are free

R-modules, the cohomology H∗(C•S) := ⊕kHk(C•

S) is determined up to graded R-module iso-

morphism by the homology H∗(C•) := ⊕kHk(C•). In view of this it might seem at first thatthere wouldn’t be much point in discussing the cohomology of a space instead of the homol-ogy, since the latter already determines the former. One significant part of the importance ofthe singular cohomology H∗(X ;R) of a space X is that, unlike the homology, it possesses anatural ring structure, not just an R-module structure, and this ring structure gives informa-tion that cannot be seen in the homology. Another feature of the cohomology that is usefulfor our purposes here is that, whereas for homology a continuous map f : X → Y induces achain map f∗ : S∗(X ) → S∗(Y ) and hence an induced map on homology H∗(X ) → H∗(Y ), forcohomology taking the transpose of f∗ gives a cochain map (i.e. a map that commutes withthe coboundary operator δ) f ∗ : S∗(Y ) → S∗(X ) and hence an induced map on cohomology

4Here for any R-modules M and N we denote HomR(M , N) the R-module consisting of R-module homomorphisms

from M to N .


H∗(Y ;R) → H∗(X ;R). Note that this induced map “goes in the same direction” as the naturalpullback operation on vector bundles. One way of constructing the characteristic classes a(E) ofa vector bundle E→ M is by identifying certain special cohomology classes a ∈ H∗(Grk(R

N ); R)

and setting a(E) = f ∗a where f : M → Grk(RN ) is a classifying map for E.

The rest of the discussion in this subsection will be purely algebraic. Let R be a PID. Recallthat, by definition, an R-module S is said to be free if S is isomorphic as an R-module to a directsum of some collection of copies of R. We will require the important algebraic fact that, since R

is a PID, any submodule of a free R-module is free (for a proof see, e.g., [La, III.7.1]).We assume throughout the following that we are given a chain complex (C•,∂ ) with the

property that each of the R-modules Ck is a free R-module (for instance this applies with (C•,∂ )equal to the singular or the cellular chain complex of some space, with coefficients in R). Ourgoal is to describe the cohomology of the cochain complex C•

Sin terms of the homology of C•.

Since by definition C kS= HomR(Ck, S), a first guess might be that we should likewise have

Hk(C•S) ∼= HomR(Hk(C•), S). We’ll see that, in general, this guess in incorrect (as you should

have found in Exercise 3.4, since HomZ(Z/2Z,Z) = 0); however there is a natural “evalua-

tion” map e : Hk(C•S)→ HomR(Hk(C•), S), defined as follows.

Note that if α ∈ C kS

with δkα = 0, and if c ∈ Ck with ∂kc = 0, then for any β ∈ C k−1S

and anyd ∈ Ck+1 we have

(α+δk−1β)(c + ∂k+1 b) = α(c) + (δkα)(b) + β(∂kc + ∂k∂k+1d)

= α(c)

by the assumptions on α and c and the fact that ∂k∂k+1 = 0. In other words, the value α(c) ∈ S

depends only on the cohomology class [α] ∈ Hk(C•) of α and the homology class [c] ∈ Hk(C•)

of c. Thus we have a well-defined map

e : Hk(C•S)→ HomR(Hk(C•), S)(1)

[α] 7→ ([c] 7→ α(c))

which is clearly an R-module homomorphism.

Lemma 3.5. The map e in (1) is surjective. Indeed, there is an R-module homomorphism ζ: HomR(Hk(C•), S)→Hk(C•

S) such that e ζ is the identity on HomR(Hk(C•), S)

Proof. For all integers i let us denote Hi = Hi(C•), Zi = ker(∂i : Ci → Ci−1), and Bi = Im(∂i+1 : Ci+1→Ci). Thus Hi = Zi/Bi . Let π: Zk→ Hk denote the quotient projection.

Let ψ ∈ HomR(Hk, S), so that ψπ ∈ HomR(Zk, S). Our intention is to construct an element

α ∈ C k = HomR(Ck, S) such that α|Zk=ψπ. Note that if we can do this, then sinceπ(Bk) = 0

we would have α ∂k+1 = 0, i.e. δkα = 0, so that α would represent a cohomology class [α] ∈Hk(C•); moreover it would follow directly from the fact that α|Zk

=ψπ and the definition of e

that e([α]) =ψ. So we have effectively reduced the problem to one of systematically extendingthe R-module homomorphism ψ π: Zk→ S to a module homomorphism α: Ck→ S.

To do this, note that we have a short exact sequence

(2) 0 // Zk// Ck

∂k // Bk−1// 0

where the first map is the inclusion. Now because R is a PID and Bk−1 is a submodule of thefree R-module Ck−1, Bk−1 is a free R-module. Thus Bk−1 has a basis bαα∈A (i.e. Bk−1 is theinternal direct sum ⊕α∈ARbα), and we can use this basis to “split” the sequence (2). Namely,choose arbitrarily elements cα ∈ Ck with ∂kcα = bα, and define q : Bk−1 → Ck to be the uniqueR-module homomorphism such that q(bα) = cα for all α. Thus ∂k q = 1Bk−1

. Moreover for any


c ∈ Ck we have c−q∂kc ∈ ker∂k = Zk, and if c ∈ Zk then c−q∂kc = c. Consequently if we defineα: Ck→ S by

α(c) = (ψ π)c − q∂kc

,

then α|Zk=ψ π, as desired.

The above immediately yields the desired map ζ: HomR(Hk(C•), S) → Hk(C•S): an explicit

formula is given by

ζ(ψ) = [ψ π (1Ck− q ∂k)].

Corollary 3.6. There is an isomorphism

Hk(C•S)∼= ker(e)⊕ HomR(Hk(C•), S)

Proof. Given the map ζ from Lemma 3.5 the idea is rather similar to the end of the proof of thatlemma: the map

[α] 7→ ([α]− ζ(e([α])), e([α]))

is the desired isomorphism (with inverse (x , y) 7→ x + ζ(y)).

So to compute Hk(C•S) it remains to determine the kernel of the map e from (1). An element

α ∈ ker(δk) has the property that e([α]) = 0 if and only if it holds that α(c) = 0 for all c ∈ Zk.

(Actually, for any α ∈ C kS

such that α|Zk= 0 we automatically have α ∈ ker(δk), since δkα =

α ∂k+1 vanishes provided that α|Bk= 0, and Bk ≤ Zk.)

In particular it holds that ker(e) = 0 if and only if every α ∈ C kS

with α|Zk= 0 has the form

α = δk−1β (i.e., α = β ∂k) for some β ∈ C k−1S

. So given α ∈ C kS

with α|Zk= 0 we consider the

problem of constructing β with α = β ∂k. Now there is a unique R-module homomorphismβ0 : Bk−1→ S such that β0(∂kc) = α(c) for all c ∈ Ck (this is well-defined since if ∂kc = ∂kc′ thenc− c′ ∈ Zk so that α(c− c′) = 0). So the problem is now one of extending β0 : Bk−1→ S to an R-module homomorphism β : Ck−1→ S. Actually, if we can just extend β0 to β1 ∈ HomR(Zk−1, S)

then this will suffice by an argument similar to one used in the proof of Lemma 3.5: we havean exact sequence

0 // Zk−1// Ck−1

∂k−1 // Bk−2// 0

where Bk−2 is free, so there is q : Bk−2 → Ck−1 with ∂k−1 q = 1Bk−2, and then any β1 ∈

HomR(Zk−1, S) extends to an element β ∈ HomR(Ck−1, S) by the formula β(c) = β1(c−q∂k−1c).In particular this argument shows that ker(e) = 0 if every R-module homomorphism Bk−1 → S

extends to an R-module homomorphism Zk−1→ S.This holds if R is a field, but when R is not a field it may not hold: for instance R, S, and Zk−1

could all be Z, and Bk−1 could be 2Z. Then there is an R-module homomorphism Bk−1 → S

which sends 2 to 1, and this does not extend to an R-module homomorphism Zk−1→ S.More precisely, we have the following:

Proposition 3.7. Where i : Bk−1→ Zk−1 is the inclusion and i∗ : HomR(Zk−1, S)→ HomR(Bk−1, S)

is the restriction map ((i∗β)(b) = β(i(b))), we have

ker(e)∼=HomR(Bk−1, S)

Im(i∗ : HomR(Zk−1, S)→ HomR(Bk−1, S)).

Proof. We have

(3) ker(e) =α ∈ C k

S|α|Zk

= 0

β ∂k|β ∈ C k−1S

.


Define an R-module homomorphism

Φ : α ∈ C kS|α|Zk

= 0 →HomR(Bk−1, S)

Im(i∗ : HomR(Zk−1, S)→ HomR(Bk−1, S))

by declaring Φ(α) to be the coset of the unique element β0 ∈ HomR(Bk−1, S) such that α= β0∂k

(as noted earlier the fact that α|Zk= 0 implies that such β0 exists; its uniqueness follows from

the fact that ∂k maps Ck surjectively to Bk−1). It suffices to show that Φ is surjective and that

ker(Φ) = β ∂k|β ∈ C k−1S.

CertainlyΦ is surjective, as for any β0 ∈ HomR(Bk−1, S) the element β0∂k ∈ C kS

has β∂k|Zk=

0 and Φ(β ∂k) = [β].If α = β ∂k where β ∈ C k−1

S, then the unique element β0 ∈ HomR(Bk−1, S) associated to α

is just the restriction of β and so belongs to Im(i∗). Thus in this case Φ(α) = 0, proving that

β ∂k|β ∈ C k−1S ≤ ker(Φ).

Conversely, if Φ(α) = 0 then there is β1 ∈ HomR(Zk−1, S) such that α = β1 ∂k. As noted acouple of paragraphs above the lemma, any element of HomR(Zk−1, S) extends to an R-modulehomomorphism Ck−1 → S (because Bk−2 is a free R-module), so choosing β ∈ HomR(Ck−1, S)

such that β |Zk−1= β1 we have α= β ∂k.

Thus by (3) Φ descends to an isomorphism from ker(e) toHomR(Bk−1,S)

Im(i∗).

We have now identified ker(e) with the cokernel of the restriction map i∗ : HomR(Zk−1, S)→HomR(Bk−1, S); hence by Corollary 3.6 we have Hk(C•

S) ∼= coker(i∗)⊕ HomR(Hk(C•), S). This

actually might seem to be contrary to what I earlier said that we’d show, namely that Hk(C•S)

should only depend on the homology H∗(C•)—the map i∗ is a map between modules that willbe different for different chain complexes having the same homology. However the cokernel ofi∗ turns out to depend only on Hk−1(C•) and S, as we will now see.

Definition 3.8. Let M be an R-module. A short free resolution of M is a short exact sequence ofR-modules

0 // Fφ // G

π // M // 0

where F and G are free R-modules.

The relevant example for us is that

0 // Bk−1i // Zk−1

π // Hk−1(C•)// 0

is a free resolution of Hk−1(C·), where i is the inclusion and π is the quotient projection. Ofcourse Bk−1 and Zk−1 are free by virtue of being submodules of the free R-module Ck−1 (usingthat R is a PID).

The exact sequence

0 // Fφ // G

π // M // 0

gives rise, for any R-module S, to a dual sequence

0 // HomR(M , S)π∗ // HomR(G, S)

φ∗ // HomR(F, S) // 0

However this latter sequence is not exact—indeed we essentially noted earlier that there are

cases where φ∗ is not surjective.5 What is true is the following.

5As you’ll show in Exercise 3.13, it is true that the rest of the sequence

0 // HomR(M , S)π∗ // HomR(G, S)

φ∗ // HomR(F, S) is exact—HomR(·, S) is a “left exact functor”.


Definition-Theorem 3.9. Let M and S be R-modules and let

0 // F0

φ0 // G0

π0 // M // 0

and

0 // F1

φ1 // G1

π1 // M // 0

be two short free resolutions of M. Then the R-modules

HomR(F0, S)

Im(φ∗0 : HomR(G0, S)→ HomR(F0, S))and

HomR(F1, S)

Im(φ∗1 : HomR(G1, S)→ HomR(F1, S))

are isomorphic. Consequently we define Ex tR(M , S) as the cokernel of the mapφ∗ : HomR(G, S)→HomR(F, S) where

0 // Fφ // G

π // M // 0

is any short free resolution of M.

Corollary 3.10 (Universal Coefficient Theorem for Cohomology). Let (C•,∂ ) be a chain complex

of R-modules where R is a PID such that Ck is a free R-module for all k ∈ Z. Then for any R-module

S and any k ∈ Z we have

Hk(C•S)∼= HomR(Hk(C•), S)⊕ Ex tR(Hk−1(C•), S).

Proof. Indeed, given Definition-Theorem 3.9, this follows directly from Corollary 3.6 and Propo-sition 3.7.

Proof of Definition-Theorem 3.9.

Lemma 3.11. There are R-module homomorphisms f : F0 → F1 and g : G0 → G1 such that the

diagram

(4) 0 // F0

φ0 //

f

G0

π0 //

g

M //

1M

0

0 // F1

φ1 // G1

π1 // M // 0

is commutative. Moreover if f ′ : F0 → F1 and g ′ : G0 → G1 are two other R-module homomor-

phisms making (4) commutative, then there is K ∈ HomR(G0, F1) such that

f − f ′ = K φ0 and g − g ′ = φ1 K .

Proof of Lemma 3.11. Let bβ0 β∈B be a basis for the free R-module G0. By the surjectivity of π1,

for each β we may choose bβ1 ∈ G1 such that π1(b

β1 ) = π0(b

β0 ), and then if we let g : G0 → G1

be the unique R-module homomorphism such that g(bβ0 ) = b

β1 then the right square of (4) will

commute.Given this g, if aα

0α∈A is a basis for the free R-module F0, then for each αwe have g(φ0aα

0) ∈

kerπ1, and so there is (in fact a unique) aα1∈ F1 such that φ1(a

α1) = g(φ0aα

0). Then letting

f : F0→ F1 be the unique R-module homomorphism such that f (aα0) = aα

1for each α results in

(4) commuting.As for the uniqueness statement, if ( f , g) and ( f ′, g ′) are two pairs of maps making (4)

commute, then in particular π1 (g − g ′) = 0, so that Im(g − g ′) ≤ Imφ1. So we may defineK : G0→ F1 by, for b ∈ G0, setting K b equal to the (unique, since φ1 is injective) element a ∈ F1

such that φ1(a) = (g − g ′)(b). K is easily seen to define a module homomorphism since the


other maps involved in the diagram are module homomorphisms, and it is immediate from thedefinition that g − g ′ = φ1 K . Meanwhile we have

φ1 ( f − f ′) = (g − g ′) φ0 = φ1 K φ0.

Since φ1 is injective this immediately implies that f − f ′ = K φ0.

Now the map f : F0 → F1 in (4) induces an adjoint map f ∗ : HomR(F1, S) → HomR(F0, S),namely f ∗α = α f for α ∈ HomR(F1, S). If α ∈ Im(φ∗

1), say α = φ∗

1β = β φ1 where

β ∈ HomR(G1, S), then

f ∗α= β φ1 f = β g φ0 = φ∗0(β g).

Thus f ∗ maps Im(φ∗1) to Im(φ∗

0), and so f ∗ descends to a map f ∗ : coker(φ∗

1) → coker(φ∗

0)

(which we will eventually show is an isomorphism, proving the theorem).For another choice of maps f ′, g ′ making (4) commute, for all α ∈ HomR(F1, S) we have

f ∗α− f ′ ∗α= α ( f − f ′) = α K φ0 ∈ Im(φ∗0).

Hence the induced maps f ∗ : coker(φ∗1) → coker(φ∗

0) and f ′

∗: coker(φ∗

1) → coker(φ∗

0) are

equal to each other. In other words, the map f ∗ : coker(φ∗1)→ coker(φ∗

0) induced by a pair of

maps f and g that make (4) commute is independent of the particular choice of maps f and g.Reversing the roles of the two resolutions of M , we obtain R-module homomorphisms p and

q making the diagram

0 // F1

φ1 //

p

G1

π1 //

q

M //

1M

0

0 // F0

φ0 // G0

π0 // M // 0

commute, and hence we get a map p∗ : coker(φ∗0) → coker(φ∗

1) which is independent of the

particular maps p and q. We then also have commutative diagrams

0 // F0

φ0 //

p f

G0

π0 //

qg

M //

1M

0

0 // F0

φ0 // G0

π0 // M // 0

and

0 // F1

φ1 //

f p

G1

π1 //

gq

M //

1M

0

0 // F1

φ1 // G1

π1 // M // 0

But the above two diagrams would also commute if the vertical maps were replaced by theidentity map, and by the independence property discussed above it hence follows that (p f )∗ =

1coker(φ∗0)and ( f p)∗ = 1coker(φ∗1)

. But by the standard functoriality property of adjoints, we

have p∗ f ∗ = ( f p)∗ and f ∗ p∗ = (p f )∗. This proves that f ∗ and p∗ are inverses to each

other, and so define isomorphisms between coker(φ∗0) and coker(φ∗

1).

The universal coefficient theorem implies in particular that, to compute the cohomology ofthe dual complex to some chain complex S∗, it suffices to compute the cohomology of anyother chain complex C∗ whose homology is the same as that of S∗. In particular, since the


cellular homology of a cell complex X is isomorphic its singular homology, the singular coho-mology of X can be computed by computing the cohomology of the dual complex of the cellularchain complex. Thus in Exercise 3.4 you in fact computed the singular cohomology groupsH∗(RPn;Z/2Z), H∗(RPn;Q), and H∗(RPn;Z)—and in principle you could even have done thiswithout first computing the (cellular or singular) homology.

Exercise 3.12. (a) Prove that if F is a free R-module and S is any R-module then Ex tR(F, S) = 0.(b) For all integers m, n≥ 2 compute Ex tZ(Z/mZ,Z/nZ).

Exercise 3.13. Let

0 // Mφ // N

π // P // 0

be a short exact sequence of R-modules, and let S be another R-module. Prove that the sequence

0 // HomR(P, S)π∗ // HomR(N , S)

φ∗ // HomR(M , S)

is exact.

Exercise 3.14. A short exact sequence of R-modules

(5) 0 // Mφ // N

π // P // 0

is said to split if there is q ∈ HomR(P, N) such that π q : P → P is the identity.(a) Prove that if P is a free R-module then the sequence (5) splits.(b) Prove that, if the sequence (5) splits, then the sequence

0 // HomR(P, S)π∗ // HomR(N , S)

φ∗ // HomR(M , S) // 0

is exact.

3.3. Basic properties of singular cohomology. As discussed earlier, if X is any space and R isan abelian group, the singular cohomology of X , H∗(X ; R), is by definition the cohomology ofthe singular cochain complex S•(X ;R), i.e. the dual complex with coefficients in R of the singu-

lar chain complex of Z-modules S•(X )6 Moreover if A ⊂ X is a subspace, the relative singular

homology with coefficients in R, H∗(X , A; R), is defined as the cohomology of the dual complex

with coefficients in R of the relative singular chain complex S•(X , A) =S•(X )

S•(A).

Since Z is a PID and each Sk(X ) is, by definition, a free Z-module, the universal coefficienttheorem applies to give isomorphisms

Hk(X ;R)∼= HomZ(Hk(X ;Z), R)⊕ Ex tZ(Hk−1(X ;Z), R).

Alternately, if the coefficient group R is a PID, then in accordance with the most recent footnotewe have an isomorphism

Hk(X ;R)∼= HomR(Hk(X ; R), R)⊕ Ex tR(Hk−1(X ; R), R).

This latter formulation is perhaps most convenient when R is a field, since in that case the Ex tR

group automatically vanishes and so the cohomology with coefficients in R is just the dual ofthe homology with coefficients in R.

6Here S•(X ) is considered as a chain complex of Z-modules, and Z and R are playing the role of the groups that

were denoted by, respectively, R and S in the notation of the last section. (I used slightly different notation in class,

which avoided the conflicting roles played by R.) If R is a ring, it would be equivalent to say that S•(X ; R) is the dual

complex with coefficients in R of S•(X )⊗ R, which in the notation of the last section would mean taking both R and S

equal to R.


It might not be immediately clear that the universal coefficient theorem applies to the relative

singular cohomology, since Sk(X , A) =Sk(X )

Sk(A)is defined as a quotient of free Z-modules, and of

course such quotients are not always free. However, Sk(X , A) is a free Z-module: it is isomorphic

to the free Z-module generated by those continuous maps σ : ∆k→ X such that σ(∆k) 6⊂ A, asyou should be able to convince yourself. Consequently the universal coefficient theorem doesapply to give an isomorphism

(6) Hk(X , A;R)∼= HomZ(Hk(X , A;Z), R)⊕ Ex tZ(Hk−1(X , A;Z), R)

Recall that if f : (X , A)→ (Y, B) is a map of pairs (i.e., f : X → Y is a continuous map, A⊂ X ,B ⊂ Y , and f (A) ⊂ B) then there is (for all k) an induced map f∗ : Sk(X , A) → Sk(Y, B) which

descends from the map f∗ : Sk(X ) → Sk(Y ) defined on simplices σ : ∆k → X by the obviousformula f∗σ = f σ. Such induced maps f∗ are functorial with respect to composition in thesense that (g f )∗ = g∗ f∗ for f : (X , A)→ (Y, B) and g : (Y, B)→ (Z , C), and moreover they are

chain maps: f∗ ∂(X ,A) = ∂ (Y,B) f∗. Moreover if f0, f1 : (X , A)→ (Y, B) are homotopic as maps

of pairs (i.e. there is a homotopy F : [0, 1]× X → Y from f0 to f1 such that F([0, 1]× A) ⊂ B),then there is a chain homotopy from ( f0)∗ to ( f1)∗, i.e., for all k a map K : Sk(X , A)→ Sk+1(Y, B)

such that ( f0)∗ − ( f1)∗ = K ∂ (X ,A) + ∂ (Y,B) K .As you likely learned in 8200, this implies that a map of pairs (X , A) → (Y, B) induces a

map H∗(X , A;Z) → H∗(Y, B;Z), and moreover that homotopic maps induce the same map onhomology. We have the corresponding result for cohomology, with the direction of the mapsreversed:

Proposition 3.15. Let R be any abelian group. To any map of pairs f : (X , A) → (Y, B) there

corresponds (for all k) a map f ∗ : Hk(Y, B; R) → Hk(X , A; R), obeying the functoriality property

(g f )∗ = f ∗ g∗ and the property that if f0 and f1 are homotopic through maps of pairs then

f ∗0= f ∗

1. Moreover the identity map 1(X ,A) : (X , A) → (X , A) has the property that 1∗

(X ,A)is the

identity on Hk(X , A;R)→ Hk(X , A;R).

Proof. Define f ∗ : Sk(Y, B;R) → Sk(X , A; R) by f ∗α = α f∗ where f∗ is the map on singularchain complexes described above the proposition. We have, generally

àg f∗α= α (g f ) = (α g) f = f ∗ g∗α,

andf ∗δ(Y,B)α= (α ∂ (Y,B) f∗ = α f∗ ∂

(X ,A) = δ(X ,A) f ∗α.

The latter identity implies that f ∗ maps kerδ(Y,B) to kerδ(X ,A) and Imδ(Y,B) to Imδ(X ,A) and so

descends to a map f ∗ : Hk(Y, B;R)→ Hk(X , A; R) for all k, and the relation àg f∗α = f ∗( g∗α)

directly implies that (g f )∗ = f ∗ g∗. Clearly the map induced by the identity is the identity.It remains to prove the homotopy invariance property. Given homotopic maps f0, f1 : (X , A)→

(Y, B) we have K : Sk(X , A)→ Sk+1(Y, B) with ( f0)∗− ( f1)∗ = K ∂ (X ,A)+ ∂ (Y,B) K . We then see

that, for β ∈ Sk(Y, B;R),

f ∗0β − f ∗

1β = β

( f0)∗ − ( f1)∗

= β K ∂ (X ,A) + β ∂ (Y,B) K = δ(X ,A)(β K) + (δ(Y,B)β) K .

So if β is a cocycle in Sk(Y, B;R) (i.e. an element of kerδ(Y,B)), then f ∗0β − f ∗

1β is a coboundary

in Sk(X , A;R) (i.e. an element of Imδ(X ,A)). Consequently where [β] denotes the cohomologyclass of β we will have f ∗

0[β] = f ∗

1[β], proving that indeed f ∗

0= f ∗

1.

Remark 3.16. Of course, as the special case where A = B = ∅, Proposition 3.15 includes thestatement that a continuous map f : X → Y induces a homomorphism of absolute cohomology

groups f ∗ : Hk(Y ;R)→ Hk(X ;R), satisfying all of the indicated properties.


Corollary 3.17. If f : (X , A)→ (Y, B) is a homotopy equivalence7 then for all coefficient groups R

and all k the induced map f ∗ : Hk(Y, B; R)→ Hk(X , A; R) is an isomorphism.

Proof. Indeed, Proposition 3.15 immediately implies that, if g is a homotopy inverse for f (asin the footnote) then f ∗ g∗ and g∗ f ∗ are the respective identities.

Remark 3.18. The universal coefficient theorem, together with the corresponding result for ho-mology (which you presumably learned in 8200) would be enough to prove that two homotopyequivalent spaces have isomorphic cohomology. However the isomorphism in the universal co-efficient theorem is not generally compatible with the maps f ∗, and so one needs the homotopyinvariance statement in Proposition 3.15 to obtain the more specific statement that the inducedmap of a homotopy equivalence is an isomorphism. (Anyway, Proposition 3.15 doesn’t use theuniversal coefficient theorem, and is easier to prove.)

Exercise 3.19. Assume that R is a nontrivial commutative ring with unity, so that we have a ringmorphism ε: Z→ R obtained by sending 1 ∈ Z to the multiplicative identity in R.

For each path component P of a space X define an element αP ∈ S0(X ; R) as follows. Since

∆0 is a one-point set, a general element of S0(X ) has the form

c =∑

x∈X

nx x

where only finitely many of the nx are nonzero. Accordingly define

αP

∑

x∈X

nx x

=∑

x∈P

ε(nx).

(a) Prove that αP is a cocycle (i.e. that δαP = 0), and that the cohomology class [αP] is a

nontrivial element of H0(X ;R).(b) Assuming that X has just finitely many path components P1, . . . , Pm, prove that

H0(X ;R) =

(m∑

i=1

ri[αPi]

r1, . . . , rm ∈ R

).

(c) For any space X let π0(X ) denote the set of path components of X . Prove that (regardless

of whether π0(X ) is finite) for any abelian group R, H0(X ; R) is isomorphic to the abelian groupof functions fromπ0(X ) to R, by a map which sends a cohomology class [α] to the function whichassigns to each path component P ⊂ X the value α(p) where p is some point in P (viewed as azero-simplex).

Recall that if (C•,∂C) and (D•,∂D) are chain complexes a chain map f : C•→ D• is a collectionof homomorphisms f : Ck→ Dk for all k such that f ∂C = ∂D f . In this case f maps ker∂C toker∂D and Im∂C to Im∂D and hence induces homomorphisms f∗ : Hk(C•)→ Hk(D•) for all k. Ashort exact sequence of chain complexes is by definition a sequence

0 // C•f // D•

g // E• // 0

where f and g are chain maps such that each of the sequences

0 // Ck

f // Dk

g // Ek// 0

7i.e., if there is g : (Y, B)→ (X , A) such that f g and g f are homotopic to the respective identities through maps

of pairs


is exact. A basic fact from elementary homological algebra (presumably familiar from 8200) isthat a short exact sequence of chain complexes induces a long exact sequence of homology groups:for certain homomorphisms d : H j+1(E•)→ H j(C•) we have an exact sequence

· · ·g∗ // Hk+1(E•)

d // Hk(C•)f∗ // Hk(D•)

g∗ // Hk(E•)d // Hk−1(C•)

f∗ // · · ·

For instance, if X is a topological space and A⊂ X , essentially by definition we have a shortexact sequence of chain complexes

(7) 0 // S•(A)i // S•(X )

π // S•(X , A) // 0

where i is the inclusion and π is the projection. Hence we get the long exact sequence forrelative homology:

· · ·π∗// Hk+1(X , A;Z)

d // Hk(A;Z)i∗ // Hk(X ;Z)

π∗ // Hk(X , A;Z)d // Hk−1(A;Z) · · ·

Meanwhile, for any abelian group R we can dualize (7) to obtain the sequence

0 // S•(X , A;R)π∗ // S•(X ; R)

i∗ // S•(A; R) // 0

The exactness of this sequence follows from Exercises 3.13 and 3.14 together with the obser-

vation made at the start of this subsection that the groups Sk(X , A) =Sk(X )

Sk(A)are free Z-modules.

Thus we have a short exact sequence of cochain complexes (under the obvious definition), which

hence8 induces a long exact sequence of cohomology groups:(8)

· · ·i∗ // Hk−1(A;R)

d // Hk(X , A;R)π∗ // Hk(X ; R)

i∗ // Hk(A; R)d // Hk+1(X , A; R) · · ·

So the exact sequence for the cohomology of a pair is just like that for homology except that all

of the maps go in the opposite direction. Note that the map i∗ : Hk(X ; R)→ Hk(A; R) is just thesame as the map induced by the inclusion ι : A→ X , as can be seen by inspecting the definitions.

At the level of cochains it is easy to understand both of the maps i∗ : Sk(X ; R)→ Sk(A; R) and

π∗ : Sk(X , A;R) → Sk(X ;R). We have Sk(X ; R) = HomZ(Sk(X ), R), and for a homomorphismα: Sk(X ) → R, i∗α: Sk(A) → R is just given by restricting α to Sk(A) (which, since it consistsof formal linear combinations of simplices in A, is obviously contained in Sk(X )). As for π∗,

note that Sk(X , A;R) = HomZ

Sk(X )

Sk(A), R

may be naturally identified with those elements of

Sk(X ;R) = HomZ(Sk(X ), R) which vanish on all simplices which are contained in A, and π∗ isprecisely this identification.

Exercise 3.20. Prove that the following are equivalent, for any nontrivial abelian group R:

(i) H0(X , A;R) = 0.(ii) The map i∗ : H0(X ;R)→ H0(A; R) is injective.

(iii) Every path component of X intersects A.

Example 3.21. Let us compute, for all n ≥ 1, the relative cohomology of the pair (Rn,Rn \ 0)with coefficients in an arbitrary abelian group R by using the long exact sequence (8). (We’llsoon see that this example is rather important in the theory of characteristic classes.) First we

8Since, abstractly, a cochain complex can be turned into a chain complex, and vice versa, by negating the grading

k, it follows formally from the corresponding result for homology that a short exact sequence of cochain complexes

induces a long exact sequence on cohomology, with the “connecting homomorphisms” d increasing grading by 1 rather

than decreasing it.


should compute the absolute cohomologies. Rn is homotopy equivalent to a point (and hencehas the same cohomology), and so there are several easy ways of seeing that therefore we have

Hk(Rn; R)∼=

§R k = 00 otherwise

(for instance one can use the corresponding computation for homology and the universal coef-ficient theorem, since Hom(Z, R)∼= R; alternatively, one could use the cellular cochain complex

for a single point). Similarly Rn \ 0 is homotopy equivalent to Sn−1, which is a cell complexwith one 0-cell, one (n − 1)-cell, and trivial boundary operator (or, if n = 1, two 0-cells andtrivial boundary operator). So cellular cohomology establishes

Hk(Rn \ 0;R)∼=

§R k = 0, n− 10 otherwise

for n> 1 and Hk(R1 \ 0; R)∼=

§R⊕ R k = 00 otherwise

Part of the exact sequence (8) reads

Hk−1(Rn;R) // Hk−1(Rn \ 0; R) // Hk(Rn,Rn \ 0; R) // Hk(Rn; R) .

If k ≥ 2 the two outermost terms are zero, and so Hk−1(Rn \ 0; R) // Hk(Rn,Rn \ 0; R)

is an isomorphism. Thus

For k ≥ 2, Hk(Rn,Rn \ 0; R)∼=

§R k = n

0 otherwise

In fact we’ll see that the above formula holds for all k, but the lower values require particularattention. The first nontrivial part of (8) is(9)

H0(Rn,Rn \ 0;R) // H0(Rn;R)i∗ // H0(Rn \ 0; R) // H1(Rn,Rn \ 0; R) // 0

(the 0 is H1(Rn;R)). Of course Rn is path-connected, while Rn \ 0 is path-connected if n > 1

and has two path components if n= 1. So from the characterization of H0 in Exercise 3.19(c),

using the fact that i∗ is the map induced by inclusion, if n> 1 then H0(Rn; R) and H0(Rn\0; R)

may both be identified with R in such a way that i∗ acts by the identity, while if n = 1 then

H0(Rn;R)∼= R and H0(Rn \ 0;R)∼= R⊕ R, with i∗ acting as the diagonal map r 7→ (r, r).

So in either case i∗ is injective, so that H0(Rn,Rn \ 0; R) = 0 (as indeed also follows from

Exercise 3.20). Moreover the exactness of (9) implies that H1(Rn,Rn \ 0; R) is isomorphic to

the cokernel of i∗; if n > 1 then i∗ is an isomorphism and so H1(Rn,Rn \ 0; R) ∼= 0, while if

n = 1 (so i∗ is the map r 7→ (r, r)) then H1(R1,R1 \ 0; R) ∼= R. This completes the proof that,for all k, n,

(10) Hk(Rn,Rn \ 0; R)∼=

§R k = n

0 otherwise

3.3.1. The Mayer-Vietoris sequence for relative cohomology. Let X be a space, with X = U ∪ V

where U and V are open. Denote by SU ,V

k(X ) the free abelian group generated by k-simplices

σ : ∆k → X with the property that either Im(σ) ⊂ U or Im(σ) ⊂ V . The boundary operator

for the singular chain complex S•(X ) maps SU ,V

k(X ) to S

U ,V

k−1(X ), since if σ has image in U or V

then so does the boundary of σ. Thus we obtain a chain complex SU ,V• (X ), with an inclusion

j : SU ,V• (X )→ S•(X ) (which is a chain map and hence induces a map on homology).

A key lemma (see for instance [H, Proposition 2.21]) from 8200 asserts that the inducedmap on homology j∗ : Hk(S

U ,V• (X ))→ Hk(X ;Z) is an isomorphism. More specifically, there is a

“subdivision map” s : S•(X )→ SU ,V• (X ) which is a chain map such that s j : SU ,V

• (X )→ SU ,V• (X )


is the identity while j s : S•(X )→ S•(X ) is chain homotopic to the identity, i.e. there are mapsK : Sk(X ) → Sk+1(X ) such that j s − 1 = ∂ K + K ∂ . Dualizing, for any abelian group

R we have cochain maps j∗ : Hom(S•(X ), R) → Hom(SU ,V• (X ), R) and s∗ : Hom(SU ,V

• (X ), R) →Hom(S•(X ), R) such that (s j)∗ = j∗s∗ is the identity and there are maps K∗ : Hom(Sk+1(X ), R)→Hom(Sk(X ), R) such that s∗ j∗ − 1= K∗ δ+δ K∗ where δ = ∂ ∗ is the coboundary operator.As in Proposition 3.15, it follows that the induced maps on cohomology:

s∗ : Hk(Hom(SU ,V• (X ), R))→ Hk(X ; R) j∗ : Hk(X ; R)→ Hk(Hom(SU ,V

• (X ), R))

are inverses to each other. In particular, for any space X with open cover X = U ∪ V and anyabelian group R and integer k,

(11) j∗ : Hk(X ;R)→ Hk(Hom(SU ,V• (X ), R)) is an isomorphism.

There is also a relative version of this statement: suppose that A⊂ X is a subspace, so that U∩A and V∩A form an open cover of A, and so we have a chain complex SU∩A,V∩A

• (A)which is both a

subcomplex of S•(A) and a subcomplex of SU ,V• (X ), with the inclusion SU∩A,V∩A

• (A)→ S•(A) being

given by the restriction of j : SU ,V• (X )→ S•(X ). By (11), j∗ : Hk(A; R)→ Hk(Hom(SU∩A,V∩A

• (A), R))

is an isomorphism for every abelian group R and integer k. Meanwhile we have the quotientcomplexes

SU ,V• (X , A) :=

SU ,V• (X )

SU∩A,V∩A• (A)

S•(X , A) =S•(X )

S•(A),

and j descends to a chain map (still denoted j) from SU ,V• (X , A) to S•(X , A).

Proposition 3.22. For all R and k the induced map on cohomology j∗ : Hk(X , A; R)→ Hk(Hom(SU ,V• (X , A), R)

is an isomorphism.

Proof. Let i : A→ X be the inclusion. Dualizing the short exact sequences corresponding to theinclusions i∗ : S•(A) → S•(X ) and i∗ : SU∩A,V∩A

• (A) → SU ,V• (X ) gives a commutative diagram of

short exact sequences of cochain complexes

0 // Hom(S•(X , A), R)π∗ //

j∗

Hom(S•(X ), R)i∗ //

j∗

Hom(S•(A), R) //

j∗

0

0 // Hom(SU ,V• (X , A), R)

π∗ // // Hom(SU ,V• (X ), R)

i∗ // Hom(SU∩A,V∩A• (A), R) // 0

,

where all maps (both horizontal and vertical) are (co)chain maps. (The exactness of the toprow has already been discussed. As for the bottom row, its exactness follows from Exercise 3.14

once we observe that SU ,V

k(X , A) is a free Z-module for all k: specifically, S

U ,V

k(X , A) is isomorphic

to the free Z-module generated by those simplices σ : ∆k→ X such that Im(σ) is not containedin A but is contained either in U or in V .) Both the top and the bottom row have associatedlong exact sequences in cohomology, and the facts that the above diagram commutes and thevertical maps are chain maps implies that the following diagram commutes (check this yourselfif you haven’t seen it):

Hk−1(X ;R)i∗ //

j∗

Hk−1(A;R)d //

j∗

Hk(X , A; R)π∗ //

j∗

Hk(X ; R)i∗ //

j∗

Hk(A;R)

j∗

Hk−1(Hom(S

U ,V• (X ), R))

i∗ //Hk−1(Hom(S

U∩A,V∩A• (A), R))

d //Hk(Hom(S

U ,V• (X , A), R))

π∗ //Hk(Hom(S

U ,V• (X ), R))

i∗ //Hk(Hom(S

U∩A,V∩A• (A), R))


Now the rows above are both exact and the diagram commutes, and the outer four maps j∗ are all

isomorphisms by (11). Hence the five-lemma implies that the innermost map j∗ : Hk(X , A; R)→Hk(Hom(SU ,V

• (X , A), R) is also an isomorphism.

Continuing to assume that we have an open cover X = U∪V and an arbitrary subspace A⊂ X ,the relative Mayer–Vietoris sequence is a consequence of the following short exact sequence ofchain complexes:

(12)

0 // S•(U ∩ V, (U ∩ V )∩ A)f // S•(U , U ∩ A)⊕ S•(V, V ∩ A)

g // SU ,V• (X , A) // 0

Here the map f is induced by the map f : S•(U ∩ V )→ S•(U)⊕ S•(V ) which sends a simplex

σ : ∆k → U ∩ V to the pair (σ,−σ) ∈ S•(U)⊕ S•(V ). Meanwhile g : S•(U , U ∩ A)⊕ S•(V, V ∩A) → SU ,V

• (X , A) is induced by the map g : S•(U) ⊕ S•(V ) → SU ,V• (X ) defined by, for simplices

σ : ∆k → U and τ: ∆k → (V ), g(σ,τ) = σ + τ. Since f and g both preserve the property ofmapping into A, they do indeed descend to maps f and g.

It is not hard to see that (12) is indeed exact. An element∑

niσi ∈ S•(U ∩ V ) (with the

σi : ∆k→ U ∩ V all distinct and ni all nonzero) which is mapped by f to an element of S•(U ∩

A)⊕S•(V ∩A) would have all of its σi contained in A and would hence belong to S•(U ∩ V ∩A);hence the induced map f : S•(U ∩ V, (U ∩ V ) ∩ A)→ S•(U , U ∩ A)⊕ S•(V, V ∩ A) is injective. Itis obvious that g f = 0, i.e. that Im( f ) ≤ ker(g) (this is just the equation σ − σ = 0). Forthe reverse inclusion, note first that any element of S•(U , U ∩ A)⊕ S•(V, V ∩ A) can be uniquelyrepresented by an element of S•(U)⊕S•(V ) of the form (

∑niσi ,

∑m jτ j), where the integers ni

and m j are all nonzero, where eachσi : ∆k→ U and each τ j : ∆

k→ V has image not contained

in A, and where the σi are all distinct from one another and the τ j are all distinct from one

another. This element of S•(U , U ∩ A)⊕ S•(V, V ∩ A) lies in the kernel of g if and only if

g∑

niσi ,∑

m jτ j

=∑

niσi +∑

m jτ j ∈ SU∩A,V∩A• (A).

But since the σi and τ j all have image not contained in A, this can occur only if∑

niσi +∑m jτ j = 0. So we must have

∑m jτ j = −

∑niσi ∈ S•(U)∩ S•(V ).

But evidently S•(U)∩ S•(V ) = S•(U ∩ V ), and so (∑

niσi ,∑

m jτ j) = f (∑

niσi) ∈ Im( f ), and

so the equivalence class of this element lies in the image of f .Finally, the definition of SU ,V

• (X ) makes it obvious that g : S•(U) ⊕ S•(V ) → SU ,V• (X ) is

surjective–essentially by definition, an element x ∈ SU ,V• (X ) has the form x = σ + τ where

σ is a linear combination of simplices in U and τ is a linear combination of simplices in V , andin this case we have x = g(σ,τ). Since g is surjective, its induced map g on the relative chaincomplex must also be surjective.

So (12) is indeed a short exact sequence of chain complexes. Taking the resulting long exactsequence on homology and using the analogue of Proposition 12 for homology gives the “Mayer-Vietoris sequence for relative homology,” which at least in the absolute case where A=∅ shouldbe familiar. We will instead need the version of this sequence for relative cohomology. Since (as

noted in the proof of Proposition 3.22) for each k the chain group SU ,V

k(X , A) is a free abelian

group, by Exercise 3.14 we have for any abelian group R a short exact sequence of cochain


complexes9

0→ Hom(SU ,V• (X , A), R)

g∗ // Hom(S•(U , U ∩ A), R)⊕ Hom(S•(V, V ∩ A), R)f ∗ // Hom(S•(U ∩ V, (U ∩ V )∩ A), R)→

Combining the long exact sequence on cohomology associated to this short exact sequence

of cochain complexes with the isomorphism j∗ : Hk(X , A; R) → Hk(Hom(SU ,V• (X , A), R)) from

Proposition 3.22 we obtain:

Theorem 3.23 (Mayer-Vietoris sequence for relative cohomology). Given a space X with an open

cover X = U ∪ V , a subspace A⊂ X , and an abelian group R, there is a long exact sequence

Hk+1(X , A;R) // · · ·

Hk(X , A;R) // Hk(U , U ∩ A;R)⊕ Hk(V, V ∩ A; R) // Hk(U ∩ V, (U ∩ V )∩ A; R)

mmZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

· · · // Hk−1(U ∩ V, (U ∩ V )∩ A; R)

mmZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

It is sometimes important to know what the horizontal maps in the above sequence are, sohere is an exercise:

Exercise 3.24. Using j∗ to identify Hk(X , A; R) with Hk(Hom(SU ,V• (X , A), R)), prove that, accord-

ing to our construction of the Mayer-Vietoris sequence as the long exact sequence associated tothe dual exact sequence of (12),

• The map Hk(X , A;R)→ Hk(U , U ∩ A; R)⊕ Hk(V, V ∩ A; R) is given by

x 7→ (i∗U

x , i∗V

x),

where iU : (U , U ∩ A)→ (X , A) and iV : (V, V ∩ A)→ (X , A) are the inclusions.

• The map Hk(U , U ∩ A;R)⊕ Hk(V, V ∩ A; R)→ Hk(U ∩ V, (U ∩ V )∩ A; R) is given by

(y, z) 7→ j∗U

y − j∗V

z,

where jU : (U ∩ V, U ∩ V ∩ A)→ (U , U ∩ A) and jV : (U ∩ V, U ∩ V ∩ A)→ (V, V ∩ A) arethe inclusions.

Given this identification of the maps, let us quickly note one consequence of the exactness of

the sequence in Theorem 3.23: Suppose we have classes y ∈ Hk(U , U ∩A; R) and z ∈ Hk(V, V ∩A;R) which restrict to the same class in Hk(U ∩ V, U ∩ V ∩ A; R) (i.e., j∗

Uy = j∗

Vz). Then there is

a class x ∈ Hk(X , A;R) which restricts both to y and to z (i.e., i∗U

x = y and i∗V

x = z—note thatthe converse to this statement is obvious: if x restricts both to y and to z then y and z musthave the same restriction to U ∩ V ). In general such a class x is not uniquely determined by y

and z; rather, given one such class x0, all others differ from x0 by an element of the image of

the map Hk−1(U ∩ V, (U ∩ V ) ∩ A;R)→ Hk(X , A; R). (One does however have uniqueness of x

in the special case that Hk−1(U ∩ V, (U ∩ V )∩ A; R) = 0.)

9In the middle term we are making use of the standard isomorphism Hom(N⊕P, R)∼= Hom(N , R)⊕Hom(P, R), given

by sending a map N ⊕ P → R to the pair consisting of its restriction to N and its restriction to P.


4. THE THOM AND EULER CLASSES

We now know enough about cohomology to begin the construction of our first example of acharacteristic class, the Euler class, for certain vector bundles.

Consider a rank-k vector bundle π: E → M . There is a distinguished copy of M , the zero-

section 0E , inside E: define s0 : M → E by sending a point m ∈ M to the zero element in

the fiber Em = π−1(m) of E over m. If U ⊂ M is an open subset with local trivialization

Φ : E|U → U ×Rk, we have Φ s0(m) = (m, 0) for all m ∈ U (since Φ restricts to each fiber Em

as a linear map to m ×Rk). So since Φ is a homomorphism s0 restricts to U is a continuousmap; thus s0 : M → E is continuous since M is covered by open sets U of the sort just described.Similarly one sees that the image 0E := s0(M) is a closed subset of E, and that s0 maps M

homeomorphically to 0E .Now define

E0 = E \ 0E ,

i.e., E0 is the complement of the zero section in E. For A ⊂ M we denote E|A = π−1(A) and

E0|A = E|A ∩ E0, and for m ∈ M we denote Em = π−1(m) and E0

m= Em ∩ E0.

It turns out to be worthwhile to investigate the relative cohomology H∗(E, E0; R) for various

coefficient rings R.10 Specifically we will work either with R= Z/2Z or R= Z. In the case when

R = Z we will additionally assume that E is an oriented vector bundle, but we will not need thisassumption when R= Z/2Z.

For each m ∈ M the inclusion im : Em → E of the fiber over m maps E0m

, and so there is aninduced map

i∗M

: H∗(E, E0; R)→ H∗(Em, E0m

; R).

Now (Em, E0m) is homeomorphic (as a pair) to (Rk,Rk \0), and we have computed the relative

cohomology of (Rk,Rk \ 0) in Example 3.21: Hk(Rk,Rk \ 0; R) is isomorphic to R, while for

j 6= k H j(Rk,Rk \ 0;R) is trivial.Let us choose once and for all (independently of the vector bundle E) a generator ω of

Hk(Rk,Rk \ 0;R). Of course, if R = Z/2Z there is no choice here since there is only onegenerator (indeed only one nonzero element). If R = Z then there are two possibilities (sinceboth 1 and −1 generate the abelian group Z), so we need to choose one or the other. We couldof course just say that we are doing so arbitrarily, though this would lead to a sign ambiguity

in later definitions. For definiteness, we can note that the simplex σ : ∆k → Rk obtained by

translating the standard simplex in Rk by a small amount (say by the vector (−ε, . . . ,−ε) for

small ε > 0) so that the origin is contained in the interior defines a relative cycle in (Rk,Rk\0)(since its boundary maps to Rk \0) and it is not too difficult to see that the relative homology

class [σ] of this cycle generates Hk(Rk,Rk \ 0;Z) (check that the boundary of σ generates

the reduced homology of Rk \ 0, and use the long exact sequence of the pair). Then let

ω ∈ Hk(Rk,Rk \ 0;Z) be the unique element which evaluates as 1 on [σ].Return now to our vector bundle π: E → M , which is arbitrary if R = Z/2Z but is assumed

oriented if R = Z. For any m ∈ M we may choose an open neighborhood U of m and a local

trivialization Φ : E|U → U ×Rk, say given by Φ(e) = (π(e),φ(e)) where φ : E|U → Rk. If R = Z

we assume this local trivialization to be compatible with the orientation that is assumed on E.

We thus have a (linear) homeomorphism of pairs φ im : (Em, E0m)→ (Rk,Rk \ 0).

Define

µm = (φ im)∗ω ∈ Hk(Em, E0

m; R).

10Perhaps a more obvious cohomology to consider would be H∗(E, 0E ; R), but since E deformation retracts to 0E by

the map (t, e) 7→ te, we have H∗(E, 0E ; R) = 0.


We claim that µm is independent of this choice of trivialization Φ. If R = Z/2Z this is obvious,

sinceµm may be alternatively characterized as the unique nonzero element of Hk(Em, E0m

;Z/2Z).

Suppose instead that R= Z. If Φ′ : e 7→ (π(e),φ′(e)) is another choice of local trivialization de-fined over an open set containing m, then since we are requiring the local trivializations to be

orientation-compatible there is g ∈ GL+(k;R) such that (φ′ im) (φ im)−1 = g, i.e., φ′ im =

g (φ im). But then since GL+(k;R) is path-connected, there is a path γ: [0, 1]→ GL+(k;R)such that γ(0) = g and γ(1) is the identity. Then (t, e) 7→ γ(t) (φ im) gives a homotopy of

maps of pairs (Em, E0m)→ (Rk,Rk \ 0). So indeed (φ im)

∗ω = (φ′ im)∗ω.

Thus to each point m in the base M of the vector bundle π: E → M we have assigned a

distinguished element µm ∈ Hk(Em, E0m

; R). (When R = Z this depends on the orientation of E:indeed it is not hard to check that reversing the orientation of E would reverse the sign of µm.)

Definition 4.1. If π: E → M is a rank-k vector bundle (assumed oriented if R = Z), a Thom

class for E with coefficients in R is a relative cohomology class τ ∈ Hk(E, E0; R) such that, forall m ∈ M ,

i∗mτ= µm

where im : Em→ E is the inclusion of the fiber over m.

Our next goal is to show that such classes exist and are unique in some generality. We startwith the following.

Lemma 4.2. Suppose M is a paracompact Hausdorff space which is contractible. If π: E → M

is any rank-k vector bundle then there exists a unique Thom class for E with coefficients in Z/2Z,and if additionally E is oriented then there exists a unique Thom class for E with coefficients in Z.

Proof. As noted in Remark 2.30, the homotopy invariance of pullbacks of vector bundles overparacompact Hausdorff spaces implies that E is trivial, and we may choose a trivialization

Ψ : E→ M ×Rk (so Ψ is a homeomorphism which is a linear isomorphism on each fiber).If E is oriented, I claim that we may choose the trivialization Ψ to be compatible with the

orientation (incidentally, the argument to follow does not depend on the topological hypotheses

on M , but only on the fact that E is trivial). Indeed, let Φα : E|Uα → Uα ×Rk be a collection

of orientation-compatible local trivializations (so the transition functions Φβ : Φ−1α are given

by (m, v) 7→ (m, gαβ(m)v) for continuous functions gαβ : Uα ∩ Uβ → GL+(k;R)). Given an

initial global trivialization Ψ for E which may not be orientation-compatible, For each α we

may define a function fα : Uα → GL(k;R) by the property that Ψ Φ−1α : Uα × R

k → Uα × Rk

is given by (m, v) 7→ (m, fα(m)v). The fact that det gαβ(m) > 0 implies that, if m ∈ Uα ∩ Uβ ,

then det fα(m) and det fβ (m) have the same sign. So there is a continuous function δ : M →

1,−1 which restricts to each Uα as m 7→det fα(m)

|det fα(m)|. Where e1, . . . , ek is the standard basis

for Rk, define A: M → GL(k;R) by A(m)e1 = δ(m)e1 and A(m)ei = ei for i ≥ 2. (So inparticular det A(m) = δ(m).) If the initial trivialization Ψ is given by e 7→ (π(e),ψ(e)), the new

trivialization Ψ ′ : E→ M ×Rk given by Ψ ′(e) = (π(e), A(π(e))ψ(e)) will be compatible with theorientation on E given by the Φα.

Having arranged Ψ to be compatible with the orientation (in the case that E is oriented), let

p2 : M ×Rk be the projection. Then for each m ∈ M , we have by definition

µm = (p2 Ψ im)∗ω = i∗

m(p2 Ψ)

∗ω.

But this precisely says that τ= (p2 Ψ)∗ω is a Thom class, proving existence (for either coeffi-

cient group Z/2Z or Z). As for uniqueness, just note that the fact that M is contractible implies

that p2 : (M ×Rk, M × (Rk \ 0))→ (Rk,Rk \ 0) is a homotopy equivalence of pairs, so since


Ψ : (E, E0)→ (M ×Rk, M × (Rk \ 0)) is a homeomorphism, p2 Ψ is a homotopy equivalence

of pairs. Meanwhile for each m ∈ M , p2 Ψ im : (Em, E0m) → (Rk,Rk \ 0) is also a homo-

topy equivalence of pairs (indeed a homeomorphism), so homotopy-inverting p2 Ψ shows that

im : (Em, E0m) → (E, E0) is a homotopy equivalence of pairs. So for any m and any coefficient

ring R there can only be one class τ ∈ Hk(E, E0; R) which restricts to (Em, E0m) as µm, which is

(more than) enough to prove uniqueness of the Thom class.

Our plan now is to make use of the Mayer–Vietoris sequence to obtain Thom classes for vectorbundles over a fairly general class of spaces. In fact this can be done in rather more generalitythan we will work in (see [MS, Chapter 10]), though doing so requires more machinery.

Definition 4.3. A good cover Uαα∈J of a topological space X is an open cover with the propertythat for each finite subset α1, . . . ,αm ⊂ J the intersection Uα1

∩ · · · ∩ Uαmis either empty or

contractible.

Theorem 4.4. Let M be a Hausdorff space in which every open subset is paracompact, and assume

that M has a finite good cover. If π: E→ M is any rank-k vector bundle then there exists a unique

Thom class for E with coefficients in Z/2Z, and if additionally E is oriented then there exists a

unique Thom class for E with coefficients in Z. Moreover we have

H j(E, E0; R) = 0

for all coefficient rings R and all j < k.

Remark 4.5. It is a fact that any smooth manifold has a good cover (see [BT, Theorem 5.1]), asdoes any simplicial complex (this is easier—just construct the open sets in the cover from thestars of the various vertices). If the manifold or simplicial complex is compact then it of coursefollows that it has a finite good cover (so this applies to Grk(R

N ), though not to the infiniteGrassmannian Grk—once we know more about the topology of these spaces we will howeverbe able to work around this problem). Manifolds and CW complexes (and hence in particularsimplicial complexes) also have the property that all of their open subsets are paracompact(this property is called “hereditary paracompactness” in the literature). Of course in the case ofmanifolds this just follows from the fact that manifolds are paracompact (see e.g. [Lee, Chapter2]) since open subsets of manifolds are manifolds; for CW complexes see [FP, Section 1.3].

Proof of Theorem 4.4. The proof is by induction on the (minimal) number r of open sets requiredto form a good cover of the space M . For the case that M admits a good cover by just oneopen set, of course that open set must be M itself and so M is contractible. So the statement

about the Thom class has been proven as Lemma 4.2. As for the statement about H j(E, E0; R)

for j < k, let Ψ : E → M × Rk be a trivialization of E, which exists since M is contractible,

paracompact, and Hausdorff. Then where p2 : M ×Rk → Rk is the projection, as noted in the

proof of Lemma 4.2 π2 Ψ : (E, E0) → (Rk,Rk \ 0) is a homotopy equivalence of pairs. So

since H j(Rk,Rk \ 0;R) = 0 for j < k and for all coefficient rings R by Example 3.21, the same

holds for H j(E, E0;R).So we may assume inductively that the theorem holds for vector bundles over all spaces that

admit open covers by at most a open sets, where a ≥ 1. Now suppose that we have a good opencover

M = U1 ∪ · · · ∪ Ua ∪ V

by a + 1 open sets, and let U = U1 ∪ · · · ∪ Ua. Now V is contractible, so the theorem holds forbundles over V . Since U = U1 ∪ · · · ∪ Ua is a good open cover of U by a open sets, the theoremholds for bundles over U . Also, we have

U ∩ V = (U1 ∩ V )∪ · · · ∪ (Ua ∩ V ),


which is a good open cover of U∩V by a open sets, so the theorem holds for bundles over U∩V .

We apply the Mayer–Vietoris sequence for relative cohomology to the pair (E, E0), writingE = (E|U)∪ (E|V ) and noting that (E|U) ∩ (E|V ) = E|U∩V . For j < k and any coefficient ring R,part of this sequence reads

H j−1(E|U∩V , E0|U∩V ;R) // H j(E, E0; R) // H j(E|U , E0|U ; R)⊕ H j(E|V , E0|V ; R)

and by induction the outer two terms are zero, so H j(E, E0; R) = 0. This proves that the secondstatement of the inductive hypothesis holds for M ; it remains to prove the statement about theThom class.

So we specialize to coefficients R= Z/2Z or R= Z, and if R= Zwe assume that E is oriented.Another part of the Mayer–Vietoris sequence reads(13)

Hk−1(E|U∩V , E0|U∩V ;R)δ // Hk(E, E0; R)

f // Hk(E|U , E0|U ; R)⊕ Hk(E|V , E0|V ; R)g // Hk(E|U∩V , E0|U∩V ;R)

By induction we have unique Thom classes τU ,τV ,τU∩V for E|U , E|V , and E|U∩V , respectively.Recall from Exercise 3.24 that f (x) = (x |U , x |V ) and g(y, z) = y |U∩V−z|U∩V , where the meaningof our restriction notation should be obvious (really we should write x |E|U , y |E|U∩V

, etc.).

If m ∈ U ∩ V , then since τU is a Thom class for E|U we have

i∗m(τU |U∩V ) = i∗

mτU = µm

and likewise since τV is a Thom class for E|V we have i∗m(τV |U∩V ) = µm. This means that both

τU |U∩V and τV |U∩V are Thom classes for E|U∩V . But by induction there is only one Thom classfor E|U∩V , so we must have τU |U∩V = τV |U∩V , i.e. g(τU ,τV ) = 0. So by the exactness of (13)

there is τ ∈ Hk(E, E0;R) such that τ|U = τU and τ|V = τV . If m ∈ M , then either m ∈ U inwhich case i∗

mτ = i∗

m(τ|U) = µm, or else m ∈ V in which case i∗

mτ = i∗

m(τ|V ) = µm. This proves

that τ is a Thom class for E. For uniqueness, if σ were another Thom class for E then by thesame argument as before σ|U would be a Thom class for E|U and σ|V would be a Thom class forE|V . But by induction the Thom classes of E|U and E|V are unique, so σ|U = τ|U and σ|V = τ|V ,

i.e. f (σ) = f (τ). Now again by induction, Hk−1(E|U∩V , E0|U∩V ; R) = 0, so the exactness of (13)shows that f is injective and so σ = τ. This proves that the Thom class is unique, completingthe induction and hence the proof.

We can now define our first example of a characteristic class. Recall the zero-section s0 : M →E; we can view this as map of pairs (M ,∅)→ (E, E0) so that there is an induced map s∗

0: H∗(E, E0; R)→

H∗(M ;R).

Definition 4.6. Given a vector bundle π: E → M with Thom class τ (with coefficients in R =

Z/2Z, or in R= Z if E is oriented), the Euler class of E is the cohomology class

e(E) = s∗0τ ∈ Hk(M ; R).

Remark 4.7. It is sometimes useful to note that the map of pairs s0 : (M ,∅) → (E, E0) factors

as (M ,∅)→ (E,∅)→ (E, E0), where the first map is a homotopy equivalence and the second

map is the inclusion. So we have induced maps Hk(E, E0; R) → Hk(E; R) → Hk(M ; R), where

the first map is the map that appears in the cohomology long exact sequence of the pair (E, E0)

and the second map is an isomorphism; the Euler class is the image of the Thom class under thiscomposition. In particular the Euler class is zero if and only if the Thom class is in the kernel of

the map Hk(E, E0;R)→ Hk(E;R) that appears in the long exact sequence of the pair.

Remark 4.8. In principle one could imagine a bundle having more than one Thom class, inwhich case the Euler class as defined above would depend on the Thom class. Since Theorem


4.4 includes a uniqueness statement for the Thom class, this issue does not arise for bundlesover suitably nice spaces (including Grassmannians of k-planes in RN ). We will shortly see howto canonically remove the possible ambiguity at least for finite-type bundles over paracompactHausdorff spaces by using classifying maps. (In fact this isn’t really an issue, since for vec-tor bundles over paracompact Hausdorff spaces there always exists a unique Thom class withappropriate coefficients, though we won’t show this in full generality.)

Example 4.9. For the trivial bundle Et r iv = M × Rk over a space M , a Thom class is given by

τ = p∗2ω ∈ Hk(M ×Rk, M × (Rk \ 0); R) where p2 is the projection. Now p2 s0 : M → Rk is

the constant map (to ~0), so (p2 s0)∗ acts trivially on H j for j > 0. Thus

e(Et r iv) = s∗p∗2ω = (p2 s0)

∗ω = 0.

Thus, contrapositively, if a vector bundle π: E→ M has nonzero Euler class then E must benontrivial. Actually more is true.

Definition 4.10. A section of a vector bundle π: E → M is a continuous map s : M → E suchthat π s = 1M .

In other words a section s assigns to each m ∈ M an element s(m) of the vector space Em, ina continuous fashion. We will say that a section s is nonvanishing if, for every m ∈ M , s(m) is anonzero element of Em.

Proposition 4.11. Let π: E→ M be any rank-k vector bundle with a nonvanishing section. Then

if E has a Thom class with coefficients in R, the Euler class e(E) ∈ Hk(M ; R) is zero.

Proof. If s : M → E is a nonvanishing section then the map S : [0, 1] × M → E defined by

S(t, m) = ts(m) defines a homotopy from the zero-section s0 to s. Hence the maps s∗0: Hk(E, E0; R)→

Hk(M ;R) and s∗ : Hk(E, E0;R)→ Hk(M ; R) are equal, and so where τ is a Thom class for E wehave e(E) = s∗τ.

The assumption that s is nonvanishing amounts to the statement that the image of s is in fact

contained in E0, so (via the composition of maps of pairs (M ,∅)→ (E0, E0)→ (E, E0)) the map

s∗ : Hk(E, E0;R)→ Hk(M ;R) splits as a composition

Hk(E, E0;R) // Hk(E0, E0; R) // Hk(M ; R)

But the group Hk(E0, E0;R) is obviously zero (use the long exact sequence of the pair (E0, E0),

or just note that Hk(E0, E0;R) is the cohomology of the zero cochain complex). So the map

s∗ : Hk(E, E0;R)→ Hk(M ;R) must be zero, and in particular e(E) = s∗τ= 0.

Now a trivial bundle M × Rk obviously has nonvanishing sections, namely m 7→ (m, ~v) for

any nonzero vector ~v ∈ Rk. But the converse is not true: starting with any rank-k vector bundleE → M , as we will discuss in detail later one can form the “direct sum” E ⊕R with the trivialrank-one bundle over M , and one can obtain nonvanishing sections of the rank-k + 1 bundleE ⊕R→ M by mapping into the second summand. It will follow from something that we willdo later that if E has nonvanishing Euler class over Z/2Z then E⊕R is nontrivial (because it hasa nontrivial Stiefel–Whitney class; this statement is false for Euler classes over Z, as can be seenfor instance when E is the tangent bundle to an even-dimensional sphere). For vector bundlesV over paracompact Hausdorff spaces, one can use a Euclidean structure on V to show that V

has a nonvanishing section if and only if V is isomorphic to a direct sum E⊕R (take E equal tothe orthogonal complement of the nonvanishing section).

Here are some examples where the Euler class is nontrivial:


Example 4.12. Let π: E → S1 be the Möbius bundle. As usual we identify S1 with R/Z, andE = [0, 1]×R/(1, t)∼ (0,−t). Now the bundle E is not orientable, so we only have Thom andEuler classes over Z/2Z, not over Z; we will now find them.

As with any vector bundle, as a topological space E is homotopy equivalent to the base of

the bundle, which in this case is S1. Thus Hk(E;Z/2Z) is isomorphic to Z/2Z for k = 0, 1

and is trivial otherwise. Meanwhile, the complement E0 of the zero section is also homotopy

equivalent to S1: the map

x 7→

§[2x , 1] 0≤ x ≤ 1/2[2x − 1,−1] 1/2≤ x ≤ 1

embeds S1 into E0; if we denote by A the image of this embedding then the map [s, t] 7→ [s, t

|t|]

gives a deformation retract of E0 to A. (You can think of A as the boundary of a copy of theclosed Möbius strip inside E.)

In particular E0, like E, is path-connected, so the restriction map H0(E;Z/2Z)→ H0(E0;Z/2Z)

is an isomorphism. Consequently the map H1(E, E0;Z/2Z) → H1(E;Z/2Z) is injective. Infact it’s an isomorphism—to see this, note that the Thom class τE of E (which exists by The-

orem 4.4) is a nontrivial element of H1(E, E0;Z/2Z) (it restricts nontrivially to the fibers), so

H1(E, E0;Z/2Z) has at least two elements, while H1(E;Z/2Z) has exactly two elements, so aninjective homomorphism from the former to the latter has to be an isomorphism. So we infer

that H1(E, E0;Z/2Z) is isomorphic to Z/2Z, with τE the unique nonzero element.

Thus the canonical map H1(E, E0;Z/2Z)→ H1(E;Z/2Z) (which is the map induced by the

inclusion (E,∅) ,→ (E, E0)) sends the Thom class τE to the generator of H1(E;Z/2Z). The Eulerclass is the image of τE under the composition

H1(E, E0;Z/2Z)→ H1(E;Z/2Z)→ H1(S1;Z/2Z)

where the second map is induced by the inclusion s0 of the zero section and is an isomorphismsince E deformation retracts to its zero section. This proves that

For the Möbius bundle π: E→ S1, e(E) is the generator of H1(S1;Z/2Z).

Example 4.13. The simplest nontrivial example of an oriented bundle is probably the tautological

complex line bundle E = γ1(C2) over CP1 = Gr1(C2).11 Thus

E = (V, ~v) ∈ Gr1(C2)×C2|~v ∈ V

and E0 ⊂ E consists of those (V, ~v) where ~v 6= 0. Now a one-dimensional complex vectorspace is uniquely determined by a single nonzero element in it, in view of which the projection

p2 : E0→ C2 \ ~0 is a homeomorphism.

As usual, E is homotopy equivalent to the base CP1 of the bundle, which of course is home-

omorphic to S2. Meanwhile, E0 is, by the previous paragraph, homotopy equivalent to S3. In

particular H1(E0;Z) = H2(E0;Z) = 0, so part of the cohomology long exact sequence of thepair reads

0 // H2(E, E0;Z) // H2(E;Z) // 0

11Incidentally, any complex vector bundle is naturally oriented bundle: in terms of structure groups the point here

is that GL(n;C) ≤ GL+(2n;R), i.e. that any real (2n)× (2n) matrix that commutes with the matrix J0 that represents

multiplication by i has positive determinant. More geometrically, given a finite-dimensional complex vector space V ,

there is a natural orientation induced on the underlying real vector space of V : if e1, . . . , en is a basis for V over C, use

the orientation on V with respect to which e1, . . . , en, ie1, . . . , ien is a positive basis. This can be seen to be independent

of the choice of complex basis by virtue of the fact that the set of complex bases for V is connected (optional exercise:

prove this, perhaps by first using Gram-Schmidt to reduce to the case of orthonormal bases and then using induction

on the dimension).


As in the previous example, whatever the Thom class τE ∈ H2(E, E0;Z) is, it must be nontrivialsince it restricts nontrivially to fibers, so by the above exact sequence τE maps to a nontrivial

element under H2(E, E0;Z) // H2(E;Z) .

Hence by Remark 4.7 the Euler class e(E) is a nontrivial element of H2(CP1;Z). In fact,

with a little more care we can see that e(E) generates H2(CP1;Z). Namely, since H2(E;Z) ∼=H2(CP1;Z) ∼= Z and since we have seen that the canonical map H2(E, E0;Z)→ H2(E;Z) is an

isomorphism, it follows that H2(E, E0;Z) is isomorphic to Z. Now the Thom class τE is charac-

terized by the fact that, for any m ∈ CP1, τE is mapped to the generator µm of H2(Em, E0m

;Z)

under restriction. Since H2(E, E0;Z) ∼= Z, τE is n times a generator of H2(E, E0;Z) for some

n ≥ 1; but then this other generator would need to map under restriction to 1

nµm, which is

only possible of n = 1 since we are working over Z. So τE generates H2(E, E;Z), and so since

e(E) = s∗0τE and we have seen that s∗

0: H2(E, E0;Z)→ H2(CP1;Z) is an isomorphism this proves

that e(E) is a generator for H2(CP1;Z).

Exercise 4.14. Viewing S2 as a subset of R3 in the usual way, the tangent bundle TS2 is given asa topological space as

TS2 = (~v, ~x) ∈ S2 ×R3|~v · ~x = 0

where · is the standard dot product, with the bundle12 projection π: TS2 → S2 given by the

projection (~v, ~x) 7→ ~v. The standard orientation on TS2 is given by saying that a basis ~x , ~y for

T~vS2 = π−1(~v) is a positive basis if and only if ~v, ~x , ~y is a positive basis for R3.

(a) Prove that the complement of the zero section (TS2)0 is homotopy equivalent to (~v, ~x ∈S2 × S2|~v · ~x = 0, which in turn is homeomorphic to the group SO(3) of orthogonal 3 × 3matrices with determinant one.

(b) It is a fact that SO(3) is homeomorphic to RP3. Given this fact, prove that the Euler class

with coefficients in Z of the oriented vector bundle TS2 is nonzero, while the Euler class withcoefficients in Z/2Z is zero.

Exercise 4.15. If φ : R→ S is any homomorphism of A-modules and if C• is any chain complexof A-modules, there is an induced map of cochain complexes φ∗ : C•

R→ C•

S(where as before

C•R= HomA(C•, R)) defined by φ∗α= φ α.(a) Prove that φ∗ is a cochain map (i.e. it commutes appropriately with the coboundary

operators δ on the cochain complexes).

(b) Letπ: E→ M be an oriented vector bundle withZ-coefficient Thom classτZ ∈ Hk(E, E0;Z).Where in the notation of part (a) we set A = R = Z, S = Z/2Z, φ : Z → Z/2Z is the quo-

tient, and C• is the relative singular chain complex S∗(E, E0), by part (a) φ∗ induces a map

φ∗: Hk(E, E0;Z)→ Hk(E, E0;Z/2Z). Prove that φ

∗τZ is a Thom class of E with coefficients in

Z/2Z.

Remark 4.16. Exercise 4.15 suggests a definition for the Thom class of an oriented bundle E→M with coefficients in an arbitrary ring with unity R: let φ : Z→ R be the additive group homo-

morphism which maps 1 to the multiplicative identity in R, and put τR = φ∗τZ ∈ Hk(E, E0; R).

The resulting Euler class eR = s∗0τR with coefficients in R likewise has eR = φ

∗eZ ∈ Hk(M ; R).

As should be expected of a characteristic class, the Euler class is functorial with respect topullbacks of bundles; indeed we will see that this functoriality is inherited from a correspondingproperty for the Thom class. In general, as I’ll leave to you to check, if f : X → Y is a continuous

12It’s not too hard to prove that this is an oriented vector bundle, but I’m not asking you to do so.


map and πW : W → Y is a vector bundle, then a vector bundle πV : V → X is isomorphic to thepullback f ∗W if and only if there is a commutative diagram

Vf //

πV

W

πW

X

f // Y

such that for each x ∈ X , f maps Vx by a linear isomorphism to Wf (x). (If V and W are

oriented, then for an oriented-bundle isomorphism V ∼= f ∗W one should additionally ask for

the restriction of f to each fiber to respect orientation.) In particular in this case we have a map

of pairs f : (V, V 0)→ (W, W 0).

Proposition 4.17. In the above situation, if τW ∈ Hk(W, W 0; R) is a Thom class for W with Euler

class e(W ) ∈ Hk(Y ;R), then f ∗τW ∈ Hk(V, V 0; R) is a Thom class with Euler class f ∗e(W ) ∈Hk(X ;R).

Proof. If m ∈ X , let U ⊂ Y be an open neighborhood of f (m) which admits a local (oriented, if

applicable) trivialization Φ : W |U → U×Rk, given in coordinates by w 7→ (πW (w),φ(w)). Then

the map Ψ : V | f −1(U)→ f −1(U)×Rk defined by Ψ(v) = (πV (V ),φ( f (v))) is a local trivialization

for U around m, and so the distinguished generator µm ∈ Hk(Vm, V 0m

; R) is given by

µm = i∗m(φ f )∗ω = ( f im)

∗(φ∗ω)

where im : Vm → V is the inclusion Meanwhile where i f (m) : Wf (m) → W is the inclusion the

distinguished generator µ f (m) ∈ Hk(Wf (m), W 0f (m)

; R) is given by µ f (m) = i∗f (m)(φ∗ω).

Let us write fm : Vm→Wf (m) for the restriction of f , so that we obviously have

i f (m) fm = f im.

Thus

µm = (i f (m) fm)∗(φ∗ω) = f ∗

mi∗f (m)φ∗ω = f ∗

mµ f (m).

Now the Thom class τW has i∗f (m)τW = µ f (m) for all m ∈ X , so we see that

i∗m( f ∗τW ) = ( f im)

∗τW = (i f (m) fm)∗τW = f ∗

mi∗f (m)τW

= f ∗mµ f (m) = µm,

confirming that f ∗τW is a Thom class for V .Where sV

0and sW

0are the zero-sections of V and W respectively we evidently have sW

0 f =

f sV0

. Hence the Euler classes are related by

e(V ) = (sV0)∗( f ∗τW ) = f ∗(sW

0)∗τW = f ∗e(W ).

Let us return to the subject of (pre-)classifying maps from Section 2. If π: E→ M is a finite-type, rank-k vector bundle over a paracompact Hausdorff space, Proposition 2.5 showed thatthere is a pre-classifying map F : E→ RN . This map induces a classifying map f : M → Grk(R

N )


sending m to F(Em); to rephrase this into the language of Proposition 4.17, we may define a

map f : E→ γk(RN ) by f (e) = ( f (π(e)), F(e)) and then we have a commutative diagram

Ef //

π

γk(RN )

M

f // Grk(RN )

with f mapping fibers isomorphically to fibers. Now since Grk(RN ) is paracompact and Haus-

dorff and admits a finite good cover, Theorem 4.4 shows that there is a unique Thom class τk,N ∈

Hk(γk(RN ),γk(RN )0;Z/2Z), with Euler class ek,N ∈ Hk(Grk(RN );Z/2Z). Hence by Proposition

4.17, f ∗τk,N ∈ Hk(E, E0;Z/2Z) is a Thom class for E, with Euler class f ∗ek,N ∈ Hk(E, E0;RN ).

Example 4.18. For N ≤ N ′ we have an inclusion map iN ,N ′ : Grk(RN ) → Grk(R

N ′) defined by

V 7→ V ×~0, and we observed in Proposition 2.19 that this is a classifying map for γk(RN ); the

lift iN ,N ′ : γk(RN ) → γk(RN ′) is given by (V, ~v) 7→

V × ~0, (~v, ~0)

. Hence i∗

N ,N ′τk,N ′ is a Thom

class with coefficients in Z/2Z for γk(RN ), with Euler class i∗N ,N ′

ek,N ′ . Now by Theorem 4.4 there

is a unique Thom class for γk(RN ), which we have denoted τk,N . So it must hold that

(14) i∗N ,N ′τk,N ′ = τk,N i∗

N ,N ′ek,N ′ = ek,N

Now suppose that, where again π: E→ M is a finite-type bundle over a paracompact Haus-dorff space, we have two pre-classifying maps F0 : E → RN0 and F1 : E → RN1 , inducing clas-

sifying maps fi : M → Grk(RNi ) and lifts fi : E → γk(RNi ). Then by Theorem 2.8 there is N ≥

maxN0, N1 such that IN0,N F and IN1,N F are homotopic through pre-classifying maps; hence

the maps iN0,N f0 and iN1,N f will be homotopic as maps of pairs (E, E0)→ (γk(RN ),γk(RN )0),

and iN0,N f0 and iN1,N f1 are homotopic as maps M → Grk(RN ). Hence we have, using (14),

f ∗0τk,N0

= f ∗0

i∗N0,Nτk,N = (iN0,N f0)

∗τk,N = (iN1,N f1)∗τk,N = f ∗

1τk,N1

and similarly

f ∗0

ek,N0= f ∗

1ek,N1

.

If Theorem 4.4 applies to M , then by Proposition 4.17 we could already have said that f ∗0τk,N0

and f ∗1τk,N1

are equal to the unique Thom class for E provided by that Theorem. However, when

Theorem 4.4 does not apply to M , we can now define the Thom class τE of E to be f ∗τk,N for any

map f : E → γk(RN ) obtained as above from a pre-classifying map F : E → RN as above, sincewhat we have just done shows that this class τE is independent of the choice of pre-classifyingmap. Likewise, we can define the Euler class of E to be e(E) = f ∗ek,N for any classifying map

f : M → Grk(RN ). Since if g : X → M is a continuous map and f : M → Grk(R

N ) is a classifyingmap for E then f g is a classifying map for g∗E, it follows that the Euler class so defined satisfiesthe important naturality property

e(g∗E) = g∗e(E).

5. THE CUP PRODUCT AND THE LERAY–HIRSCH THEOREM

While the universal coefficient theorem shows that, for purely algebraic reasons, the co-homology groups (with arbitrary coefficients) of a space X are explicitly determined by thehomology with integer coefficients, an important fact about cohomology is that, provided thatone works with coefficients in a ring, it also carries a ring structure, and this ring structure is


not determined by the homology. So cohomology is a stronger invariant than homology: thereare many examples of pairs of spaces with isomorphic homology groups but nonisomorphic

cohomology rings (S2 × S2 and the connected sum CP2#CP2 is one example of such a pair).To describe the multiplicative operation (called the cup product) for this ring structure, we

first introduce the maps between simplices upon which it is based. For notational simplicity,

we will regard all of the standard simplices ∆k as subsets of the same space R∞. We denoteby v0 the zero-vector in R∞ and v1, . . . , vk, . . . the standard basis vectors, so that the standardk-simplex is

∆k =

(k∑

i=0

t i vi

t i ≥ 0,

k∑

i=0

t1 = 1

)

As alluded to at the start of Section 3, we have face maps

φkj: ∆k→∆k+1 (0≤ j ≤ k+ 1)

k∑

i=0

t i vi 7→

j−1∑

i=0

t i vi +

k∑

i= j

t i vi+1

(so φkj

embeds ∆k as that boundary face of ∆k+1 which does not contain the jth vertex of

∆k+1). Of course, the boundary operator is defined by setting ∂ σ =

∑(−1) jσ φ j for a map

σ : ∆k→ X .We now introduce new maps

fp,p+q : ∆p→∆p+q

p∑

i=0

t i vi 7→

p∑

i=0

t i vi

and

bq,p+q : ∆q→∆p+q

q∑

i=0

t i vi 7→

p+q∑

i=p

t i−pvi

Thus fp,p+q embeds ∆p into ∆p+q as the “front p-face,” while bq,p+q embeds ∆q into ∆p+q as the

“back q-face.” The images of these two maps intersect in a single point, namely vp (which is

both the last vertex in the front p-face and the first vertex in the back q-face).The following identities relating these maps is left as a recommended exercise to the reader

(none of them should be difficult):

Proposition 5.1. The face maps φkj, front maps fp,p+q, and back maps bq,p+q obey the following:

fp,p = 1∆p bq,q = 1∆q

fp,p+1 = φp

p+1 bq,q+1 = φq

0

fp+q,p+q+r fp,p+q = fp,p+q+r as maps ∆p→∆p+q+r

bq+r,p+q+r br,q+r = br,p+q+r as maps ∆r →∆p+q+r

fp+q,p+q+r bq,p+q = bq+r,p+q+r fq,q+r as maps ∆q→∆p+q+r

φp+q

j fp,p+q =

fp+1,p+q+1 φ

p

j0≤ j ≤ p+ 1

fp,p+q+1 p+ 1≤ j ≤ p+ q+ 1


φp+q

j bq,p+q =

bq,p+q+1 0≤ j ≤ p

bq+1,p+q+1 φq

j−pp ≤ j ≤ p+ q+ 1

We now define the cup product on the singular cochain complex S•(X ; R) = Hom(S•(X ); R)

where R is an arbitrary commutative ring with unity. This takes the form of a map

∪ : Sp(X ; R)× Sq(X ; R)→ Sp+q(X ; R)

(α,β) 7→ α∪ β

To define α∪β we have to give the value (in R) of (α∪β)(σ)whereσ : ∆p+q→ X is an arbitrarycontinuous map. This is given by the following formula:

(α∪ β)(σ) = α(σ fp,p+q)β(σ bq,p+q).

(This makes sense, since σ fp,p+q and σ bq,p+q are maps into X with domains ∆p and ∆q

respectively, so α(σ fp,p+q) and β(σ bq,p+q) are elements of R, which can then be multipliedtogether since we have assumed that R is a ring and not just an abelian group.)

It should be obvious that we have distributive laws α ∪ (β0 + β1) = α ∪ β0 + α ∪ β1 and(α0+α1)∪β = α0∪β +α1∪β , essentially inherited from the distributive laws in R. Moreover:

Proposition 5.2. ∪ is associative: for α ∈ Sp(X ; R), β ∈ Sq(X ; R), γ ∈ S r(X ; R) we have

(α∪ β)∪ γ= α∪ (β ∪ γ).

Proof. This follows fairly quickly from the third, fourth, and fifth lines of Proposition 5.1: forany map σ : ∆p+q+r → X we have

(α∪ β)∪ γ(σ) = (α∪ β)(σ fp+q,p+q+r)γ(σ br,p+q+r)

= α(σ fp+q,p+q+r fp,p+q)β(σ fp+q,p+q+r bq,p+q+r)γ(σ br,p+q+r)

= α(σ fp,p+q+r)β(σ bq+r,p+q+r fq,q+r)γ(σ bq+r,p+q+r br,q+r)

= α(σ fp,p+q+r) · (β ∪ γ)(σ bq+r,p+q+r)

=α∪ (β ∪ γ)

(σ)

The above shows that ∪ makes S∗(X ; R) = ⊕∞p=0

Sp(X ; R) into a graded ring (i.e. a ring A

which decomposes as a direct sum as A = ⊕p∈NAp where Ap · Aq ⊂ Ap+q). Moreover since we

assume that R has a multiplicative identity there is also a multiplicative identity in S∗(X ; R),

which we will denote by 1: namely the element 1 ∈ S0(X ; R) defined by the property that for

each 0-simplex σ : ∆0 → X (i.e., each point in X ) we have 1(σ) = 1 (so for a general 0-chain∑ni x i we have 1(

∑ni x i) =

∑ni). The fact that 1∪α = α∪ 1= α for any α ∈ Sp(X ; R) follows

directly from the first line of Proposition 5.1. Since 1 assigns the same value to every point (andin particular assigns the same value to any two points in the same path component), it followsas in Exercise 3.19 that

(15) δ1= 0.

So we have made S∗(X ;R) into a graded ring with unity. (It is not commutative in any sense,as you can easily check.) This is not what was promised earlier, of course—in order to make thecohomology into a graded ring using ∪ we need to show that ∪ interacts appropriately with the

coboundary operator δ : Sk(X ;R)→ Sk+1(X ; R). In fact we have the following “Leibniz rule”:

Proposition 5.3. For α ∈ Sp(X ;R) and β ∈ Sq(X ; R),

δ(α∪ β) = (δα)∪ β + (−1)pα∪ (δβ).


Proof. We need to show that both sides evaluate in the same way on an arbitrary continuous

map σ : ∆p+q+1→ X . We have:

δ(α∪ β)

(σ) = (α∪ β)(∂ σ) =

p+q+1∑

j=0

(−1) j(α∪ β)(σ φp+q

j)

=

p+q+1∑

j=0

(−1) jα(σ φp+q

j fp,p+q)β(σ φ

p+q

j bq,p+q)

=

p∑

j=0

(−1) jα(σ fp+1,p+q+1 φp

j)β(σ bq,p+q+1) +

p+q+1∑

j=p+1

(−1) jα(σ fp,p+q+1)β(σ bp+1,p+q+1 φq

j−p)

= α

p∑

j=0

(−1) j(σ fp+1,p+q+1) φp

j

!β(σ bq,p+q+1)

+ (−1)pα(σ fp,p+q+1)β

q+1∑

k=1

(−1)k(σ bq+1,p+q+1) φq

k

!

Now we have

∂ (σ fp+1,p+q+1) =

p+1∑

j=0

(−1) j(σ fp+1,p+q+1) φp

j

and

∂ (σ bq+1,p+q+1) =

q+1∑

k=0

(−1)k(σ bq+1,p+q+1) φq

k

(so these each differ from expressions above by one term). We thus obtain:δ(α∪ β)

(σ) = α(∂ (σ fp+1,p+q+1))β(σ bq,p+q+1) + (−1)pα(σ fp,p+q+1)β(∂ (σ bq+1,p+q+1))

− (−1)p+1α(σ fp+1,p+q+1 φp

p+1)β(σ bq,p+q+1)− (−1)pα(σ fp,p+q+1)β(σ bq+1,p+q+1 φq

0)

= (δα)(σ fp+1,p+q+1)β(σ bq,p+q+1) + (−1)pα(σ fp,p+q+1)(δβ)(σ bq+1,p+q+1)

− (−1)p+1α(σ fp,p+q+1)β(σ bq,p+q+1)− (−1)pα(σ fp,p+q+1)β(σ bq,p+q+1)

=(δα)∪ β + (−1)pα∪ (δβ)

(σ)

where in the second-to-last line we have used lines 2-4 of Proposition 5.1.

Corollary 5.4. The map ∪ : Sp(X ;R) × Sq(X ; R) → Sp+q(X ; R) gives rise to a well-defined map

∪ : H p(X ;R)× Hq(X ;R)→)H p+q(X ;R) by means of the formula [α]∪ [β] = [α∪ β] for cocycles

α and β . This operation makes H∗(X ;R) = ⊕∞k=0

Hk(X ; R) into a graded ring, with multiplicative

identity given by 1 := [1].

Proof. If δα = δβ = 0 then δ(α∪ β) = 0∪ β + (−1)pα∪ 0 = 0 by (5.2), so whenever α and βare cocycles α∪ β is also a cocycle.

To show that ∪ is well-defined as an operation on cohomology we must show that, for anytwo cocycles α ∈ Sp(X ;R) and β ∈ Sq(X ; R) the cohomology class [α∪β] of α∪β depends only

on the cohomology classes of α and β . Indeed by Proposition 5.2 we have, for η ∈ Sp−1(X ; R)

and ζ ∈ Sq−1(X ;R), δ(α∪ζ) = (−1)pα∪δζ (since δα= 0), and δ(η∪(β+δζ)) = (δη)∪(βδζ)(since δ(β +δζ) = 0). Hence

(α+δη)∪ (β +δζ) = α∪ β +δ(−1)pα∪ ζ+η∪ (β +δζ)


and so (α + δη) ∪ (β + δζ) represents the same cohomology classas α ∪ β . So indeed ∪ iswell-defined on cohomology.

Given that ∪ is well-defined, the ring axioms for H∗(X ; R) are straightforwardly inherited

from those of S∗(X ;R). In view of (15), we have a well-defined cohomology class 1 = [1],

which is a multiplicative identity in H∗(X ; R) since 1 is a multiplicative identity in S∗(X ; R).

Remark 5.5. One can show ([H, Theorem 3.14]) that H∗(X ; R) is a graded-commutative ring inthe sense that for x ∈ H p(X ;R) and y ∈ Hq(X ; R) one has x ∪ y = (−1)pq y ∪ x . This is perhapssomewhat surprising since there is not a corresponding identity on S∗(X ; R)—the most that one

can say is that if α and β are cocycles then α∪ β − (−1)pqβ ∪α is a coboundary.13

Remark 5.6. The same discussion as above goes through without change for the relative coho-mology H∗(X , A;R) where A ⊂ X , except that there is generally no unit when A 6= ∅. Indeedrecall that the relative singular cochain complex S∗(X , A : R) can be regarded as the subcom-plex consisting of those cochains on X which vanish on all simplices contained in A, and this

condition is clearly preserved by the cup product operation (but is not obeyed by 1 ∈ S∗(X ; R)).In fact a stronger statement is true: if one has a subsets A⊂ X and cochains α,β ∈ S∗(X ; R)

such that β vanishes on all simplices in A, then α ∪ β also vanishes on all simplices in A. As aresult we have a well-defined cup product operation

∪ : H∗(X ;R)× H∗(X , A; R)→ H∗(X , A; R).

This makes H∗(X , A;R) into a module over the ring H∗(X ; R).

As may be intuitively obvious, the ring structure on H∗(X ; R) behaves well with respect tothe homomorphisms induced by continuous maps.

Proposition 5.7. For any continuous map of pairs f : (X , A)→ (Y, B) the induced map f ∗ : H∗(Y, B; R)→H∗(X , A;R) is a ring homomorphism. If A= B = ∅ then f ∗(1) = 1.

Proof. For α ∈ Sp(Y ;R), β ∈ Sq(Y ;R), and σ : ∆p+q→ X we have

( f ∗(α∪ β))(σ) = (α∪ β)( f σ) = α(( f σ) fp,p+q)β(( f σ) bq,p+q)

= α( f (σ fp,p+q))β( f (σ bq,p+q)) = ( f∗α)(σ fp,p+q) · ( f

∗β)(σ bq,p+q) =( f ∗α)∪ ( f ∗β)

(σ).

Thus we have (even on cochain level) f ∗(α∪β) = ( f ∗α)∪( f ∗β), so the induced map f ∗ : H∗(Y, B; R)→H∗(X , A;R) is a ring homomorphism.

As for the units, it is obvious from the definitions that f ∗1 = 1, and so upon passing tocohomology we have f ∗1= 1.

Exercise 5.8. Identify the torus with T = R2/Z2, and define one-cycles γ1,γ2 : ∆1 → T byγ1(t) = [(t, 0)] and γ2(t) = [(0, t)]. As you learned in 8200, H2(T ;Z) ∼= Z, and the homologyclasses [γ1], [γ2] form a basis for H1(T,Z). So by the universal coefficient theorem there are

α,β ∈ H1(T ;Z) such that α([γ1]) = β([γ2]) = 1 and α([γ2]) = β([γ1]) = 0.(a) Prove directly from the definition of the cup product that α ∪ α = β ∪ β = 0 while

α ∪ β 6= 0. (Suggestion: Construct a singular two-cycle which generates H2(T ;Z) by dividingthe square into two triangles.)

(b) Prove that every continuous map f : S2→ T has degree zero (i.e. induces the zero mapon H2; hint: functoriality of the cup product).

13In fact, there are certain cohomology operations called Steenrod squares whose nontriviality reflects the impossi-

bility of creating a cochain-level definition of the cup product which is exactly graded-commutative. You may be familiar

with de Rham cohomology (which for a smooth manifold X is isomorphic to H∗(X ;R)) in which the multiplication is

graded commutative on cochain level—however nothing like this can be constructed using coefficients in Z or Z/2Z.


Exercise 5.9. Let Σ denote the compact surface of genus two. Prove directly from the definition

of the cup product that the bilinear pairing∪ : H1(Σ;Z)×H1(Σ;Z)→ H2(Σ;Z) is nondegenerate,

i.e. that for any nonzero α ∈ H1(Σ;Z) there is β ∈ H1(Σ;Z) such that α∪ β 6= 0. Deduce fromthis that, where T is the torus, every continuous map g : T → Σ has degree zero.

(More generally, a similar argument shows that if g < h then all maps from the genus-gsurface to the genus-h surface have degree zero.)

5.1. Fiber bundles and their cohomology. A rank-k vector bundle E→ M can be thought of

as a family of copies of Rk parametrized by M . There is a generalization of this where Rk isreplaced by an arbitrary topological space:

Definition 5.10. If M and F are topological spaces, a fiber bundle over M with fiber F consists ofa continuous map π: P → M where P is another topological space, such that each m ∈ M has a

neighborhood U and a “local trivialization” Φ : π−1(U)→ U × F where Φ is a homeomorphism

which map each “fiber” Px := π−1(x) to x × F .

Thus a rank-k vector bundle is a particular kind of fiber bundle with fiber Rk (for the general

fiber bundle overRk one would not require the local trivializations to respect a linear structure).We will find it useful to consider a relative version of the above definition:

Definition 5.11. If M , F, G are topological spaces with G ⊂ F , a fiber bundle pair over M with fiber(F, G) is a pair of topological spaces (P,Q) where Q ⊂ P together with a fiber bundle π: P → M

such that the local trivializations Φ : π−1(U)→ U × F for P map π−1(U)∩Q homeomorphicallyto U × G.

For example of π: E → M is a rank-k vector bundle and E0 is the complement of the zero-

section, then the local trivializations for E map π−1(U)∩E0 homeomorphically to U×(Rk\0).Thus (E, E0) is a fiber bundle pair with fiber (Rk,Rk \ 0).

Remark 5.12. If M is a contractible paracompact Hausdorff space, then every fiber bundle pair(P,Q) over M with fiber (F, G) is trivial, i.e. there is a homeomorphism Ψ : P → M × F whichmaps Q to M ×G and each fiber (Px ,Q x) homeomorphically to (x× F, x×G). The proof ofthis proceeds exactly as in Section 2.4: one can define the pullback of a fiber bundle in the sameway as the pullback of a vector bundle, and the proof of Theorem 2.24 extends to show that thepullbacks of a fiber bundle by two homotopic maps are isomorphic, from which the conclusionfollows just as in Proposition 2.30.

The Leray–Hirsch theorem, stated below, will determine the cohomology of fiber bundle pairsthat satisfy the following property:

Definition 5.13. Let R be a commutative ring with unity. We say that the fiber bundle pair(P,Q) over M satisfies the Leray–Hirsch property over R if there exist cohomology classes f j ∈Hn j (P,Q;R) (with j ranging over some index set J) such that, for each m ∈ M , where im : (Pm,Qm)→(P,Q) denotes the fiber over m,

i∗m

f j j∈J is a basis for the R-module H∗(Pm,Qm; R).

Example 5.14. Assuming that H∗(F, G;R) is a free R-module, say generated by x j ∈ Hn j (F, G; R)

( j ∈ J), the trivial bundle (M × F, M × G) over M satisfies the Leray–Hirsch property over R:where π2 : (M × F, M × G) → (F, G) is the projection we can just set f j = π

∗2x j . (Of course,

H∗(F, G;R) will automatically be a free R-module if R is a field.)

Example 5.15. If π: E→ M is a finite-type rank-k vector bundle over a paracompact Hausdorff

space and if either R= Z and E is oriented or else R= Z/2Z, then the fiber bundle pair (E, E0)


over M satisfies the Leray–Hirsch property over R. Indeed, in this case the bundle E has a Thom

class τ ∈ Hk(E, E0;R) by results of Section 4. By Example 3.21, H∗(Rk,Rk \ 0; R) is a freerank-one R-module generated by the element denoted µm in Section 4, and the definition of aThom class is precisely that i∗

mτ= µm for all m.

Example 5.16. Consider the quotient map π: Cn+1 \ ~0 → CPn where n≥ 1; this is easily seen

to be a fiber bundle with fiber C \ 0. (So we are letting P = Cn+1 \ ~0 and Q = ∅ in this

example.) If m ∈ M then H1(Pm;R)∼= R, while since Cn+1\~0 is homotopy equivalent to S2n+1,

H1(P;R) = ∅. So for all m the restriction i∗m

: H∗(P; R) → H∗(Pm; R) fails to be surjective, inview of which P does not satisfy the Leray–Hirsch property.

Example 5.17. Let π: E → S1 denote the Möbius bundle and consider the fiber bundle pair

(E, E0) but now with coefficients in Z. As in Example 4.12, both E and E0 are homotopy equiv-

alent to S1. It is not hard to check that the inclusion maps a generator for H1(E0;Z) to two

times a generator of H1(E;Z) (the former generator can be represented by the boundary ofthe closed Möbius strip and the latter generator can be represented by the zero section). Tak-

ing adjoints we see that, under appropriate identifications of H1(E;Z) and H1(E0;Z) with Z,

i∗ : H1(E;Z) → H1(E0;Z) is given by multiplication by 2. But then consideration of the long

exact sequence of the pair shows that H1(E, E0;Z) must be isomorphic to ker(i∗), and hence is

zero. So since the relative cohomology H1(R,R \ 0;Z) is nonzero it follows that (E, E0) mustnot satisfy the Leray–Hirsch property over Z (even though it does satisfy it over Z/2Z).

Suppose that the fiber bundle pair (P,Q) over M satisfies the Leray–Hirsch property with co-efficients in R, with classes f j ∈ Hn j (F, G; R) as in Definition 5.13. Now we have a ring structure

on H∗(M ;R); we can then form the graded (left) H∗(M ; R)-module H∗(M ; R)[x j] where we

set the grading of each generator x j equal to n j (i.e. to the grading of f j). Thus a general ele-

ment of H∗(M ;R)[x j] is given by a sum∑

j c j x j where c j ∈ H∗(M ; R) and only finitely many

c j are nonzero, and such an element has grading k provided that c j ∈ Hk−n j (M ; R).

We can moreover define a map

Φ : H∗(M ; R)[x j]→ H∗(P,Q; R)(16) ∑

j

c j x j 7→∑

j

(π∗c j)∪ f j

(this makes sense, since we have π∗c j ∈ H∗(P; R) and f j ∈ H∗(P,Q; R), so as explained in

the last section they have a well-defined cup product in H∗(P,Q; R), making H∗(P,Q; R) intoa H∗(P;R)-module). The ring homomorphism π∗ : H∗(M ; R) → H∗(P; R) induces a naturalH∗(M ;R)-module structure on H∗(P,Q; R), and it is easy to see that Φ is a H∗(M ; R)-modulehomomorphism. (It is not a ring homomorphism, since we have not even made its domain intoa ring.)

Theorem 5.18 (Leray–Hirsch). Let M be a Hausdorff space in which every open subset is para-

compact, and assume that M has a finite good cover. Let (P,Q) be a fiber bundle pair over M which

satisfies the Leray–Hirsch property over R. Then the map Φ : H∗(M ; R)[x j] → H∗(P,Q; R) of

(16) is an isomorphism of H∗(M ;R)-modules.

Remark 5.19. In fact, this result holds for fiber bundles over rather more general spaces B thanthose mentioned in the theorem; see for instance [H, Theorem 4D.8].

Remark 5.20. As a special case, in view of Example 5.14 we see that if M is as in Theorem 5.18and if F is a space whose cohomology is a free R-module then H∗(M× F ; R) is isomorphic to the

tensor product of H∗(M ;R) and H∗(F ;R): elements of H(M × F ; R) can be written uniquely as


finite sums∑π∗

1ci ∪π

∗2

fi where ci ∈ H∗(M ; R) and fi ∈ H∗(F ; R). This is (a special case of) theKünneth theorem for cohomology.

Proof of Theorem 5.18. Similarly to the proof of Theorem 4.4, we use induction on the numberof sets required to form a good cover of M .

If M has a good cover by just one open set, then of course this open set is all of M andso M is contractible. So the cohomology of M is isomorphic to that of a point, and so is a

rank-1 R-module generated by 1 ∈ H0(M ; R). So in this case the map Φ is just the R-modulehomomorphism Φ : R[x j] → H∗(P,Q; R) which sends x j to π∗1 ∪ f j = f j . Given that Φ has

this form, the Leray–Hirsch condition amounts to the statement that, for each m ∈ M , thecomposition i∗

mΦ : R[x j]→ H∗(Pm,Qm; R) is an R-module isomorphism.

Meanwhile by Remark 5.12, the bundle pair (P,Q) is trivial, so there is a homeomorphismΨ : (P,Q)→ (M × F, M × G) mapping (Pm,Qm) to (m × F, m × G). Since M is contractible,the map (M × F, M × G) → (m × F, m × G) which is the identity on the second factor is ahomotopy equivalence, in view of which the inclusion im : (Pm,Qm)→ (P,Q) is also a homotopyequivalence. So the fact that i∗

mΦ : H∗(M ; R)→ H∗(Pm,Qm; R) is an isomorphism implies that

Φ is an isomorphism. This proves the result in the case that M admits a good open cover by justone set.

Now assume the theorem to be proven whenever the base of the bundle admits a good opencover by at most a open sets, where a ≥ 1, and suppose that M admits a good open cover bya+ 1 sets—say this good open cover is

M = U1 ∪ · · · ∪ Ua ∪ V.

As in the proof of Theorem 4.4, write U = U1 ∪ · · · ∪ Ua, and where π: P → M is the bundle

projection for any W ⊂ M write P|W = π−1(W ) and Q|W = P|W ∩ Q. We observe that the

inductive hypothesis implies that the conclusion of the theorem holds for the fiber bundle pairsover U , (P|V ,Q|V ) over V , and (P|U∩V ,Q|U∩V ) over U ∩ V (in the last case this follows from theobservation that U ∩ V = (U1 ∩ V )∪ · · · ∪ (Ua ∩ V ) gives a good cover of U ∩ V by a sets).

The Leray–Hirsch property gives a set of classes f j ∈ H∗(P,Q; R) which restrict to a basis

for the cohomology of the fibers; these classes restrict to sets of classes f Uj∈ H∗(P|U ,Q|U ; R),

f Vj∈ H∗(P|V ,Q|V ;R), and f U∩V

j∈ H∗(P|U∩V ,Q|U∩V ; R) as in the Leray–Hirsch property. So

by the inductive hypothesis the corresponding maps ΦU : H∗(U; R)[x j] → H∗(P|U ,Q|U ; R),

ΦV : H∗(V ;R)[x j] → H∗(P|V ,Q|V ;R), and ΦU∩V : H∗(U ∩ V ; R)[x j] → H∗(P|U∩V ,Q|U∩V ; R)

(given for example by ΦU(∑

c j x j) =∑(π|P|U )

∗c j ∪ f Uj

) are all isomorphisms.

From the naturality of the construction of the Mayer–Vietoris sequence, for each k the fol-lowing diagram of Mayer–Vietoris sequences commutes:(17)

Hk−1(P|U ;R)⊕ Hk−1(P|V ;R) // Hk−1(P|U∩V ; R) // Hk(P; R) // Hk(P|U ; R)⊕ Hk(P|V ; R) // · · ·

Hk−1(U;R)⊕ Hk−1(V ;R)

π∗⊕π∗

OO

// Hk−1(U ∩ V ; R)

π∗

OO

// Hk(M ; R) //

π∗

OO

Hk(U; R)⊕ Hk(V ; R)

π∗⊕π∗

OO

// · · ·

For any subset W introduce the notation ÒH∗(W ) for the graded H∗(M ; R)-module ÒH∗(W ) =H∗(W ;R)[x j]. Note that this module is isomorphic to a direct sum of #x j-many copies of

H∗(W ;R), each shifted in degree by the degree of the corresponding x j . We can trivially extendthe Mayer–Vietoris sequence associated to the decomposition M = U ∩ V to a sequence

ÒHk−1(U ∩ V ) // ÒHk(M) // ÒHk(U)⊕ ÒHk(V ) // ÒHk(U ∩ V )


(with each map sending x j to x j and otherwise acting as before); this is a direct sum of exact

sequences and therefore is exact. We then have a diagram(18)

Hk−1(P|U ,Q|U ; R)⊕ Hk−1(P|V ,Q|V ; R)//Hk−1(P|U∩V ,Q|U∩V ; R)

//Hk(P,Q; R)

//Hk(P|U ,Q|U ; R)⊕ Hk(P|V ,Q|V ; R)

//Hk(P|U∩V ,Q|U∩V ; R)

ÒHk−1(U;R)⊕ ÒHk−1(V ;R)

ΦU⊕ΦV

OO

// ÒHk−1(U ∩ V ; R)

ΦU∩V

OO

// ÒHk(M ; R) //

Φ

OO

ÒHk(U; R)⊕ Hk(V ; R)

ΦU⊕ΦV

OO

// ÒHk(U ∩ V )

ΦU∩V

OO

Using the facts that the classes f j , f Uj

, f Vj

, and f U∩Vj

are intertwined by the inclusion maps

that appear in the top line above, together with the fact that (17) commutes, one can see from alittle diagram chasing (left as a recommended exercise to the reader) that (18) also commutes.By the inductive hypothesis, the outer four vertical maps in (18) are all isomorphisms, and sothe five lemma shows that Φ is an isomorphism. This completes the induction and hence theproof.

Specializing to the case of Example 5.15, we have:

Corollary 5.21 (Thom Isomorphism Theorem). Let M be a Hausdorff space in which every open

subset is paracompact and which admits a finite good cover. Let π: E→ M be a vector bundle with

Thom class τE ∈ Hk(E, E0;R) where either R = Z and E is oriented or R = Z/2Z. Then for all

integers j the map

Φ : H j(M ; R)→ H j+k(E, E0; R)

c 7→ (π∗c)∪τE

is an isomorphism.

Of course this extends Theorem 4.4, which showed that H j+k(E, E0; R) = 0 for j < 0, and

produced a nonzero element (namely the Thom class) in Hk(E, E0; R). Note that the Thom classτE is the image of the unit 1 under the Thom isomorphism Φ.

Recall that the Euler class of E is defined by e(E) = ( j s0)∗τE where s0 : M → E is the

zero section and j : (E,∅) → (E, E0) (in the past we have just written s0 instead of j s0, but

for clarity here it is better to record the two steps of the inclusion of M into (E, E0) separately).Now we have isomorphisms s∗

0: H∗(E;R)→ H∗(M ; R) (since s0 is a homotopy equivalence), and

Φ : H∗−k(M ;R)→ H∗(E, E0;R) (the Thom isomorphism). At the same time, where i : E0→ E isthe inclusion, we have the long exact sequence of the pair

· · · // H j−1(E0;R) // H j(E, E0; R)j∗ // H j(E; R)

i∗ // H j(E, E0; R) // · · ·

By means of the isomorphisms s∗0

and Φ, this yields an sequence

(19) · · · // H j−1(E0;R) // H j−k(M ; R)s∗0 j

∗Φ// H j(M ; R)

i∗(s∗0)−1

// H j(E0; R) // · · ·

Now s0 : M → E has homotopy inverse given by the bundle projection π: E → M , so (s∗0)−1 =

π∗, and i∗ (s∗0)−1 = (π i)∗. So the map H j(M ; R)→ H j(E0; R) is just the map induced by the

restriction of the bundle projection π to E0.

Meanwhile for c ∈ H j−k(M ;R) we have

s∗0 j∗ Φ(c) = ( j s0)

∗(π∗c)∪τE

= (π j s0)

∗c ∪ ( j s0)∗τE .

But π js0 is the identity on M , and ( js0)∗τE = e(E) by definition. So the second map in (19)

is simply the map · ∪ e(E): H j−k(M ;R)→ H j(M ; R) given by c 7→ c ∪ e(E), i.e. by cup product

with the Euler class.


Summing up, we have proven:

Corollary 5.22. Let M be a Hausdorff space in which every open subset is paracompact and which

admits a finite good cover. Let π: E→ M be a vector bundle with Euler class e ∈ Hk(M ; R) where

either R= Z and E is oriented or R = Z/2Z. Then there is an exact sequence (the Gysin sequence)

· · · // H j−1(E0;R) // H j−k(M ;R)·∪e // H j(M ; R)

π∗ // H j(E0; R) // H j−k+1(M ; R)·∪e // · · ·

where E0 is the complement of the zero-section in E.

Using the Gysin sequence, we can determine the cohomology rings of RPn (over Z/2Z) andof CPn (over Z):

Example 5.23. Let π: E→ RPn be the tautological rank-one line bundle over RPn = Gr1(Rn+1).

We then have

E0 = (V, ~v) ∈ RPn × (Rn+1 \ ~0)|~v ∈ V ∼= Rn+1 \ ~0

(the projection onto the second factor is a homeomorphism, since a one-dimensional subspace

of Rn+1 is uniquely determined by a single nonzero vector in it). Let e ∈ H1(RPn;Z/2Z) be the

Euler class of E. For 1 ≤ j ≤ n− 1 we have H j(E0;Z/2Z) = 0 since E0 deformation retracts toSn, so part of the Gysin sequence reads

0 // H j(RPn;Z/2Z)·∪e // H j+1(RPn;Z/2Z)

Here both H j(RPn;Z/2Z) and H j+1(RPn;Z/2Z) are isomorphic to Z/2Z (again we are assum-

ing 1≤ j ≤ n−1), so the fact that ·∪ e : H j(RPn;Z/2Z)→ H j+1(RPn;Z/2Z) is injective impliesthat it is an isomorphism.

It readily follows by induction that, for 1≤ k ≤ n, the k-fold cup product e∪k ∈ Hk(RPn;Z/2Z)is the unique nonzero element of Hk(RPn;Z/2Z). Of course H0(RPn;Z/2Z) has its unique

nonzero element given by the unit 1, while for k /∈ 0, . . . , n, Hk(RPn;Z/2Z) = 0. Thus, as a

graded ring

H∗(RPn;Z/2Z) =(Z/2Z)[e]

⟨en+1⟩

where e ∈ H1(RPn;Z/2Z) is the Euler class of the tautological line bundle.

Example 5.24. Now let π: E → CPn be the tautological complex line bundle over CPn =

Gr1(Cn+1); since any complex vector bundle is naturally oriented (and in our case the rank

of E considered as a real vector bundle is 2), we have an Euler class e ∈ H2(CPn;Z). Now E0

is homeomorphic to Cn+1 \ ~0, which is homotopy equivalent to S2n+1. In particular H j(E0;Z)for 1≤ j ≤ 2n. So part of the Gysin sequence reads

0 // H j−2(CPn;Z)·∪e // H j(CPn;Z) // 0

for all 2 ≤ j ≤ 2n. In other words, for 2 ≤ j ≤ 2n, cup product with e defines an isomorphism

H j−2(CPn;Z)→ H j(CPn;Z).Now you likely learned in 8200 that CPn has a cell decomposition consisting precisely of one

2k cell for 1≤ k ≤ n. (This cell fk : D2k→ CPn is given by

fk(z1, . . . , zk) = [1−

k∑

i=1

|zi |2 : z1 : z2 : · · · : zk : 0 : · · · : 0]

where we identify D2k with the closed unit disk in Ck and we use homogeneous coordinates on

CPn. So the image of fk is the standard copy of CPk in CPn, and the restriction of fk to the


boundary gives the Hopf map S2k−1→ CPk−1.) By grading considerations, the cellular boundary

and coboundary operators must vanish, and so H j(CPn;Z) is isomorphic to Z if 0≤ j ≤ 2n andj is even, and is zero otherwise.

So just as in the case of RPn, it follows by induction that, for 1 ≤ k ≤ n, the k-fold cup

product e∪k generates H2k(CPn;Z). Thus

H∗(CPn;Z) =Z[e]

⟨en+1⟩

where e ∈ H2(CPn;Z) is the Euler class of the tautological complex line bundle over CPn.

6. STIEFEL–WHITNEY AND CHERN CLASSES

We are almost finally ready to define the Stiefel–Whitney and Chern classes of a (respec-tively, real or complex) vector bundle; before we do this we need to introduce the notion of theprojectivization of a vector bundle.

Let π: E→ M be a rank-k vector bundle, and as before let E0 denote the complement of thezero section in E. We can then define the topological space

P(E) =E0

v ∼ λv for v ∈ E0,λ ∈ R \ 0

where of course we are using the vector space structures on the fibers of E when we write “λv.”Since v and λv are always in the same fiber for v ∈ E,λ ∈ R, the map π: E→ M descends to amap π: P(E)→ M defined by π([v]) = π(v). It follows immediately from the definition of thequotient topology that π is continuous.

Suppose that Φ : E|U → U × Rk is a local trivialization for the vector bundle E, given incoordinates by Φ(v) = (π(v),φ(v)). Since φ restricts to the fiber Em as a linear isomorphism to

Rk for each m ∈ U , we have an induced map

Ψ : P(E)|U → U ×RPk−1

[v] 7→ (π(v), [φ(v)])

where we write P(E)|U = π−1(U). It is not hard to see that this map is a homeomorphism: First,

the fact that Φ is bijective readily implies that Ψ is also bijective. Now let p1 : E0|U → P(E)|U and

p2 : U × (Rk \ ~0)→ U ×RPk−1 be the quotient projections. Then for any set W ⊂ P(E)|U note

that p−12(Ψ(W )) = Φ(p−1

1(W )). So since Φ|E0|U

maps E0|U homeomorphically to U × (Rk \ ~0)

it follows from the definition of the quotient topology that Ψ(W ) is open if and only if W is,proving that Ψ is a homeomorphism.

So since Ψ maps each fiber P(E)m to m × RPk−1, it follows that π: P(E) → M is a fiber

bundle with fiber RPk−1 (and with local trivializations given by the maps Ψ). This fiber bundleis called the (real) projectivization of E

Similarly, if π: E→ M is a rank-k complex vector bundle, then we obtain the complex projec-

tivization of E, given by

PC(E) =E0

v ∼ λv for λ ∈ C \ 0,

with the obvious projection induced from π; this is a fiber bundle with fiber CPk−1.


6.1. Stiefel–Whitney classes. Recall from earlier that, if π: E → M is a vector bundle, a pre-

classifying map for E is a continuous map F : E→ RN (for some N) which restricts to each fiberEm as a linear injection. Such a map evidently gives rise to a “projectivized pre-classifying map”

P(F): P(E)→ RPN−1

[v] 7→ [F(v)]

Proposition 6.1. Let F : E→ RN be a pre-classifying map, inducing the projectivized pre-classifying

map P(F): P(E)→ RPN−1. Define

xF = (P(F))∗e(γ1(RN )) ∈ H1(P(E);Z/2Z).

Then for each m ∈ M, where im : P(Em) ,→ P(E) is the inclusion of the fiber, we have

H∗(P(Em);Z/2Z) = spanZ/2Zi∗m

1, i∗m

xF , . . . , i∗m

x k−1F.

Proof. This follows quickly from the following lemma:

Lemma 6.2. Let A: Rk → RN be an injective linear map and define A: RPk−1 → RPN−1 by

A([v]) = [Av]. Then A∗e(γ1(RN )) generates H1(RPk−1;Z/2Z).

Proof of Lemma 6.2. We have a commutative diagram

γ1(Rk)A //

γ1(RN )

RPk−1

A // RPN−1

where A is defined by A(V, ~v) = (AV, A~v) (recalling that γ1(Rk) consists of pairs (V, ~v) ∈ RPk−1×

Rk with ~v ∈ V ). The map A sends the fiber over a point V ∈ RPk−1 by a linear isomorphism to

the fiber over AV . So by Proposition 4.17, we have e(γ1(Rk)) = A∗e(γ1(RN )). By Example 5.23,

e(γ1(Rk)) generates H1(RPk−1;Z/2Z).

Given this lemma, note that for any m ∈ M , P(F) im : P(Em)→ RPN−1 is the projectivizationof an injective linear map from the k-dimensional real vector space Em to RN . So by Lemma

6.2, the class (P(F) im)∗e(γ1(RN )) = i∗

mxF generates H1(P(Em);Z/2Z) (where P(Em) is home-

omorphic to RPk−1). Then by Example 5.23, H∗(P(Em);Z/2Z) is the (Z/2Z)-span of the classes

1, i∗m

xF , . . . , (i∗m

xF )k−1. Since i∗

mis a unital ring homomorphism, Proposition 6.1 follows.

Proposition 6.3. If F0 : E → RN0 and F1 : E → RN1 are two pre-classifying maps for the same

vector bundle then xF0= xF1

(where xF jare defined as in Proposition 6.1).

Proof. By Theorem 2.8, there is N ≥ maxN0, N1 such that the pre-classifying maps IN0,N ′ F0

and IN1,N ′ F1 are isotopic (i.e., homotopic through pre-classifying maps), where IN j ,N′(~v) =

(~v, ~0). Where iN j ,N′ : RPN j−1→ RPN−1 is defined by iN j ,N

′([v]) = [(v, ~0)], we evidently have, for

j = 0, 1,P(IN j ,N

′ F) = iN j ,N′ P(F).

Moreover the isotopy from IN0,N ′ F0 to IN1,N ′ F1 gives rise to a homotopy from iN0,N ′ P(F) to

iN1,N ′ P(F). Now as noted in (14), we have i∗N j ,N

′e(γ1(RN )) = e(γ1(RN j )). So

xF0= P(F0)

∗e(γ1(RN0)) = P(F0)∗i∗

N0,N ′e(γ1(RN ′)) = (iN0,N ′ P(F0))

∗e(γ1(RN ′))

= (iN1,N ′ P(F1))∗e(γ1(RN ′)) = P(F1)

∗i∗N1,N ′

e(γ1(RN ′)) = xF1


Definition-Theorem 6.4. Let π: E→ M be a finite-type rank-k vector bundle over any paracom-

pact Hausdorff space M such that the Leray–Hirsch Theorem 5.18 holds for fiber bundles over M.

Then there are unique classes wi(E) ∈ H i(M ;Z/2Z) (0≤ i ≤ k) (the Stiefel–Whitney classes of E)

such that w0(E) = 1 and, for any pre-classifying map F : E→ RN , the class xF ∈ H1(P(F);Z/2Z)obeys the equation

k∑

i=0

π∗wi(E)∪ x k−iF= 0

Proof. The bundle E admits a pre-classifying map F by Proposition 2.5, and the class xF isindependent of the choice of F by Proposition 6.3. Theorem 5.18 and Proposition 6.1 then

combine to show that any class y ∈ Hk(P(E);Z/2Z) can be written uniquely in the form

y =

k∑

i=1

π∗ci ∪ x k−iF

where ci ∈ H i(E;Z/2Z). In particular this applies to the class y = −x kF, and then the Stiefel–

Whitney classes are given by w0(E) = 1 and wi(E) = ci .

Example 6.5. Suppose that π: E → M is a trivial rank-k vector bundle. Then the second com-

ponent of a trivialization Φ : E → M × Rk gives a pre-classifying map F : E → Rk (which is

an isomorphism on each fiber, not just a linear injection as usual). Now xF = P(F)∗e(γ1(Rk)),

and of course we have e(γ1(Rk))k = 0 since Hk(RPk−1;Z/2Z) = 0. Hence x kF= 0. Thus the

Stiefel–Whitney classes wi(E) are equal to 0 for i ≥ 1.

Example 6.6. Let π: E → RPn be the tautological line bundle over RPn, so E = (V, ~v) ∈RPn ×Rn+1|~v ∈ V. Then

P(E) = (V, [~v]) ∈ RPn ×RPn|[~v] = V

is the diagonal in the product RPn×RPn; the fiber bundle projection π: P(E)→ RPn just sends

(V, V ) to V and is a homeomorphism (this should make sense: P(E) is an RP0-bundle, and RP0

is a single point).

There is a pre-classifying map F : E → Rn+1 defined by F(V, ~v) = ~v, and this projectivizesto the map P(F): P(E)→ RPn defined by P(F)(V, V ) = V . In other words, using π to identifyP(E) with RPn, P(F) is just the identity.

So continuing to use π to identify P(E) with RPn, the Stiefel–Whitney classes w0(E) ∈

H0(M ;Z/2Z) and w1(E) ∈ H1(M ;Z/2Z) are determined by the properties that w0(E) = 1and

w0(E)∪ e(γ1(Rn+1)) + w1(E)∪ 1= 0

Thus

w1(E) = e(γ1(Rn+1))

(there is no need for a sign as we are working over Z/2Z).

The above example may seem slightly silly, but it turns out that this one special case, togetherwith some general formal properties proven below, is enough to determine all of the Stiefel–Whitney classes of all finite-type bundles over paracompact Hausdorff spaces.

Definition 6.7. The total Stiefel–Whitney class of a rank-k vector bundle as in Definition–Theorem6.4 is the class

w(E) = w0(E) + w1(E) + · · ·+ wk(E) ∈ H∗(M ;Z/2Z)


We will see later that the total Stiefel–Whitney class is useful for keeping track of informationabout how bundles behave under direct sum. Note that the trivial bundle has total Stiefel–Whitney class w(E) = 1.

Proposition 6.8. Let f : X → M be a continuous map and let πE : E → M and πV : V → X

be vector bundles as in Definition–Theorem 6.4 such that there is a map f : V → E making the

diagram

Vf //

πV

E

πE

X

f // M

commute, with f |Vxa linear isomorphism to E f (x). Then, for all i,

wi(V ) = f ∗wi(E)

In particular

(i) wi(E) depends only on the isomorphism type of E.

(ii) The Stiefel–Whitney classes satisfy the naturality property

wi( f∗E) = f ∗wi(E)

Proof. The assumption implies that there is a projectivized map P( f ): P(V )→ P(E) leading toa commutative diagram

P(V )P( f ) //

πV

P(E)

πE

X

f // M

If F : E → RN is a pre-classifying map, then F f : V → RN is also a pre-classifying map, with

projectivization P(F)P( f ), so that by definition (and by the functoriality of pullbacks) we have

xF f = P( f )∗xF .

Hence

0= P( f )∗

n∑

k=0

π∗Ewi(E)∪ x k−i

F

=

n∑

k=0

(P( f )∗π∗Ewi(E))∪ x k−i

F f

=

n∑

k=0

(π∗V

f ∗wi(E))∪ x k−i

F f

So by the uniqueness part of Definition–Theorem 6.4, we must have wi(V ) = f ∗wi(E).As for the two statements at the end, (i) is the special case in which f is the identity, and

(ii) is immediate, since V = f ∗E fits into a commutative diagram of the form described in theproposition.

In particular, it follows that, if f : M → Grk(RN ) is a classifying map for a vector bundle

π: E→ M then wi(E) = f ∗wi(γk(RN )). This suggests another way of defining Stiefel–Whitney

classes: first define them just for the tautological bundles γk(RN )→ Grk(RN ) as before, and then

define them for an arbitrary finite-type vector bundle E→ M over any paracompact Hausdorff

space M by setting wi(E) = f ∗wi(γk(RN )) for any classifying map f . This conveniently gets

around the issue that we have only proven the Leray–Hirsch theorem for a limited class of spaces,


since this limited class does include the Grassmannians. Indeed, if one (correctly) assumes theLeray–Hirsch theorem and hence the construction of Stiefel–Whitney classes for the infiniteGrassmannian Grk = Grk(R

∞), then one could drop the finite-type assumption on E and define

wi(E) = f ∗wi(γk)where γk→ Grk is the tautological bundle and f : M → Grk is any classifying

map.

Exercise 6.9. Let π: E→ S1 be any rank-k vector bundle. Cover S1 by the two open sets

U0 = (x , y) ∈ R2|x2 + y2 = 1, y > −1/2 U1 = (x , y) ∈ R2|x2 + y2 = 1, y < 1/2

By Remark 2.30, since U0 and U1 are contractible there are trivializations Φ0 : E|U0→ U0 ×R

k

and Φ : E|U1→ U1 × R

k. The map Φ1 Φ−10

: (U0 × U1) × Rk → (U0 ∩ U1) × R

k is given by

(x , v) 7→ (x , g01(x)v) for some map g01 : U0 ∩ U1 → GL(k;R). Note that both U0 ∩ U1 andGL(k;R) have two path components.

(a) Prove that if g01 maps the two path components of U0 ∩ U1 to the same path componentof GL(k;R), then E is trivial.

(b) Prove that if g01 maps the two path compenents of U0∩U1 to different path componentsof GL(k;R), then w1(E) 6= 0. (Suggestion: first show that any two vector bundles satisfying thehypothesis are isomorphic, and then prove the statement for just one such bundle—the Whitneyproduct formula (Theorem 6.16 below) might help for latter.)

(c) Now let π: V → M be any vector bundle, where M is assumed paracompact, Haus-dorff, and locally path-connected. Prove that V is orientable if and only if, for every continuous

γ: S1 → M , γ∗V is orientable. (Hint: A bundle is orientable if and only if its classifying mapslift to maps to the oriented Grassmannian.)

(d) Where π: V → M is as in part (c), prove that V is orientable if and only if w1(V ) = 0.

6.2. Chern classes. A very similar construction, when combined with Example 5.24 rather

than 5.23, gives rise to the Chern classes ci(E) ∈ H2i(M ;Z) of a finite-type rank-k complexvector bundle π: E → M over a space M to which the Leray–Hirsch theorem applies. Recallthat such a bundle admits a “complex pre-classifying map” F : E → CN (which is continuous

and complex-linear on each fiber); this projectivizes to a map PC(F): PC(E)→ CPN−1, whereπC

: PC(E) → M is the complex projectivization of E. We summarize the construction of theChern classes in the following; the proofs of the various statements are left to the reader (theyare directly analogous to the corresponding statements for Stiefel–Whitney classes, proven inthe last subsection):

Definition-Theorem 6.10. Under the above assumptions, the class zF ∈ H2(PC(E);Z) defined by

zF = −PC(F)∗e(γ1(CN )) is independent of the choice of complex pre-classifying map F and has the

property that, for each m ∈ M, one has

H∗(PC(Em);Z) = spanZi∗m

1, i∗m

zF , . . . , i∗m

zk−1F

Consequently by the Leray–Hirsch theorem there are unique cohomology classes (the Chern classes

of E) ci(E) ∈ H2i(M ;Z) with c0(E) = 1 and

k∑

i=0

π∗C

ci(E)∪ zk−iF= 0

The classes ci(E) depend only on the complex-isomorphism type of E and obey the naturality prop-

erty ci( f∗E) = f ∗ci(E) for a continuous map f : X → M.

Note the negative sign in the definition of zF ; it is included so as to avoid a negative sign inthe conclusion of the following example:


Example 6.11. Let π: E→ CPn be the tautological complex line bundle over CPn. Then just asin Example 6.6 one has

PC(E) = (V, [~v]) ∈ CPn ×CPn|[~v] = V,

i.e. PC(E) is the diagonal in the product CPn ×CPn, with the bundle projection given by pro-

jection onto the first factor. There is a complex pre-classifying map F : E → Cn+1 given byprojection to the second factor, and the projectivization of this pre-classifying map is again themap (V, V ) 7→ V from PC(E) to CPn. So, using the projection to identify PC(E) with CPn, theChern classes c0(E) and c1(E) are characterized by c0(E) = 1 and

1∪ (−e(E)) + c1(E)∪ 1= 0,

i.e.,

c1(γ1(Cn)) = e(γ1(Cn)) ∈ H2(CPn;Z)

As with Stiefel–Whitney classes, we define the total Chern class of a rank-k complex vectorbundle as c(E) = c0(E) + · · ·+ ck(E) ∈ H∗(M ;Z).

One could alternately define the Chern classes by first just defining the Chern classes ci(γk(CN )) ∈

H2i(Grk(CN );Z) of the tautological bundles and then putting ci(E) = f ∗ci(γ

k(CN )) for any com-plex classifying map f : M → Grk(C

N ).

Exercise 6.12. For each d ∈ Z define a complex line bundle π: Ed → CP1 by

Ed =[z0 : z1], (w0, w1)

∈ CP1 ×C2|zd

1w0 = zd

0w1

if d ≥ 0 and

Ed =[z0 : z1], (w0, w1)

∈ CP1 ×C2|z−d

1w0 = z−d

0w1

if d < 0 (with the bundle projection π given by projection to the first factor).

(a) Determine the first Chern classes c1(Ed), and in particular show that for each c ∈ H2(CP1;Z)there is a unique value of d with c1(Ed) = c.

(b) Prove that every complex line bundle π: E→ CP1 is isomorphic to Ed for some d. (Hint:By Remark 2.30 the restrictions of E to small neighborhoods of the northern and southern

hemispheres are trivial. The transition map Φ1 Φ−10

associated to a pair of trivializations overneighborhoods of the northern and southern hemispheres gives a map from a neighborhood ofthe equator to GL(1;C) = C \ 0. The correct value of d is determined by the homotopy classof this map.)

6.3. Whitney Sums. We now give details for a construction alluded to earlier of the “directsum” (sometimes called “Whitney sum”) of two vector bundles. Let πV : V → M and πW : W →M be vector bundles of ranks k and l, respectively, over the same space M . Define

V ⊕W = (v, w) ∈ V ×W |πV (v) = πW (w).

We have an obvious continuous map πV⊕W : V ⊕W → M given as the composition of the pro-jection to V with πV : V → M (or equivalently as the composition of the projection to W withπW ). M is covered by open sets U with local trivializations

ΦV : V |U → U ×RkΦW : W |U → U ×Rl

v 7→ (πV (v),φV (v)) f 7→ (πW (w),φW (w))

which gives rise to a continuous bijection Φ : π−1V⊕W(U) → U × Rk+l defined by Φ(v, w) =

(πV (v),φV (v),φW (w)). Each fiber (V ⊕W )m = π−1V⊕W(m) is in obvious bijection with the

vector space Vm ⊕Wm, and if m ∈ U then Φ maps each (V ⊕W )m by a linear isomorphism to

Rk+l . We have Φ−1(m, v, w) = (Φ−1V(m, v),Φ−1

W(m, w)), so Φ−1 is continuous.


This suffices to prove that the maps Φ : (V ⊕W )|U → U ×Rk+l form a system of local trivial-izations, thus making V⊕W into a rank-(k+ l) vector bundle. Evidently if V and W are complexvector bundles then so too is V ⊕W .

Example 6.13. An important context in which direct sums of bundles naturally appear is thefollowing. Suppose that M is a smooth manifold, and suppose that N is an embedded subman-ifold of M . There are then three natural vector bundles over N : one has the tangent bundleT N ; the restriction T M |N of the tangent bundle of M to N ; and the normal bundle νN ,M , whose

fiber at n ∈ N can be naturally identified withTn M

TnN(see for instance [U2, Example 2.4]). It is

fairly easy to see from the definitions that there is an isomorphism of vector bundles

T M |N∼= T N ⊕ νN ,M .

Somewhat more generally, if g : N → M is an immersion, one can still form the normal

bundle νg → N to g with fiber over n isomorphic toTg(n)M

g∗TnN: just use the fact that the restriction

of an immersion to a sufficiently small open set is an embedding and use such small open setsto construct local trivializations. In this case one has a direct sum decomposition

g∗T M ∼= T N ⊕ νg .

We will soon see that, using these direct sum splittings, Stiefel–Whitney classes can be usedto give obstructions to the existence of certain classes of immersions or embeddings.

We will now work toward establishing formulas that represent the characteristic classes ofthe direct sum V ⊕W of two vector bundles V and W to the characteristic classes of V and W .The following lemma, which is of independent interest, turns out to be relevant:

Proposition 6.14. Let R be a commutative ring with unity, let X be a space, and let i1 : U1→ X ,

i2 : U2 → X be inclusions of open sets, with X = U1 ∪ U2. Suppose that c1, c2 ∈ H∗(X ; R) be two

classes such that i∗1c1 = 0 and i∗

2c2 = 0. Then c1 ∪ c2 = 0.

Proof. For j = 1, 2 consider the long exact sequence of the pair (X , U j), part of which reads

H∗(X , U j;R)π∗j // H∗(X ; R)

i∗j // H∗(U j; R)

So the assumption on c j implies that c j ∈ Im(π∗j: H∗(X , U j; R)→ H∗(X ; R)), i.e., that the coho-

mology class c j can be represented can be represented by a cocycle α j ∈ S∗(X ; R)which vanisheson all simplicies with image contained in U j .

But then, as is obvious from the definition of the cup product, the cocycle α1∪α2 vanishes onall simplices whose image is contained either in U1 or in U2, i.e. the cocycle α1 ∪ α2 evaluatestrivially on all elements of the subcomplex SU1,U2

• (X ) described at the start of Section 3.3.1. By

(11), this implies that the cohomology class c1 ∪ c2 = [α1 ∪α2] is zero.

Exercise 6.15. If R is a commutative ring with unity, the R-cuplength of a space X , denotedcl(X ;R), is the supremum of all positive integers k such that there are classes ci ∈ Hni (X ; R)

(1≤ i ≤ k), with each ni > 0, such that c1∪· · ·∪ck 6= 0. Prove that any open cover X = U1∪· · ·Ul

of X by contractible open sets must have cardinality l ≥ cl(X ; R) + 1. Show also that there is anopen cover of RPn by exactly l = cl(RPn;Z/2Z) + 1 many contractible open sets.

Recall that the total Stiefel-Whitney class of a rank-k vector bundle π: E → M is the (in-

homogeneous) cohomology class w(E) =∑k

i=0wi(E) ∈ H∗(M ;Z/2Z). (Of course, as always,

w0(E) = 1.)


Theorem 6.16 (Whitney Product Formula). Let πV : V → M and πW : W → M be two (finite-

type) vector bundles over the paracompact Hausdorff space M. Then we have

w(V ⊕W ) = w(V )w(W ).

Proof. The maps v 7→ (v, 0) and w 7→ (0, w) projectivize to embeddings jV : P(V )→ P(V ⊕W )

and jW : P(W )→ P(V ⊕W ), respectively; we use these embeddings to identify P(V ) and P(W )as subsets of P(V ⊕W ). Define open sets

UV = [(v, w)] ∈ P(V ⊕W )|v 6= 0 UW = [(v, w)] ∈ (V ⊕W )|w 6= 0

Note that P(V⊕V ) = UV ∪UW . Also, the homotopy (t, [(v, w)]) 7→ [(v, tw)] gives a deformationretraction of UV to P(V ), and likewise (t, [v, w]) 7→ [(t v, w)] gives a deformation retraction ofUW to P(W ). So where iV : UV → P(V ⊕W ) and iW : UW → P(V ⊕W ) are the inclusions, forany class c ∈ H∗(P(V ⊕W );Z/2Z) we have i∗

Vc = 0 if and only if j∗

Vc = 0, and i∗

Wc = 0 if and

only if j∗W

c = 0. In particular it follows from Proposition 6.14 that

(20) If j∗V

c1 = j∗W

c2 = 0, then c1 ∪ c2 = 0

Now let F0 : V → RN0 and F1 : W → RN1 be two pre-classifying maps. We then obtain apre-classifying map F : V ⊕W → RN0+N1 by setting F(v, w) = (F0(v), F1(w)). Note that the

compositions P(F) jV : P(V )→ RPN0+N1−1 and P(F) jW : P(W )→ RPN0+N1−1 are projectivizedpre-classifying maps for V an W .

Now let x = P(F)∗e(γ1(RN0+N1)). Where πV : P(V ) → M , πW : P(W ) → M , and π: P(V ⊕W ) → M are the bundle projections, we hence have π jV = πV , π jW = πW , and, by thedefinition of the Stiefel–Whitney classes

(21)

k∑

i=0

π∗V

wi(V )∪ j∗V

x k−i = 0 ∈ H∗(V ;Z/2Z)

and

(22)

l∑

i=0

π∗W

wi(W )∪ j∗W

x l−i = 0 ∈ H∗(P(W );Z/2Z)

(here we denote the rank of V by k and the rank of W by l).Now define

cV =

k∑

i=0

π∗wi(V )∪ x k−i ∈ H∗(P(V ⊕W );Z/2Z)

and

cW =

l∑

i=0

π∗wi(W )∪ x l−i ∈ H∗(P(V ⊕W );Z/2Z)

From (21) and (22) we see that j∗V

cV = j∗W

cW = 0. Hence by (20) cV ∪ cW = 0, i.e.,

0=

k∑

i=0

π∗wi(V )∪ x k−i

!∪

l∑

i=0

π∗wi(W )∪ x l−i

!

=

k+l∑

m=0

π∗

m∑

i=0

wi(V )∪ wm−i(W )

x k+l−m


(we have used Remark 5.5 and the fact that we are working over Z/2Z to commute x with some

of the w j(W ) in the last equation). The coefficient on x k+l above is w0(V )∪ w0(W ) = 1, so by

the definition of the Stiefel–Whitney classes this shows that

wm(V ⊕W ) =

m∑

i=0

wi(V )∪ wm−i(W ),

which is more concisely expressed by the formula for the total Stiefel–Whitney classes

w(V ⊕W ) = w(V )∪ w(W )

Similarly for the total Chern classes:

Theorem 6.17. Let πV : V → M and πW : W → M be two finite-type complex vector bundles

over the paracompact Hausdorff space M. Then their total Chern classes c(V ) =∑

ci(V ), c(W ) =∑ci(W ) obey

c(V ⊕W ) = c(V )∪ c(W )

The proof of Theorem 6.17 is exactly the same as that of Theorem 6.16 and so is left tothe reader—just replace real projectivizations and the classes xF of Definition-Theorem 6.4 bycomplex projectivizations and the classes zF of Definition-Theorem 6.10. (All of the relevantcup products are of even-degree classes, so they are still commutative by Remark 5.5.)

The Whitney product formula is especially helpful for gaining information about subbundlesof vector bundles. First we define these:

Definition 6.18. Let π: E → M be a rank-n real vector bundle. A rank-k subbundle is a sub-set V ⊂ E such that there is an open cover Uα of M and a collection of local trivializations

Φα : E|Uα → Uα ×Rn such that E|Uα ∩ V = Φ−1

α (Uα ×Rk × ~0).

Obviously V is in this case a vector bundle in its own right, with local trivializations given bythe Φα|E|Uα∩V .

Proposition 6.19. If M is a paracompact Hausdorff space and π: E→ M is a vector bundle over

M, then for any subbundle V ⊂ E there is another subbundle W ⊂ E such that E ∼= V ⊕W.

Proof. (Sketch) The main point is that E admits an orthogonal structure (i.e. a continuouslyvarying family of inner products on the fibers Em). Indeed, if

Φα : E|Uα → Uα ×Rn

e 7→ (π(e),φα(e))

is a system of local trivializations, since M is paracompact and Hausdorff we may find a partitionof unity χα subordinate to the cover Uα of M . Then for m ∈ M and e1, e2 ∈ Em define

⟨e1, e2⟩=∑

α

χα(m)φα(e1) ·φα(e2)

where · denotes the standard dot product on Rn. Since a convex combination of inner productsis still an inner product, ⟨·, ·⟩ defines an inner product on each fiber Em.

We accordingly define W to be the orthogonal complement to V with respect to ⟨·, ·⟩:

W = e ∈ E|⟨e, v⟩= 0 for all v ∈ Vπ(e)

It is not difficult to check (see [MS, Theorem 3.3] for details) that W is a subbundle of E, andthat E ∼= V ⊕W .


One way of constructing a subbundle is the following. Suppose that we have sections s1, . . . , sk : M →E (where E is a rank-n vector bundle) with the property that, for each m ∈ M , the elementss1(m), . . . , sk(m) are linearly independent. Define

V =

(k∑

i=1

cisi(m)

ci ∈ R, m ∈ M

)

Let us show that V is a subbundle. If m ∈ M choose a local trivialization Φ : E|U → U×Rn (givenby Φ(e) = (π(e),φ(e))). Let Am be the orthogonal complement of φ(s1(m)), . . . ,φ(sk(m)) inRn. By continuity, since Am ⊕ spanφ(s1(m)), . . . ,φ(sk(m))= R

n, we will have

(23) Am ⊕ spanφ(s1(x)), . . . ,φ(sk(x))= Rn

for all x in some neighborhood O ⊂ U of m. Moreover the map px : Rn→ Rk given by composingthe projection to spanφ(s1(x)), . . . ,φ(sk(x)) determined by the direct sum decomposition(23) with the map

∑cisi(x) 7→ (c1, . . . , ck) is continuous as a function of x . Likewise the function

qx : Rn → Rn−k defined by composing the projection onto Am determined by (23) with some

fixed isomorphism from Am to Rn−k is continuous in x . Therefore we can define a new localtrivialization Φ′ for E over the neighborhood O of m by

Φ′(e) = (π(e), px(φ(e)), qx(φ(e)))

This trivialization maps V ∩ E|O to O × Rk. So since E|M is covered by the domains of suchtrivializations it follows that V is a subbundle of E.

Moreover it is obvious that in this situation V is a trivial bundle: a global trivialization

Ψ : V → M ×Rk is given by

Ψ

k∑

i=1

cisi(m)

!= (m, c1, . . . , ck)

Consequently we obtain the following consequence of the Whitney product formula, which mayshed some additional light on the meaning of the Stiefel–Whitney classes.

Corollary 6.20. Let M be a paracompact Hausdorff space. Suppose that the rank-n vector bundle

π: E→ M admits sections s1, . . . , sk : M → E such that s1(m), . . . , sk(m) is linearly independent

for every m. Then

wn−k+1(E) = · · · · · ·= wn−1(E) = wn(E) = 0

Proof. We have a trivial, rank-k subbundle V ⊂ E by the above discussion (V is the span of thesections si). By Proposition 6.19 we may write E ∼= V ⊕W for some rank-(n− k) vector bundleW . Now since V is trivial, the its total Stiefel–Whitney class is given by w(V ) = 1. So

w(E) = w(V )∪ w(W ) = w(W ) = 1+ w1(W ) + · · ·+ wn−k(W ),

as the rank of W is only n− k.

The special case where k = 1 shows that a rank-n vector bundle E with a nonvanishingsection must have wn(E) = 0. Of course we earlier saw in Proposition 4.11 that such a bundlealso has e(E) = 0; Theorem 6.24 below will show that this is not a coincidence.

The same argument shows that if a rank-n complex vector bundle π: E → M has sectionss1, . . . , sk which are everywhere linearly independent over C, then we must have cn−k+1(E) =

· · ·= cn(E) = 0.


Exercise 6.21. (a) Let M be a paracompact Hausdorff space and let π: E → M be a vectorbundle with projectivization π: P(E) → M . Prove that there is a rank-1 subbundle V ⊂ π∗E.(Hint: Your V should be somewhat similar to the tautological line bundle over RPn.)

(b) Prove the splitting principle: If M is a paracompact Hausdorff space and π: E → M is arank-k vector bundle then there is a space N , a map p : N → M such that p∗ : H∗(M ;Z/2Z)→H∗(N ;Z/2Z) is injective, and rank-1 vector bundles V1, . . . , Vk over N such that p∗E = V1⊕· · ·⊕Vk. (Throughout your proof you may freely assume without verifying it that all spaces that youconstruct are paracompact and Hausdorff and have the property that the Leray–Hirsch theoremapplies to fiber bundles over them.)

Exercise 6.21 (and the fact that the space N in (b) can be taken to be paracompact and Haus-dorff) allows one to show that the Stiefel–Whitney classes of vector bundles over paracompactHausdorff spaces are uniquely characterized by the following axioms (each of which we haveverified based on our construction):

• For each finite-type rank-k vector bundle π: E → M over any paracompact Hausdorff

space M we have classes wi(E) ∈ H i(M ;Z/2Z) for 0 ≤ i ≤ k, such that w0(E) = 1. Wewrite w(E) = w0(E) + w1(E) + · · ·+ wk(E).• (Naturality) If f : X → Y is a continuous map between paracompact Hausdorff spaces,

then for any vector bundle π: E→ Y

w( f ∗E) = f ∗w(E)

• (Whitney Product) If V and W are two vector bundles over M then

w(V ⊕W ) = w(V )∪ w(W )

• (Normalization) For the tautological line bundle E = γ1(Rn) over RPn−1, w1(E) =

e(E).

Indeed the naturality and normalization axioms suffice to determine w(V ) for any line bundle

V → M , since such a bundle is isomorphic to f ∗γ1(Rn) for some n and some f : M → RPn−1. Sothe Whitney Product axiom determines w(E) whenever E splits as a direct sum of line bundles.Although not every vector bundle splits as a direct sum of line bundles (see Exercise 6.26), ifp : N → M is as in Exercise 6.21(b) then p∗w(E) = w(p∗E) is determined by the axioms, andhence w(E) is as well since p∗ is injective.

Again, a similar set of axioms may be used to characterize the Chern classes of complex vectorbundles.

Of course, this argument proves uniqueness of the Stiefel–Whitney classes subject to theabove axioms, but does not by itself prove existence (since, for instance, a vector bundle mightsplit as a sum of line bundles in many different ways)—for existence we needed an explicitconstruction such as the one given earlier in this section based on the Leray–Hirsch theorem. Adifferent construction of the Stiefel–Whitney classes is given in [MS] using Steenrod squares;the uniqueness argument shows that the Stiefel–Whitney classes defined there are the same asours.

6.4. Relationships between the characteristic classes. If π: E → M is a rank-k complexvector bundle we now have three types of characteristic classes for E: the Euler class e(E) ∈H2k(M ;Z); the Stiefel–Whitney classes wi(E) ∈ H i(M ;Z/2Z); and the Chern classes ci(E) ∈H2i(M ;Z). As we will see, these are not completely independent of each other. We begin bycomparing the wi and ci . Of course these classes live in different cohomology groups (withZ/2Z coefficients for one and Z coefficients for the other); however as discussed in Exercise4.15 there is a natural map H j(M ;Z)→ H j(M ;Z/2Z) induced by the quotient map Z→ Z/2Z.In general, for a class x ∈ H j(M ;Z), we will write x for the image of x under this map, so


x ∈ H j(M ;Z/2Z). So in particular the classes w2i and ci(E) both belong to the cohomology

group H2i(M ;Z/2Z).To relate these, we first prove the following:

Lemma 6.22. For any integer N ≥ 1 define p : RP2N−1 → CPN−1 to be the map which sends

a one-dimensional real subspace V of R2N = CN to the one-dimensional complex subspace of CN

spanned by any nonzero element of V . Then we have

e(γ1(R2N ))∪ e(γ1(R2N )) = p∗e(γ1(CN ))

Proof. If N = 1 both sides are zero (they belong to H2(RP1;Z/2Z)) so we assume N ≥ 2.It is not difficult (and left to the reader) to show that the map p is a fiber bundle with fiber

RP1 (the fiber over W ≤ CN consists of the one-real-dimensional subspaces of the two-real-

dimensional subspace W of CN , which is a copy of RP1). Moreover this fiber bundle satisfies

the Leray–Hirsch property: the class e := e(γ1(R2N )) restricts to each RP1-fiber as the Euler

class of the tautological line bundle over (that copy of) RP1, so the cohomology of the fiber

over a point m ∈ CPN−1 is the Z/2Z module generated by i∗m

1 and i∗m

e.So by the Leray–Hirsch theorem we can write

e ∪ e = p∗ y2 ∪ 1+ p∗ y1 ∪ e

for some classes y1 ∈ H1(CPN−1;Z/2Z) and y2 ∈ H2(CPN−1;Z/2Z). But of course H1(CPN−1;Z/2Z) =0, so y1 = 0. Meanwhile e ∪ e 6= 0 by Example 5.23 (since N ≥ 2), while the only nonzero ele-

ment of H2(CPN−1;Z/2Z) is e(γ1(CN )) by (the Z/2Z-coefficient version of) Example 5.24. So

we indeed havee ∪ e = p∗e(γ1(CN ))

This leads to the following relationship between the Stiefel–Whitney and Chern classes:

Theorem 6.23. If π: E → M is a rank-k complex vector bundle with Stiefel–Whitney classes

wi(E) ∈ H i(M ;Z/2Z) and mod 2 Chern classes ci(E) ∈ H2i(M ;Z/2Z) then we have

w2i(E) = ci(E) w2i+1(E) = 0

for all natural numbers i.

Proof. Let F : E→ CN be a complex pre-classifying map. Of course, viewing CN as R2N , we can

also view F as an ordinary (real) pre-classifying map to R2N .

Now there is a quotient map pE : P(E)→ PC(E), since the equivalence relation on E0 defining

PC(E) is a refinement of (i.e., contains) the equivalence relation on E0 defining P(E), and clearlyπC pE = π: P(E)→ M . Moreover we have a commutative diagram

P(E)P(F) //

pE

RP2N−1

p

PC(E)

PC(F) // CPN−1

Using straightforward functoriality properties of the mod 2 reduction process of Exercise 4.15(in particular the fact that it defines a ring homomorphism, by virtue of the fact that Z→ Z/2Zis a ring homomorphism), we see from the above diagram that the classes xF = P(E)

∗e(γ1(R2N ))

from Lemma 6.2 and zF= PC(E)

∗e(γ1(CN )) from Definition-Theorem 6.10 obey

xF ∪ xF = p∗Ez

F


Now the definition of the Chern classes gives

k∑

i=0

π∗C

ci(E)∪ zk−iF= 0

Applying p∗E

to this equation gives, since πC pE = π,

k∑

i=0

π∗ci(E)∪ x2k−2iF

= 0

So since the w j(E) are the unique classes with w0(E) = 1 satisfying an equation

2k∑

j=0

π∗w j(E)∪ x2k− j

F

the result follows.

Now we turn to the Euler class:

Theorem 6.24. Let π: E→ M be a finite-type rank-k vector bundle over a paracompact Hausdorff

space. Then the (mod 2) Euler class e(E) is given by

e(E) = wk(E) ∈ Hk(M ;Z/2Z)

Proof. By pulling back by classifying maps and using the naturality of w1 and e we can assumethat M = Grk(R

N )—in particular we can assume that the Leray–Hirsch theorem applies tobundles over M .

Let R be the trivial rank-1 bundle over M , and form the Whitney sum E⊕R. So E⊕Rmay beidentified with E×R, with the bundle projection πE⊕R : E⊕R→ M just given by projection to E

followed by π. The fiber of E⊕R over a point m is just Em⊕R. There is a special non-vanishing

section s1 : M → E⊕R, defined by setting s1(m) equal to the element (0, 1) of (E⊕R)m = Em⊕R.Now form the projectivization πE⊕R : P(E ⊕R)→ M . A convenient thing about P(E ⊕R) for

our purposes is that it contains both a copy of P(E) (embedded by the map [e] 7→ [(e, 0)]) anda copy of M (embedded by the map m 7→ [s1(m)] where s1 : M → E ⊕R is defined in the lastparagraph). For the rest of the proof we use these embeddings to regard P(E) and M as closedsubsets of P(E ⊕R); note that these closed subsets are disjoint.

In fact, the whole bundle E embeds into P(E ⊕R), via the map j : E→ P(E ⊕R) defined by

j(e) = [(e, 1)]

We see that

j(E) = P(E ⊕R) \ P(E) j(E0) = P(E ⊕R) \ (P(E)∪M)

Now the Excision Theorem from 8200 implies that the inclusion of pairs

(P(E ⊕R) \ P(E),P(E ⊕R) \ (P(E)∪M)) ,→ (P(E ⊕R),P(E ⊕R) \M)

induces a chain homotopy equivalence of relative singular chain complexes, and hence an iso-

morphism on cohomology. So since j : (E, E0)→ (P(E ⊕R) \ P(E),P(E ⊕R) \ (P(E)∪M)) is ahomeomorphism of pairs it follows that

(24) j∗ : H∗(P(E ⊕R),P(E ⊕R) \M ;Z/2Z)→ H∗(E, E0;Z/2Z) is an isomorphism


In particular, H2k(P(E ⊕R),P(E ⊕R) \M ;Z/2Z)∼= Z/2Z, with generator ( j∗)−1(τ) where τis the Thom class of E. Moreover for any m ∈ M the inclusions of fibers give a commutativediagram

H∗(P(E ⊕R),P(E ⊕R) \M ;Z/2Z)j∗ //

H∗(E, E0;Z/2Z)

H∗(P(E ⊕R)m,P(E ⊕R)m \ [0, 1];Z/2Z) // H∗(Em, E0

m;Z/2Z)

where the bottom row is again an isomorphism by excision. Consequently ( j∗)−1(τ) restricts to

each (P(E⊕R)m,P(E⊕R)m\[0, 1]) as a generator for H2k(P(E⊕R)m,P(E⊕R)m\[0, 1];Z/2Z)Note also that P(E ⊕ R) \ M deformation retracts to P(E) by the homotopy (t, [(e, z)]) 7→

[(e, tz)]. In particular a class c ∈ H∗(P(E ⊕ R);Z/2Z) restricts to zero in P(E) if and only ifit restricts to zero in P(E ⊕ R) \ M . Restricting this deformation retraction to the fiber over

m ∈ M gives a deformation retraction of P(E ⊕R)m \ [0, 1] to P(E)m∼= RPk−1. In particular

Hk(P(E ⊕R)m \ [0, 1];Z/2Z) = 0.

Let F0 : E → RN be a pre-classifying map for E. Define F : E ⊕ R → RN+1 by F(e, z) =

(F0(e), z); clearly F is a pre-classifying map for E ⊕ R, and it also restricts to E as anotherpre-classifying map for E. As in Definition-Theorem 6.10, define

x = P(F)∗e(γ1(RN+1)) ∈ H2(P(E ⊕R);Z/2Z)

Note that P(F) restricts to our copy M ⊂ P(E ⊕R) as the constant map to [(~0, 1)], and so

x |M = 0.

Also, whereπ: P(E)→ M is the bundle projection, the Stiefel–Whitney classes wi(E) ∈ H i(M ;Z/2Z)

(1≤ i ≤ k) are by definition the unique classes in H i(M ;Z/2Z) such that

x k|P(E) +

k∑

i=1

π∗wi(E)∪ x k−1|P(E) = 0

Now consider the commutative diagram of long exact sequences of pairs

Hk(P(E ⊕R),P(E ⊕R) \M ;Z/2Z)p∗ //

i∗m

Hk(P(E ⊕R);Z/2Z)i∗ //

i∗m

Hk(P(E ⊕R) \M ;Z/2Z)

Hk(P(E ⊕R)m,P(E ⊕R)m \ [0, 1];Z/2Z) // Hk(P(E ⊕R)m;Z/2Z) // Hk(P(E ⊕R)m \ [0, 1];Z/2Z)

where p∗ is induced by the inclusion p : (P(E⊕R),∅)→ (P(E⊕R),P(E⊕R)\M). As noted earlier,

the group on the bottom right is zero—it may be identified with Hk(RPk−1;Z/2Z)—so the first

map on the bottom is surjective. The group in the bottom middle is isomorphic to H2k(RPk;Z)∼=Z/2Z. Now we saw earlier that i∗

m( j∗)−1τ generates Hk(P(E⊕R)m,P(E⊕R)m\[0, 1];Z/2Z), so

its image under the bottom left map must generate Hk(P(E⊕R)m;Z/2Z). So since the diagram

commutes, p∗( j∗)−1τ ∈ H2k(P(E ⊕ R);Z) restricts to each fiber P(E ⊕ R)m as a generator for

Hk(P(E ⊕R)m;Z/2Z).

By the Leray–Hirsch theorem the general element of Hk(P(E ⊕ R);Z/2Z) takes the form∑k

j=0π∗

E⊕R y j ∪ x k− j where y j ∈ H j(M ;Z/2Z). Such a class restricts to each fiber P(E ⊕R)m∼=


RPk as a generator if and only if y0 = 1. So we can write

p∗( j∗)−1τ= x k +

k∑

j=1

π∗E⊕R y j ∪ x k− j

But the exactness of our sequence implies that i∗p∗( j∗)−1τ = 0 ∈ Hk(P(E);Z/2Z). As notedabove, this forces y j = w j(E) by the definition of the Steifel–Whitney classes of E. Now wehave a commutative diagram

(P(E ⊕R),∅)

p

))SSSSS

SSSS

SSSS

S

(M ,∅)

88pppppppppp

s0 &&NNNN

NNNN

NN(P(E ⊕R),P(E ⊕R)|M )

(E, E0)

j

55kkkkkkkkkkkkkkk

where the unlabeled map is the inclusion of M into P(E ⊕R). By definition e(E) = s∗0τ, and so

we obtain

e(E) =p∗( j∗)−1τ

|M =

k∑

j=0

π∗E⊕Rw j(E)x

k− j

!M

But as noted earlier x |M = 0, so the only term that contributes is that where j = k, and thefiber bundle projection πE⊕R restricts to M ⊂ P(E ⊕ R) as the identity; thus we indeed have

e(E) = wk(E).

There is a similar result for the Z-coefficient Euler class and the Chern class, whose proof weomit:

Theorem 6.25. Let π: E→ M be a finite-type rank-k complex vector bundle over a paracompact

Hausdorff space. Then the (oriented) Euler class e(E) is given by

e(E) = ck(E) ∈ H2k(M ;Z)

The proof of this is almost identical to that of Theorem 6.24, just replacing real projectiviza-tions by complex projectivizations everywhere. The only real difference comes near the end ofthe proof: since for Theorem 6.25 we are working with coefficients in Z and not Z/2Z, there

are two different generators for the top-degree cohomology of a fiber of a CPk-bundle, and sothe argument that we gave to say that the class denoted y0 in the above proof was equal to1 only applies in the Chern class case to show that y0 = ±1. From this one can see just as inthe rest of the proof that e(E) = ±ck(E) with the sign ± determined by the property that our

distinguished generatorω ∈ H2k(Ck,Ck \0;Z) (chosen long ago in the definition of the Thomclass) is sent by the the composition

H2k(Ck,Ck \ 0;Z)→ H2k(CPk,CPk \CPk−1;Z)→ H2k(CPk;Z)

(where the first map is the excision isomorphism) to

±−e(γ1(Ck))

k

In particular this sign depends only on k, not on the bundle or on the base space. Example 6.11shows that the sign is positive when k = 1, and then one can prove that it is positive for all k


by appealing to product formulas for Whitney sums both for the Euler class ([MS, 9.6]) and forthe Chern class.

Exercise 6.26. Let π: E → CP1 be the tautological complex line bundle, but regard E as a realrank-2 vector bundle. Prove that there does not exist a rank-1 subbundle V ⊂ E. (Hint: Showthat if V existed then w1(V ) would be a nonzero element of H1(CP1;Z/2Z).)

7. IMMERSIONS AND EMBEDDINGS

We can now start to put Stiefel–Whitney classes to work in the study of smooth manifolds.Recall that a basic example of a vector bundle is the tangent bundle T M of a smooth manifoldM . If φ : U → Rm is a smooth coordinate chart for M and p ∈ U , the tangent space Tp M

is the vector space with basis

§∂

∂ x1

p

, . . . , ∂

∂ xm

p

ª. This then gives rise to a local trivialization

Φ : T M |U → U ×Rm, defined by

Φ

m∑

i=1

ci

∂

∂ x i

p

= (p, c1, . . . , cm).

(The transition maps between two such local trivializations are given by differentiating the

smooth transition functions φβ φ−1α of the atlas for m, and hence are continuous (indeed

smooth), and so we get a well-defined topology on T M by declaring such trivializations Φ to behomeomorphisms, and with this topology the projection π: T M → M defines a vector bundle.)

If M and N are smooth manifolds and if f : M → N is a smooth map then for each p ∈ M

we have a linear map f∗ : TpM → T f (p)N obtained by differentiating f . Recall that a smooth

map f : M → N is called an immersion if for every p ∈ M the derivative f∗ : TpM → T f (p)N is

injective. Using the inverse function theorem, one can show that if f : M → N is an immersion,then for each p ∈ M there is a coordinate chart (x1, . . . , xm) : U → Rm for M around p and acoordinate chart (y1, . . . , yn) : V → Rn for N around f (p) in terms of which f takes the form

f (x1, . . . , xm) = (x1, . . . , xm, 0, . . . , 0). (In particular U ⊂ f −1(V ))We have a local trivialization for T N around f (p) associated to the coordinates (y1, . . . , yn);

pulling this back gives a local trivialization Ψ : ( f ∗T N)|U → U×Rn for the bundle f ∗T N aroundp ∈ M . At each point q ∈ M , f∗ defines an injection TqM → ( f ∗T N)q, and in fact we see that f∗identifies T M |U with Ψ−1(U ×Rm × ~0). So f∗ identifies T M with a rank-m subbundle of therank-n vector bundle T N |M . Combined with Proposition 6.19 (which applies since all smoothmanifolds are paracompact and Hausdorff) this proves:

Proposition 7.1. If f : M → N is an immersion between smooth manifolds M and N with

dim M = m and dim N = n, then there is a rank-(n − m) vector bundle ν f → M (called the

normal bundle of f ) such that

f ∗T N ∼= T M ⊕ ν f

Consider the following problem:

Problem 7.2. Given a smooth n-manifold N, a smooth m-manifold M, and a smooth map f0 : M →N, determine whether there is an immersion f : M → N which is homotopic to f .

There is an obvious necessary condition here, namely that n ≥ m. Less trivially, Whitneyshowed in 1936 that the desired immersion always exists if n ≥ 2m (in fact, what he reallyshowed is that immersions form a residual subset of the space of smooth maps from M to N

with respect to a certain topology, so that in particular a random perturbation of f0 should beexpected to be an immersion—see [Hi, Section 3.2]).


Proposition 7.1 and the Whitney product formula have implications for Problem 7.2. Indeed,suppose that an immersion f : M → N exists which is homotopic to f0. Then where w denotestotal Stiefel–Whitney class we have

f ∗0

w(T N) = f ∗w(T N) = w( f ∗T N) = w(T M ⊕ ν f ) = w(T M)∪ w(ν f ).

Here is a useful observation:

Proposition 7.3. Let M be a topological space such that there exists m ∈ Z+ with H i(M ;Z/2Z) = 0for all i > m, and let π: E → M be a vector bundle with total Stiefel–Whitney class w(E). Then

there is a unique class w(E) = w0(E) + · · · + wm(E), where w i(E) ∈ H i(M ;Z/2Z), such that

w(E)∪ w(E) = 1.

(The classes w i(E) are called the dual Stiefel-Whitney classes of E.)

Proof. We show by induction on k that there is a unique collection of classes w i(E) ∈ H i(M ;Z/2Z)for 0≤ i ≤ k such that, for all i ≤ k, the grading-i part of (w0(E) + · · ·+ wk(E))∪ w(E) is 1 fori = 0 and 0 otherwise.

For k = 0 this is obvious: since w0(E) = 1 the grading-0 part of w0(E)∪ w(E) is just w0(E),and so the unique w0(E) satisfying the requirement is w0(E) = 1.

Assume the statement for k−1 where k ≥ 1, so that we have identified elements w0(E), . . . , wk−1(E).

Now for an arbitrary collection of elements x i ∈ H i(M ;Z/2Z), for all i ≤ k−1 the grading i partof (x0+ · · ·+ xk−1+ xk)∪w(E) is the same as the grading-i part of (x0+ · · ·+ xk−1)∪w(E). Thusthe requirement will be satisfied for all i ≤ k − 1 if and only if we set x i = w i(E) for i ≤ k − 1.Meanwhile the grading k part of (w0(E) + · · ·+ wk−1(E) + xk)∪ w(E) is given by

xk +

k∑

j=1

wk− j(E)∪ w j(E)

so that the requirement for i = k is satisfied if and only if we put xk = wk(E) :=∑k

j=1wk− j(E)∪

w j(E).

This completes the induction; applying this in particular when k = m gives a class w(E) withw(E)∪ w(E) = 1 (since the degree-i part is automatically zero for i > m by the assumption onM).

It is a general fact that a smooth m-dimensional manifold is homotopy equivalent to a cell

complex all of whose cells have dimension at most m, so H i(M ;Z/2Z) = 0 for i > m; thusProposition 7.3 applies to our manifold M .

Corollary 7.4. Suppose that f0 : M → N is a smooth map which is homotopic to some immersion

f : M → N. Then the normal bundle ν f → M must have total Stiefel–Whitney class given by

w(ν f ) = w(T M)∪ f ∗0

w(T N).

Proof. By the Whitney product formula we have

f ∗0

w(T N) = w( f ∗T N) = w(T M)∪ w(ν f ).

Multiplying both sides on the left by w(T M) immediately gives the result.

Now we are taking M , N , and f0 is given, so let us regard w(T N), w(T M), and the mapf ∗0

: H∗(N ;Z/2Z)→ H∗(M ;Z/2Z) as known (in actual examples it may or may not be easy to

compute them). Then the class w(T M)∪ f ∗0

w(T N) is known, and so the normal bundle of anyhypothetical immersion homotopic to f must have its total Stiefel–Whitney classes equal to this


known cohomology class. Moreover the rank of the vector bundle ν f → M is n−m. So if we

can show that there does not exist any rank-(n−m) vector bundle with total Stiefel–Whitneyclass equal to w(T M) ∪ f ∗

0w(T N) then it will follow that there does not exist any immersion

homotopic to f0.One conceivable reason why there might not exist such a vector bundle is that the total

Stiefel–Whitney class of a rank-(n − m) vector bundle is a sum of terms of grading at mostn−m. Thus if, for some i > n−m, the grading-i part of w(T M) ∪ f ∗

0w(T N) is nonzero, then

there is no immersion homotopic to f0. If n ≥ 2m, of course, this issue does not arise, since

then whenever i > n − m we will have i > m and so H i(M ;Z/2Z) = 0. This is consistentwith Whitney’s theorem that when n≥ 2m every smooth map from M to N is homotopic to animmersion. But if n< 2m then we obtain the constraint

w(T M)∪ f ∗

0w(T N)

n−m+1

= · · ·=w(T M)∪ f ∗

0w(T N)

m= 0

which must be satisfied in order for there to exist an immersion homotopic to f0. (Here fora general cohomology class y we denote by (y)i its grading-i part.) So if we compute theseclasses and show that one of them is nonzero then we will have proven that no such immersionexists. We will do just that in some concrete cases later on, but first we explain how one can getsome slightly sharper constraints on embeddings than we have on immersions.

7.1. Embeddings and the normal Euler class. Recall that a smooth map f : M → N is anembedding if it is an immersion and moreover is a homeomorphism to its image f (M). So inthis case f (M) is a submanifold of N ; we may as well assume that M already is a submanifoldof N and that our embedding is just the inclusion of M into N . So since pullback by inclusionis the same as restriction, we have a splitting

T N |M = T M ⊕ νM ,N

where νM ,N is a rank-(n−m) vector bundle over M . The following is an important and standardresult in basic smooth manifold theory:

Theorem 7.5. (Tubular Neighborhood Theorem,[Lee, 10.17]) If M ⊂ N is any submanifold then

there is a neighborhood U ⊂ N of M and a diffeomorphism T : νM ,N → U which maps the zero

section of 0M ⊂ νM ,N to M ⊂ U by the identity.

At the same time that Whitney proved that any m-manifold immerses in to any n-manifold ifn≥ 2m, he also proved that any m-manifold embeds in any n-manifold if n≥ 2m+1 (with imagewhich is closed as a subset). Consistently with this, we will now see that there is an additionalconstraint on the characteristic classes of the manifolds in the case of a closed embedding, goingjust one step beyond those that arise for immersions that we described above.

Theorem 7.6. Let M ⊂ N be a submanifold which is closed as a subset of N. Let m = dim M and

n= dim N. Then there is a class c ∈ Hn−m(N ;Z/2Z) such that c|M = e(νM ,N ) ∈ Hn−m(M ;Z/2Z).

If additionally νM ,N is oriented, with oriented Euler class eZ ∈ Hn−m(M ;Z), then there is cZ ∈

Hn−m(M ;Z) with cZ|M = eZ.


Proof. Consider the following commutative diagram of maps of pairs, where T is as in Theorem7.5, s0 : M → νM ,N is the zero-section and the other maps are inclusions:

(N ,∅)i // (N , N \M)

(M ,∅)

88qqqqqqqqqqq

s0 &&MMMM

MMMM

MM

(νM ,N ,ν0M ,N)

T // (U , U \M)

j

OO

Whereτ ∈ Hn−m(νM ,N ,ν0M ,N

;Z/2Z) is the (Z/2Z)-Thom class, we have by definition e(νM ,N ) =

s∗0τ. Now by excision the map j∗ : H∗(N , N \ M ;Z/2Z) → H∗(U , U \ M ;Z/2Z) is an isomor-

phism (check this—we are using here that M ⊂ N is closed). Of course T induces an isomor-

phism on cohomology since it is a homeomorphism of pairs. So the class c = i∗( j∗)−1(T ∗)−1τ ∈Hn−m(N ;Z/2Z) restricts to M as s∗

0τ= e(νM ,N ).

The argument in the Z-coefficient case when νM ,N is oriented is the same.

Remark 7.7. The proof of Theorem 7.6 canonically associates to any closed m-dimensional sub-manifold M ⊂ N a cohomology class cM ∈ Hn−m(n;Z/2Z); if M and N are, additionally, ori-ented then we obtain a natural orientation on νM ,N and so we get an integer-coefficient class

cMZ

. When N is compact this can be understood in terms of Poincaré duality which gives an

isomorphism between Hm(N ;Z/2Z) and Hn−m(N ;Z/2Z) and, if N is oriented, an isomorphismbetween Hm(N ;Z) and Hn−m(N ;Z): the submanifold M determines a degree-m homology class[M] in N (always over Z/2Z, and if M is oriented also over Z), and the class cM that we haveconstructed turns out to be the image of [M] under the Poincaré duality isomorphism. Notethat cM , as constructed in the proof, lies in the image of the map Hn−m(N , N −M)→ Hn−m(N),and hence vanishes on restriction to N \ M . So cM evaluates as zero on cycles in N which aredisjoint from M ; more generally, the evaluation of cM on an (n−m)-cycle P can be interpretedin terms of the number of intersections of P with M .

This quickly leads to an obstruction to embedding some manifolds in others:

Corollary 7.8. Suppose that the m-dimensional smooth manifold M is a closed submanifold of the

(m+ 1)-dimensional smooth manifold N. Then where i : M → N is the inclusion,

w1(T M) ∈ Im(i∗ : H1(N ;Z/2Z)→ H1(M ;Z/2Z))

Proof. If νM ,N is the normal bundle, which by the assumption on dimensions has rank one, wesee from the Whitney product formula that

w(T N)|M = w(T M)∪ (1+ w1(νM ,N ))

By Theorems 7.6 and 6.24 we can write w1(νM ,N ) = c|M for some c ∈ H1(N ;Z/2Z). So equatingdegree-one terms gives

w1(T N)|M = w1(T M) + c|M ,

i.e. w1(T M) = i∗(w1(T N)− c).

Here is a striking consequence:

Corollary 7.9. Let M be any m-dimensional smooth manifold such that T M is not orientable. If M

embeds as a closed submanifold of some (m+1)-dimensional manifold N, then H1(N ;Z/2Z) 6= 0.

In particular M does not embed as a closed submanifold of Rm+1.


Proof. According to Exercise 6.9, w1(T M) 6= 0. So by Corollary 7.8 the map i∗ : H1(N ;Z/2Z)→H1(M ;Z/2Z) has nontrivial image, and in particular H1(N ;Z/2Z) 6= 0.

A smooth manifold is called orientable if its tangent bundle is orientable; this coincides withany other notion of orientability that you may have learned. In particular Corollary 7.9 appliesto the nonorientable surfaces such as the Klein bottle and the real projective plane, proving

that they do not embed in R3, a famous result that you have probably heard of before. Higher-dimensional examples of nonorientable manifolds include the even-dimensional real projective

spaces RP2k; we will soon see that for some values of k one can prove a much stronger result.Now a nonorientable manifold M does embed as a closed codimension-one submanifold of

some smooth manifolds N , notably N = M × R or N = M × S1. And of course the compact

orientable surfaces embed in R3, and Sm embeds in Rm+1. This begins to illustrate that theproblem of which manifolds embed into which others depends in a somewhat subtle way onthe topology both of the source and of the target, rather than just depending on the dimension.

7.2. The tangent bundle of real projective space. We will now compute the Stiefel–Whitneyclasses of the tangent bundle to RPm. Using the arguments described above, this will quicklyallow us to get some restrictions on embeddings and immersions of RPm into Euclidean space.

Let γ → RPm denote the tautological line bundle. (So γ = γ1(Rm+1) in previous nota-tion.) We thus have w(γ) = 1 + e, where e = e(γ) generates H∗(RPm;Z/2Z) as a unital ring

(more specifically, by Example 5.23 we have H∗(RPm;Z/2Z) =Z/2Z[e]

⟨em+1⟩). We intend to compute

w(TRPm); to do this we will relate TRPm to γ.The starting geometric observation is that, viewing an element ℓ ∈ RPm as a line through the

origin inRm+1, there is a natural identification of the tangent space TℓRPm with the vector space

Hom(ℓ,ℓ⊥) of linear maps from ℓ to its orthogonal complement. To describe this identification,

we will associate to each A ∈ Hom(ℓ,ℓ⊥) a curve cA : R → RPm with cA(0) = ℓ. Then A willcorrespond to the tangent vector c′

A(0) ∈ TℓRPm.

Specifically, we let

cA(t) = ~v + tA~v|~v ∈ ℓ

It is clear that cA(t) is a one-dimensional subspace of Rm+1 for all A and t (since A has image

in ℓ⊥ we never have tA~v = −~v for nonzero ~v ∈ ℓ). Let us study cA in terms of a (conveniently-

chosen) coordinate chart for RPm. Choose an orthonormal basis e1, . . . , em+1 for Rm+1 such

that em+1 generates the given one-dimensional subspace ℓ. Hom(ℓ,ℓ⊥) is then identified with

Rm by associating to A ∈ Hom(ℓ,ℓ⊥) the tuple (A1, . . . , Am) where Aem+1 =∑m

i=1Aiei . One

choice of local coordinate patch around ℓ is given by letting the domain be U = [x1 : . . . :

xm+1]|xm+1 6= 0 and defining φ : U → Rm by φ([x1 : · · · : xm+1]) =

x1

xm+1

,x2

xm+1

, . . . ,xm

xm+1

. We

see then that

φ(c(A1,...,Am)(t)) = φ

[tA1 : · · · : tAm : 1]

= (tA1, . . . , tAm).

So

φ∗(c′(A1,...,Am)

(0)) = (A1, . . . , Am).

This proves both that the curve cA is smooth for each A, and that the map A 7→ c′A(0) gives an

isomorphism from Hom(ℓ,ℓ⊥) to TℓRPm.

Recall that γ= (ℓ, ~v) ∈ RPm ×Rm+1|~v ∈ ℓ. Define likewise

γ⊥ = (ℓ, ~v) ∈ RPm ×Rm+1|~v ∈ ℓ⊥

One can construct local trivializations for γ⊥ in much the same way that they were constructed

for γ (or for a quicker argument, note that there is a homeomorphism RPm→ Grm(Rm+1)which


sends ℓ to ℓ⊥, and γ⊥ is the pullback of γm(Rm+1) by this homeomorphism). Now in generalif V and W are two vector bundles over the same space M it is straightforward to construct avector bundle structure on

Hom(V, W ) = (x , A)|A∈ Hom(Vx , Wx)

(over an open set U on which V |U is identified by a local trivialization with U ×Rk and W |U is

identified with U×Rl , the restriction Hom(V, W )|U is identified with U×Rl,k where Rl,k denotesthe space of l × k real matrices).

Considering the case where V = γ and W = γ⊥, the construction described two paragraphsago gives a map

Φ : Hom(γ,γ⊥)→ TRPm

(ℓ, A) 7→ c′A(0)

and it follows directly from the construction that Φ is a bundle isomorphism. Thus

(25) TRPm ∼= Hom(γ,γ⊥)

This still may not seem like enough to determine the Stiefel–Whitney classes, but after some

further manipulation we will see that it is. For any positive integer k let Rk be the trivial rank-kvector bundle over RPm. Note that

γ⊥ ⊕ γ= (ℓ, ~v) ∈ RPm ×Rm+1|~v ∈ ℓ⊥ ⊕ ℓ= RPm ×Rm+1 = Rm+1.

Moreover it is straightforward to see that, very generally for any vector bundles V, W1, W2 overthe same space, we have Hom(V, W1 ⊕W2)

∼= Hom(V, W1)⊕ Hom(V, W2). So we see that

Hom(γ,γ⊥)⊕ Hom(γ,γ)∼= Hom(γ,Rm+1)

∼= Hom(γ,R)⊕(m+1)(26)

If V is a real vector bundle over a paracompact Hausdorff space, one can choose an orthogonalstructure ⟨·, ·⟩ on V , and then the map which sends a vector ~v ∈ Vx to the linear functional onVx defined by ~w 7→ ⟨~v, ~w⟩ defines an isomorphism from V to Hom(V,R). So (26) gives

Hom(γ,γ⊥)⊕ Hom(γ,γ)∼= γ⊕(m+1)

But the bundle Hom(γ,γ) is trivial: the map which sends the element (ℓ, t) ∈ RPm ×R = R tothe endomorphism of γℓ given by multiplication by t is an isomorphism. Thus we have:

Proposition 7.10.

TRPm ⊕R ∼= γ⊕(m+1)

Consequently

w(TRPm) = (1+ e)m+1 =

m∑

i−0

m+ 1

i

ei

Proof. The first line has just been proven in view of (25). For the second, since w(R) = 1,applications of the Whitney product formula give

w(TRPm) = w(TRPM )∪ w(R) = w(TRPm ⊕R) = w(γ⊕(m+1)) = w(γ)m+1

and so the first equality follows since w(γ) = 1 + e. The second equality follows from the

binomial formula and the fact that em+1 = 0.

Corollary 7.11. We have w(TRPm) = 1 if and only if, for some integer j, m= 2 j − 1.


Proof. For any x ∈ H∗(RPm;Z/2Z) we have (1+ x)2 = 1+ x2 (as we are working in an algebraover Z/2Z, so that 2x = 0). From this it follows by induction on j that

(1+ e)2j

= 1+ e2 j

.

If m= 2 j − 1 we thus have

w(TRPm) = (1+ e)2j

= 1+ e2 j

= 1+ em+1 = 1

(as em+1 = 0), proving the backward implication.

For the forward implication, if m is not of the form 2 j −1 then there is an odd integer r such

that r > 1 and m+ 1= 2 j r for some j ∈ N. Then

w(TRPm) = (1+ e)2j r = (1+ e2 j

)r =

r∑

i=0

r

i

e2 j i

In particular

w2 j (TRPm) = re2 j

= e2 j

6= 0

where the second equality follows from r being odd (as we are working over Z/2Z) and the

reason that e2 j

6= 0 is that the fact that r > 1 implies that 2 j = m+1

r< m+ 1.

We consider the problem of embedding or immersing RPm into Rn for various values of m

and n. Whitney showed in 1944 that any smooth m-manifold may be embedded in R2m and

(for m> 1) immersed into R2m−1. For certain values of m this is best possible:

Theorem 7.12. Let f : RP2 j

→ Rn be an immersion. Then n ≥ 2 j+1 − 1. If f is an embedding

then n≥ 2 j+1.

Proof. We have a splitting f ∗TRn ∼= TRP2 j

⊕ ν f where ν f → RP2 j

has rank n− 2 j . Of course

w( f ∗TRn) = 1 since TRn is trivial. Meanwhile

w(TRP2 j

) = (1+ e)2j+1 = (1+ e)2

j

(1+ e) = (1+ e2 j

)(1+ e)

= 1+ e+ e2 j

So we have

(1+ e+ e2 j

)∪ w(ν f ) = 1

Evaluating grading-2 j terms shows that

e2 j

+ e ∪ w2 j−1(ν f ) + w2 j (ν f ) = 0

So one or both of w2 j−1(ν f ) and w2 j (ν f ) must be nonzero. But since ν f has rank n − 2 j , we

have wi(ν f ) = 0 for i > n− 2 j . So we must have 2 j − 1≤ n− 2 j , i.e. n≥ 2 j+1 − 1.

If f is an embedding then wn−2 j (ν f ) is the restriction of a class in H∗(Rn;Z/2Z) by Theorems

7.6 and 6.24, and hence is zero—thus when f is an embedding wi(ν f ) = 0 for i > n− 2 j − 1.

So in this case 2 j − 1≤ n− 2 j − 1, i.e. n≥ 2 j+1.

There is an interesting connection between the tangent bundle to RPn and algebra. If V is avector space over a field K , a division algebra structure on V is a bilinear map

p : V × V → V

(v, w) 7→ v ·w


with no zero-divisors, i.e., v · w 6= 0 whenever v 6= 0 and w 6= 0. We take K = R and assumeV to be finite-dimensional, so we may as well set V = Rn for some n.

Proposition 7.13. Suppose that p : Rn×Rn→ Rn is a division algebra structure. Then the vector

bundle TRPn−1 ∼= Hom(γ,γ⊥) is trivial.

Proof. Let e1, . . . , en be the standard basis for Rn. The map v 7→ v · en is a linear map from Rn

to Rn with trivial kernel, so it has an inverse, say A. (Thus (Av) · en = v for all v.) Then for allnonzero v ∈ V , (Av) · e1, . . . (Av) · en is a basis for Rn, with last entry equal to v.

For ℓ ∈ RPn−1 let Πℓ denote the orthogonal projection Rn+1→ ℓ⊥. Then if v ∈ ℓ \ ~0 we see

that Πℓ((Av) · e1), . . . ,Πℓ((Av) · en−1) is a basis for ℓ⊥. So for i = 1, . . . , n− 1 and ℓ ∈ RPn−1

we may define a linear map si(ℓ): ℓ→ ℓ⊥ by si(ℓ)(v) = Πℓ((Av) · ei); the maps s1(ℓ), . . . , sn−1(ℓ)

are linearly independent elements of Hom(ℓ,ℓ⊥) (and hence form a basis for it).

We have thus constructed a set of n − 1 sections si : RPn−1 → Hom(γ,γ⊥) (they are easily

checked to be continuous) which are linearly independent at every ℓ ∈ RPn−1. Since the rank of

Hom(γ,γ⊥) is only n− 1, it follows from the discussion above Corollary 6.20 that Hom(γ,γ⊥)is trivial.

Corollary 7.14. If there exists a division algebra structure on Rn then n= 2 j for some j.

Proof. By what have just proven, TRPn−1 would have to be trivial, and hence we would have

w(TRPn−1) = 1. So the conclusion follows from Corollary 7.11.

For n = 1, 2, 4 there are familiar division algebra structures on Rn given by multiplication inthe real numbers, the complex numbers, and the quaternions, respectively. A division algebra

structure on R8, called octonion multiplication, was discovered in the 1840s by Cayley andGraves. It turns out that this is a complete list—it follows from work of Bott and Milnor that for

n /∈ 1, 2, 4, 8 the tangent bundle TRPn−1 is nontrivial (indeed TSn−1, which is the pullback

of TRPn−1 by the covering map, is nontrivial), though when n is a power of two this is notdetected by Stiefel–Whitney classes.

8. ORIENTATIONS AND THE FUNDAMENTAL CLASS

Throughout this section we will work with coefficients in a ring R which is either Z or Z/2Z.Recall from Example 3.21 the following calculation of relative cohomology groups:

H i(Rm,Rm \ ~0; R) =

§R i = m

0 otherwise

In Section 4 we chose once and for all a generator ω ∈ Hm(Rm,Rm \ ~0; R) (of course whenR = Z/2Z ω is just the unique nonzero element, but when R = Z there were two possiblechoices of generator, and we chose one).

Considering instead relative homology, either by using a similar argument to Example 3.21or by using the conclusion of Example 3.21 together with the Universal Coefficient Theorem,one sees that we likewise have

Hi(Rm,Rm \ ~0; R) =

§R i = m

0 otherwise

Moreover the Universal Coefficient Theorem shows that the evaluation map gives an isomor-

phism from Hm(Rm,Rm \ ~0;R) to HomR(Hm(Rm,Rm \ ~0; R), R) ∼= HomR(R, R). In particular

there is a unique oRm ∈ Hm(Rm,Rm \ ~0; R) such that ⟨ω,oRm⟩ = 1, and then this element oRm


generates Hm(Rm,Rm \ ~0;R). (As is essentially indicated at the start of Section 4, oRm is rep-

resented by the singular chain that is obtained by translating the standard m-simplex ∆m ⊂ Rm

slightly so that the origin lies in its interior.)

Now suppose that U , V are open subsets of Rm such that ~0 ∈ U ⊂ V , and let jUV : U → V bethe inclusion. By the Excision Theorem, the induced map on relative homology jUV

∗ : H∗(U , U \

~0;R)→ H∗(V, V \ ~0;R) is an isomorphism.

Applying this with V = Rm shows that, for any open subset U ⊂ Rm with ~0 ∈ U , we have

Hi(U , U \ ~0; R) =

§R i = m

0 otherwise

with Hm(U , U \ ~0;R) generated by the unique element oU such that jURm

∗ oU = oRm .

If now ~0 ⊂ U ⊂ V ⊂ Rm, since we have jURm

= jVRm

jUV , we obtain

jVRm

∗ oV = oRm = jURm

∗ oU = jVRm

∗ jUV∗ oU

and so, since jVRm

∗ is an isomorphism

(27) jUV∗ oU = oV

In fact, the classes oU behave consistently under more general maps than inclusions:

Proposition 8.1. Let U and V be open subsets of Rm with ~0 ∈ U ∩ V , and let f : U → V be a

smooth injective map such that f (~0) = ~0, so that we have an induced map f∗ : H∗(U , U \~0; R)→

H∗(V, V \ 0;R). Assume moreover that the derivative of f at ~0 obeys (d f )~0 ∈ GL+m(R). Then

f∗oU = oV

Proof. Since jVRm

∗ oV = oRm and since jVRm

∗ is injective, by replacing f with jVRm

f we may aswell assume that V = Rm.

For r > 0 let Bm(r) = ~x ∈ Rm|‖~x‖ < r be the ball of radius r around the origin. Choose

r so that Bm(r) ⊂ U , and abbreviate A = (d f )~0. Then Taylor’s Theorem shows that there is aconstant C so that, whenever ‖~x‖ ≤ r, we have

‖ f (~x)− A~x‖ ≤ C‖~x‖2

Consequently (using that A is invertible to obtain a lower bound ‖A~x‖ ≥ c‖~x‖ for some c > 0)we may choose ε > 0 so that

(28) ‖ f (~x)− A~x‖ ≤1

3‖A~x‖ whenever ‖x‖ ≤ 3ε

Now choose a smooth function η: R→ [0, 1] such that η(s) = 0 for s ≤ 2ε and η(s) = 1 fors ≥ 3ε, and define g : U → Rm by

g(~x) = A~x +η(‖~x‖)( f (~x)− A~x)

Thus g(~x) = A~x for ‖x‖ ≤ 2ε and g(~x) = f (~x) for ‖~x‖ ≥ 3ε. Also, if we define

G(t, ~x) = (1− t)g(~x) + t f (~x)

= (1− t)A~x + (1− t)η(‖~x‖)( f (~x)− A~x) + t f (~x)

= A~x +t + (1− t)η(‖~x‖)

( f (~x)− A~x)

then G gives a homotopy from g to f , with G(t, ~x) = f (~x) for ‖~x‖ ≥ 3ε. Further, since 0 ≤

t+(1− t)η(s)≤ 1 for all s ∈ R, t ∈ [0, 1] we see that, for ‖~x‖ ≤ 3ε, ‖G(t, ~x)−A~x‖ ≤ 1

3‖A~x‖ and


so ‖G(t, ~x)‖ ≥ 2

3‖A~x‖. These facts together imply that G(t, ~x) = 0 only when ~x = 0. Thus G

gives a homotopy from g to f through maps of pairs (U , U\~0)→ (Rm,Rm\~0). Consequently

(29) g∗oU = f∗oU

Now since A∈ GL+m(R), which is path-connected, there exists a continuous path γ: [ε, 2ε]→

GL+m(R) such that γ(ε) is the identity and γ(2ε) = A. Define a map H : [ε, 2ε]× U → Rm by

H(δ, ~x) =

γ(δ)~x ‖~x‖ ≤ δγ(‖~x‖)~x δ ≤ ‖~x‖ ≤ 2εg(~x) ‖~x‖ ≥ 2ε

Since when ‖~x‖ = 2ε we have γ(‖~x‖) = A and g(~x) = A~x , H is a continuous map. Clearly

H(δ, ~x) = ~0 only when ~x = 0, since the same statement holds for g(~x) and since each γ(s) isan invertible matrix. So where h(~x) = H(ε, ~x), H is a homotopy from h to g through maps of

pairs (U , U \ ~0)→ (Rm,Rm \ ~0). So

(30) h∗oU = g∗oU

The map h: U → Rm has the property that h(~x) = ~x for ‖~x‖ ≤ ε, i.e., h jBm(ε)U = jBm(ε)Rm

.Thus

h∗ jBm(ε)UoBm(ε) = jBm(ε)Rm

∗ oBm(ε) = oRm

But jBm(ε)U∗ oBm(ε) = oU , so this shows that h∗oU = oRm . Combined with (29) and (30), this

completes the proof (as in the first paragraph we reduced to the case that V = Rm).

Proposition 8.1 will allow us to identify an analogue of the classes oU in the homologyHm(M , M \p;Z) where M is an oriented m-dimensional smooth manifold and p ∈ M . (As wewill see, there is also a version with Z/2Z coefficients, which is easier to construct and does notrequire M to be oriented.)

Recall that, in our conventions, an orientation of a smooth manifold M is an orientation (inthe sense of vector bundles) of its tangent bundle T M . Suppose that M is a smooth orientedm-dimensional manifold. We can then choose a smooth atlas φα : Uα → R

m of coordinatecharts for m—thus the Uα ⊂ M are open and each φα is a homeomorphism to its image (which

is open in Rm) and the transition functions φβ φ−1α : φα(Uα∩Uβ )→ φβ (Uα∩Uβ) are smooth.

For simplicity let us suppose that the Uα are all connected.Now the φα determine local trivializations Φα : T M |Uα → Uα ×R

m: if p ∈ Uα then we have

a natural identification of Tφα(p)Rm with Rm, and in terms of this identification Φα is given by

Φα(v) =p, (φα)∗(v)

for v ∈ Tp M where (φα)∗ : TpM → Tφα(p)R

m ∼= Rm is the derivative

of φα at p. The orientation of T M gives us an orientation of each TpM , and for all p ∈ Uα,

(φα)∗ : TpM → Rm is a linear isomorphism, so it is either orientation-preserving or orientation-

reversing (where we use the standard orientation onRm). We claim that (since Uα is connected)either (φα)∗ is orientation-preserving for all p ∈ Uα or else (φα)∗ is orientation-reversing for allp ∈ Uα. Indeed, around any p ∈ Uα there is some local trivialization Ψ : T M |V → V ×Rm which

is consistent with the orientation on T M . We have a transition functionΨΦ−1α : (Uα∩V )×Rm→

(Uα ∩ V )×Rm, given by (q, ~v) 7→ (q, g(q)~v) where g : Uα ∩ V → GLm(R) is continuous. Then,for any q ∈ Uα∩V , (φα)∗ : Tp M → Rm is orientation-preserving if and only if det(g(q))> 0 and

orientation-reversing if and only if det(g(q))< 0. From this it follows that the subsets of Uα onwhich (φα)∗ is, respectively, orientation-preserving or orientation-reversing are both open. Sosince Uα is connected one of these sets is empty and the other is all of Uα, confirming the claim.

This almost suffices to prove:


Proposition 8.2. Let M be an oriented smooth m-manifold. Then there is an atlas ψα : Uα →Rm for M such that, for each α and each p ∈ Uα, the map (ψα)∗ : TpM → Tψα(p)R

m ∼= Rm is

orientation-preserving (with respect to the given orientation on TpM and the standard orientation

on Rm).

Proof. Start with an atlas φα : Uα → Rm where each Uα is connected, as in the discussion

above the statement of the proposition. Then for each α, either (φα)∗ is orientation preservingat all p ∈ Uα, or else (φα)∗ is orientation-reversing at all p ∈ Uα. In the first case, let ψα = φα.In the second case, choose some m×m matrix B with det B = −1 and define ψα(q) = Bφα(q).Since the composition of two orientation-reversing maps is orientation-preserving, (ψα)∗ willbe orientation-preserving at all p ∈ Uα in either case.

Definition 8.3. Let M be an oriented smooth manifold and let p ∈ M . An oriented chart around

p is a diffeomorphism φ : U → φ(U) where U ⊂ M and φ(U) ⊂ Rm are open subsets such that

φ(p) = ~0 and φ∗ : Tp M → Rm is orientation-preserving.

It follows from Proposition 8.2 that oriented charts around p always exist: given an atlasψα : Uα → R

m as in Proposition 8.2 choose α such that p ∈ Uα and define φ : Uα → φ(U) ⊂Rm by φ(q) =ψα(q)−ψα(p).

Of course, oriented charts are far from unique; however, ifψ: V →ψ(V ) is another oriented

chart around p we can consider the transition functionψφ−1 : φ(U∩V )→ψ(U∩V ). Evidently

we have ψ φ−1(~0) = ~0, and moreover by the chain rule d(ψ φ−1)~0 is the composition

Rmφ−1∗ // Tp M

ψ∗ // Rm

where both maps are orientation-preserving. Thus d(ψ φ−1)~0 is an orientation-preservinglinear operator on Rm, i.e. it belongs to GL+

m(R). This makes Proposition 8.1 relevant.

Theorem 8.4. Let M be a smooth oriented m-manifold and let p ∈ M. Then

Hi(M , M \ p;Z)∼=

§Z i = m

0 otherwise

and there is a generator oM ,p for Hm(M , M \ p;Z) which is uniquely specified by the following

prescription: Choose any oriented chart φ : U → φ(U) ⊂ Rm around p, and let

oM ,p = jU M∗ (φ

−1)∗oφ(U)

where jU M : U → M is the inclusion and oφ(U) ∈ Hm(φ(U),φ(U)\ ~0;Z) is the class described at

the start of this section.

Proof. If φ : U → φ(U) is an oriented chart around p, we have a sequence of maps

H∗(φ(U),φ(U) \ ~0;Z)(φ−1)∗ // H∗(U , U \ p;Z)

jU M∗ // H∗(M , M \ p;Z)

where the first map is an isomorphism becauseφ−1 is a homeomorphism of pairs and the secondmap is an isomorphism by excision. So it follows from our earlier work in Rm that

Hi(M , M \ p;Z)∼=

§Z i = m

0 otherwise

with Hm(M , M \ p;Z) generated by the class

oM ,p(φ, U) := jU M∗ (φ

−1)∗oφ(U).

It remains only to prove that this class is independent of the choice of oriented chart around p.


First we observe that if W ⊂ U is an open subset containing p, then we have

oM ,p(φ|W , W ) = jW M∗ (φ−1|φ(W ))∗oφ(W ) = ( j

U M φ−1 jφ(W )φ(U))∗oφ(W )

= jU M∗ (φ

−1)∗oφ(U) = oM ,p(φ, U),

where we have used (27) to see that jφ(W )φ(U)∗ oφ(W ) = oφ(U).

Now suppose thatφ : U → φ(U) andψ: V →ψ(V ) are two different oriented charts aroundp. We must show that oM ,p(U ,φ) = oM ,p(V,ψ). By the observation that we have just made, wehave

oM ,p(φ, U) = oM ,p(φ|U∩V , U ∩ V ) oM ,p(ψ, V ) = oM ,p(ψ|U∩V , U ∩ V )

But applying Proposition 8.1 to ψ φ−1 : φ(U ∩ V )→ψ(U ∩ V ) gives

oM ,p(ψ|U∩V , U ∩ V ) = jU∩V M∗ ((ψ|U∩V )

−1)∗oψ(U∩V )

= jU∩V∗ ((ψ|U∩V )

−1)∗(ψ|U∩V (φ|U∩V )−1)∗oφ(U∩V )

= jU∩V M∗ ((φ|U∩V )

−1)∗oφ(U∩V ) = oM ,p(φ|U∩V , U ∩ V )

So indeed we have oM ,p(U ,φ) = oM ,p(V,ψ) for any two oriented charts φ : U → φ(U) and

ψ: V →ψ(V ) around p.

As alluded to earlier there is an (easier) analogue to Theorem 8.4 for Z/2Z coefficients,which does not require an orientation for M . (Consistently with earlier terminology, if M isa not-necessarily-oriented smooth manifold and p ∈ M a chart around p is a diffeomorphism

φ : U → φ(U) where U ⊂ M and φ(U) ⊂ Rm are open subsets and φ(p) = ~0.)

Theorem 8.5. Let M be a smooth m-manifold and let p ∈ M. Then

Hi(M , M \ p;Z/2Z)∼=

§Z/2Z i = m

0 otherwise

and there is a generator oM ,p for Hm(M , M \ p;Z) which is uniquely specified by the following

prescription: Choose any chart φ : U → φ(U) ⊂ Rm around p, and let

oM ,p = jU M∗ (φ

−1)∗oφ(U)

where jU M : U → M is the inclusion and oφ(U) ∈ Hm(φ(U),φ(U)\~0;Z/2Z) is the class described

at the start of this section.

Proof. For any chart φ : U → φ(U) around p the composition

jU M∗ φ

−1∗ : H∗(φ(U),φ(U) \ ~0;Z/2Z)→ H∗(M , M \ p;Z/2Z)

is an isomorphism by excision, so the formula for H∗(M , M \ p;Z/2Z) follows from the cal-

culations at the start of the section. Furthermore oM ,p = jU M∗ (φ

−1)∗oφ(U) is the unique nonzero

element of H∗(M , M\p;Z/2Z), so it must be independent of the choice of chartφ : U → φ(U)(since any other chart ψ: V →ψ(V ) would also have jV M

∗ (ψ−1)∗oψ(V ) 6= 0).

Now we consider the effect of replacing the singleton p by a more general compact subsetK ⊂ M .

Where R= Z or Z/2Z, define

H∗(M |K) = H∗(M , M \ K; R)

(this is sometimes called the local homology of M at K). Note that if L ⊂ K ⊂ M with L and K

compact, then the inclusion iK L : (M , M \ K) ⊂ (M , M \ L) induces a map

iK L∗ : H∗(M |K)→ H∗(M |L).


In particular this applies with L = p where p ∈ K (in which case we will write iKp instead of

iKp).

Definition 8.6. Let R = Z/2Z or Z, and let M be a smooth m-manifold, which we assume to beoriented if R= Z. Let K ⊂ M be compact. A fundamental class for M at K with coefficients in R

is a class oM ,K ∈ Hm(M |K) such that for all p ∈ K we have

iKp∗ oM ,K = oM ,p

Of course, for any nonempty K , if oM ,K exists it must be nonzero, since oM ,p 6= 0 for allp ∈ K . Actually, the case in which we are most interested is where M is a compact smoothm-manifold and K = M . Then the local homology H∗(M |M ) of M at M is just the absolutehomology H∗(M ;R), and a fundamental class oM ,M for M at M would give a nonzero element

of Hm(M ;R) (more commonly this element is just denoted [M]). The reason for consideringthe more general situation is that in order to construct the fundamental class we will need topatch together constructions made in various coordinate charts, and the coordinate charts arenoncompact even if M is compact.

Theorem 8.7. Let R= Z/2Z or Z, let M be a smooth m-manifold which is oriented if R= Z. Then

there exists a unique fundamental class oM ,K for M at K. Moreover, for i > m, Hi(M |K) = 0.

Proof. We divide the proof (which is somewhat similar to but more complicated than our earilierproof of Theorem 4.4 on the existence of Thom classes) into a series of eight claims.

Claim 8.8. If L ⊂ K ⊂ M with K and L compact and if oM ,K is a fundamental class for M at K

then iK L∗ oM ,K is a fundamental class for M at L.

Proof of Claim 8.8. For p ∈ L the various inclusions obviously obey iKp = iLp iK L , and therefore

iLp∗ (i

K L∗ oM ,K) = (i

Lp iK L)∗oM ,K = iKp∗ oM ,K = oM ,p,

so iK L∗ oM ,K indeed satisfies the defining property for a fundamental class for M at L.

Claim 8.9. If K1 and K2 are compact subsets of M such that Theorem 8.7 holds for the pairs(M , K1), (M , K2), and (M , K1 ∩ K2), then Theorem 8.7 also holds for (M , K1 ∪ K2).

Proof of Claim 8.9. Just like in Theorem 4.4 we use the Mayer–Vietoris sequence (but now forrelative homology instead of relative cohomology). In view of De Morgan’s laws and the generaldefinition H∗(M |K) = H∗(M , M \ K;R), part of the Mayer–Vietoris sequence reads(31)

Hi+1(M |K1∩K2) // Hi(M |K1∪K2

)(i

K1∪K2,K1∗ ,i

K1∪K2,K2∗ )// Hi(M |K1

)⊕ Hi(M |K2)

iK1,K1∩K2∗ −i

K2,K1∩K2∗ // Hi(M |K1∩K2

)

Suppose first that i > m. Then by assumption the first and third terms in the above se-quence are zero, whence so is the second term. This proves the part of Theorem 4.4 stating thatHi(M |K1∪K2

) = 0 for i > m, so we now turn to the uniqueness and existence of the fundamental

class.Set i = m in the above Mayer–Vietoris sequence. Then in particular the first term vanishes

and so the map

(32) (iK1∪K2,K1

∗ , iK1∪K2,K2

∗ ): Hm(M |K1∪K2)→ Hm(M |K1

)⊕ Hm(M |K2)

is injective. Suppose now that c ∈ Hm(M |K1∪K2) is a fundamental class for M at K1∪K2. Now by

assumption there are unique fundamental classes oM ,K j∈ Hm(M |K j

) for j = 1, 2, so by Claim 1 it

must hold that iK1∪K2,K j

∗ c = oM ,K jfor j = 1, 2. But the injectivity of (32) shows that there can be


at most one class c obeying this property. This proves the uniqueness of any fundamental classfor M at K1 ∪ K2.

As for existence, by assumption there is a unique fundamental class oM ,K1∩K2for M at K1∩K2.

So in view of Claim 8.8 we have iK j ,K1∩K2

∗ oM ,K j= oM ,K1∩K2

for j = 1, 2. Therefore (oM ,K1,oM ,K2

)

lies in the kernel of the last map in the Mayer–Vietoris sequence 31, and hence in the image of

the map (32). So there is a class oM ,K1∪K2∈ Hm(M |K1∪K2

) such that iK1∪K2,K j

∗ oM ,K1∪K2= oM ,K j

for

j = 1, 2. To see that this is indeed the desired fundamental class, just note that if p ∈ K1 ∪ K2

then p ∈ K j for some j ∈ 1, 2 and so

iK1∪K2,p∗ oM ,K1∪K2

= iK j p

∗ iK1∪K2,K j

∗ oM ,K1∪K2= i

K j p

∗ oM ,K j= oM ,p,

as desired.

Claim 8.10. Let K1, . . . , Kr be compact subsets of M such that Theorem 8.7 holds for the pair(M , K j1

∩· · ·∩K jl) for each intersection K j1

∩· · ·∩K jlof a subcollection of the K j . Then Theorem

8.7 also holds for (M , K1 ∪ · · · ∪ Kr).

Proof of Claim 8.10. We use induction on r together with Claim 8.9. For r = 1 the statement istrivial. Assuming the statement for r and supposing that K = K1∪· · ·Kr+1 is a union of compactsets such that the theorem holds for all intersections of subcollections of the K j , let us write

L = K1 ∪ · · · ∪ Kr . The theorem is known to hold for Kr+1 (by assumption), for L (by inductivehypothesis), and also (again by the inductive hypothesis) for L∩Kr+1 = (K1∩Kr+1)∩· · ·∩ (Kr ∩Kr+1). So by Claim 8.9 it holds for K = L ∪ Kr+1 as well.

Claim 8.11. Suppose that K ⊂ U ⊂ M with K compact and U open, so that in particular U isalso a smooth m-manifold containing the compact set K . Then Theorem 8.7 holds for (U , K) ifand only if Theorem 8.7 holds for (M , K).

Proof of Claim 8.11. It follows directly from excision that the inclusion jU M : (U , U \ K) →(M , M \ K) induces an isomorphism on homology, so in particular Hi(U |K) = 0 for i > m ifand only if Hi(M |K) = 0 for i > m.

For p ∈ U , let ψ: V → ψ(V ) be a chart (oriented if R = Z) around p; replacing V by itsintersection with U , if necessary, we may assume that V ⊂ U . We then have, essentially bydefinition,

oU ,p = jV U∗ (ψ

−1)∗oψ(V ) oM ,p = jV M∗ (ψ

−1)∗oψ(V )

Since jV M = jU M jV U it follows directly that jU M∗ oU ,p = oM ,p for all p ∈ U .

In view of the commutative diagram

H∗(U |K)jU M∗ //

iKp∗

H∗(M |K)

iKp∗

H∗(U |p)

jU M∗ // H∗(M |p)

in which the bottom line sends oU ,p to oM ,p for all p ∈ K ⊂ U , it follows directly from the

definition of a fundamental class that oU ,K is a fundamental class for U at K if and only if

jU M∗ oU ,K is a fundamental class for M at K . This obviously proves the forward implication of the

claim, and it also proves the backward implication in view of the fact that the top line of theabove commutative diagram is an isomorphism.


Claim 8.12. It suffices to prove Theorem 8.7 in the special case that M = Rm.

Proof of Claim 8.12. Assume given an arbitrary pair K ⊂ M as in the statement of the theorem.For each p ∈ K let φp : Up → φp(Up) be a coordinate chart (taken from an oriented atlas as in

Proposition 8.2 if R = Z) around p, and choose a neighborhood Vp of p such that Vp ⊂ Up. We

then have an open cover Vpp∈K of K , so for some p1, . . . , pr ∈ K we have K ⊂ Vp1∪ · · · ∪ Vpr

.

Write Ki = Vpi∩ K . Then K = K1 ∪ · · · ∪ Kr , where each Ki is a compact set contained in the

domain of a coordinate chart φpi: Upi→ φpi

(Upi). Of course, each intersection Ki1

∩ · · · ∩ Kilis

also contained in Upi1.

Therefore, in view of Claim 8.10, it suffices to prove Theorem 8.7 in the case where thecompact set K is contained in the domain of a coordinate chart φ : U → φ(U) (which may betaken to be an oriented coordinate chart in the case that R = Z). Furthermore, Claim 8.11shows that in this situation the theorem holds for the pair (M , K) if and only if the theoremholds for (U , K).

It follows directly from the definitions (and a brief diagram chase that is left to the reader)that the theorem holds for (U , K) if and only if it holds for the pair (φ(U),φ(K)), where nowφ(U) is an open subset of Rm. But then another application of Claim 8.11 shows that if thetheorem holds for (Rm,φ(K)) then it also holds for (φ(U),φ(K)). So this indeed reduces us tothe case that M = Rm.

Claim 8.13. Theorem 8.7 holds when M = Rm and K is a rectangular prism, i.e. a (possiblyempty) set of the form (x1, . . . , xm) ∈ R

m|(∀i)(ai ≤ x i ≤ bi) where ai , bi ∈ R are arbitrary.

Proof of Claim 8.13. The case where K is empty is trivial, since then H∗(M |K) = H∗(M , M \M ;R) = 0, and 0 serves as a fundamental class since there are no p ∈ K on which to checkthe required property. So we assume K 6= ∅.

Let B be an open ball which contains K , centered at some point p0 ∈ K . Let φp0: Rm\p0 →

Rm \B be the map which is the identity outside of B and which sends each point x ∈ B \ p0 tothe point on the boundary of B which lies on the ray from p0 to x . (So φp0

is the terminal map

of a deformation retraction of Rm \ p0 to Rn \ B.)For any p ∈ K let hp : Rm → Rm be a homeomorphism which is the identity on Rn \ B and

which, for each point q on the boundary of B, sends the line segment from p0 to q to theline segment from p to q. (So in particular h(p) = q, and we can and do arrange that hp is

differentiable away from the boundary of B and that (dh)p0∈ GL+

m(R)). It then follows from

Proposition 8.1 and the definition of oM ,p in Theorems 8.4,8.5 that

(hp)∗oRm,p0= oRm,p

(strictly speaking to apply Proposition 8.1 we should replace hp by a function which is smooth

eveywhere and not just away from the boundary of B, but since we can find such a map which ishomotopic to hp through maps of pairs (Rm,Rm \p0)→ (R

m,Rm \p) the conclusion applies

to hp as well, as homotopic maps induce the same maps on homology.) Define φp = φp0 h−1

p.

Thus φp retracts Rm \ p to Rm \ B by pushing points outward along line segments from p to

the boundary of B.Regard φp as having range Rm \ K ⊃ Rm \ B. Where iKp : Rm \ K → Rn \ p denotes the

inclusion, iKp φp : Rm \ p → Rm \ p is homotopic to the identity, with a homotopy given

by a family of maps which push points of Rm \ p just part of the way along segments from p

to the boundary of B. Likewise φp iKp : Rm \ K → Rm \ K is homotopic to the identity: since

K is convex, once a line segment from p to the boundary of B exits K , it remains outside ofK (otherwise a line segment from p to another point of K would pass through Rm \ K), so the


homotopy given by progressively pushing points of Rm \ K outward along such line segmentspreserves the set Rm \ K .

Thus all of the maps φp : Rm \ p → Rm \ K are homotopy equivalences, and so induce

isomorphisms H∗(Rm \ p;R)→ H∗(R

m \ K; R). Since Hi(Rm) = 0 for i ≥ 1, the long exact

sequences of the pair show that the boundary maps ∂p : Hi(Rm|p) → Hi−1(R

m \ p; R) and

∂K : Hi(Rm|K)→ Hi−1(R

m\K;R) are isomorphisms for i ≥ 2 and injective for i = 1. In particularfor i > m we have Hi(R

m|K)∼= Hi(R

m|p0) = 0, proving part of the theorem.

For i = m, we can then define (for each p ∈ K) an isomorphism φp : Hm(Rm|p)→ Hm(R

m|K)

by φp(x) = ∂−1K(φp)∗(∂p x).14

Define

oRm,K = φp0oRm,p0

∈ Hm(Rm|K).

Thus

∂KoRm,K = (φp0)∗∂p0

oRm,p0.

Now for any p ∈ K the maps on homology induced by the inclusion iKp : (Rm,Rm \ K) →(Rm,Rm \ p) commute with the maps appearing in the long exact sequences of the pairs(Rm,Rm \ K) and (Rm,Rm \ p), and a similar statement applies to the maps on homology

induced by h−1p

: (Rm,Rm \ p)→ (Rm,Rm \ p0). Therefore

∂p iKp∗ oRm,K = iKp

∗ ∂KoRm,K = iKp∗ (φp0

)∗(∂p0oRm,p0

)

= iKp∗ (φp0

)∗(∂p0h−1

p∗ oRm,p)

= iKp∗ (φp0

h−1p)∗∂poRm,p

= iKp∗ (φp)∗∂poRm,p

= ∂poRm,p

where in the last line we have used that φp and iKp are homotopy inverses. Since ∂p is injective

it follows that iKp∗ oRm,K = oRm,p for all p ∈ K , i.e. that oRm,K is a fundamental class for Rm at K .

As for uniqueness, since for all p ∈ K we have isomorphisms Hm(Rm|K)∼= Hm(R

m|p)∼= R

where R= Z/2Z or Z, we just note that either of our possible rings R has the property that anynontrivial homomorphism R→ R is injective. (So the nontrivial element oM ,p cannot have more

than one preimage under iKp∗ .)

Claim 8.14. Theorem 8.7 holds when M = Rm and K is a union of finitely many rectangularprisms.

Proof. This follows directly from Claims 8.13 and 8.10, together with the fact that an intersectionof finitely many rectangular prisms is a (possibly empty) rectangular prism.

Claim 8.15. Theorem 8.7 holds when M = Rm and K is an arbitrary compact subset of Rm.

Proof. Existence of the fundamental class is easy: let L be a large rectangular prism that containsK , so that we have a fundamental class oRm,L for Rm at L by Claim 8.13. Then by Claim 8.8,

iLK∗ oRm,L is a fundamental class for Rm at K .

For the other statements, the key point is that for any chain in Rm \ K we can find a finiteunion of rectangular prisms L disjoint from the given chain.

14A slightly different formulation is required when m = 1—for instance one can work with reduced homology,

or just observe quite directly that (φp)∗ maps the image of ∂p isomorphically to the image of ∂K and define again

φp = ∂−1K (φp)∗ ∂p with (φp)∗ now regarded as a map Im(∂p)→ Im(∂K ).


More specifically, suppose that x ∈ Hi(M |K). There is then a chain c ∈ Si(Rm; R) which

descends to a relative cycle in Si(Rm,Rm \ K; R) = Si(R

m)/Si(Rm \ K) whose homology class is

x . That c is a relative cycle means that the boundary ∂ c is given by ∂ c =∑

j n jσ j for some

n j ∈ R and σ j : ∆i−1→ Rm \ K .

Since the various (finitely many) Im(σ j) are compact, around any point p ∈ K we can find

a neighborhood Vp whose closure Vp is a rectangular prism which is disjoint from ∪ j Im(σ j).

The collection Vpp∈K then has a finite subcover Vp1, . . . , Vpr

. Let L = Vp1∪ · · · ∪ Vpr

. Then

K ⊂ L and L is a compact subset of Rm which is disjoint from each Im(σ j), and by Claim 8.14,

Theorem 8.7 applies to the pair (Rm, L).Since L is disjoint from each Im(ci), our chain c descends to a relative cycle in Si(R

m,Rm \L;R), representing a class [c] ∈ Hi(M |L). But since c, when considered as a relative cycle in(Rm,Rm \ K), was chosen to be a representative of the given class x , we have iLK

∗ [c] = x .

If i > m, then by Claim 8.14 we have Hi(M |L) = 0, so that [c] = 0 and so x = 0. Since x

was an arbitrary element of Hi(M |K), this completes the proof that Hi(M |K) = 0 for i > m.Finally we prove the uniqueness of the fundamental class of M at K . Suppose that oRm,K ,o′

Rm,K

are two fundamental classes of M at K , and let x = oRm,K − o′Rm,K

. Then x ∈ Hm(M |K) has

the property that iKp∗ x = 0 for all p ∈ K; we will be done if we show that x = 0. Let L

and c be as above, so that in particular iLK∗ [c] = x . Note also that, by the construction of L,

for each point q of L there is a rectangular prism contained in L which contains both q andsome point of K . Now if p ∈ K , then iLp

∗ [c] = iKp∗ iLK∗ [c] = 0. We claim that, more generally,

if q ∈ L then iLq∗ [c] = 0. Indeed, choose p ∈ K and a rectangular prism P ⊂ L such that

p, q ∈ P. Since p ∈ K , iLp∗ [c] = iPp

∗ (iLP∗ [c]) = 0. Also, as noted in the proof of Claim 8.13,

iPp∗ : Hm(R

m|P)→ Hm(Rm|p) and iPq

∗ : Hm(Rm|P)→ Hm(R

m|q) are isomorphisms. So the fact

that iPp∗ (i

LP∗ [c]) = 0 implies that iLq

∗ [c] = iPq∗ (i

LP∗ [c]) = 0.

Thus [c] ∈ Hm(Rm|L) has the property that iLq

∗ [c] = 0 for all q ∈ L. But this implies that,

where oRm,L is a fundamental class for Rm at L, oRm,L +[c] is also a fundamental class for Rm at

L. So by the uniqueness part of Theorem 8.7 applied to (Rm, L), [c] = 0. So since iLK∗ [c] = x ,

x = 0 and oRm,K = o′Rm,K

.

The theorem now follows from Claims 8.12 and 8.15.

8.1. Manifolds with boundary. For any positive integer m we define the left half-space

Hm = (x1, . . . , xm) ∈ Rm|x1 ≤ 0,

endowed with its subspace topology inherited from Rm. Recall that a smooth m-dimensionalmanifold with boundary is a second countable Hausdorff space M equipped with an atlasφα : Uα → φα(Uα) where each φα is a homeomorphism from an open subset Uα ⊂ M toan open subset φα(Uα) ⊂ Hm, where M = ∪αUα, and where the transition functions φβ

φ−1α : φα(Uα ∩ Uβ)→ φβ (Uα ∩ Uβ) are smooth. (Here a function f : U → V where U , V ⊂ Hm

are open subsets is said to be smooth provided that there is an open subset U ′ ⊂ Rm such thatU ⊂ U ′ and a smooth function f ′ : U ′→ Rm such that f ′|U = f .)

Note that, contrary to what one might imagine on grammatical grounds, a manifold withboundary is not a type of manifold. Rather, a manifold is a type of manifold with boundary.

We denote

∂ Hm = (x1, . . . , xm) ∈ Hm|x1 = 0

If M is a smooth m-manifold with boundary (with atlas φα as above) let

∂M = x ∈ M |(∃α)(φα(x) ∈ ∂ Hm)


One can see that a homeomorphism between two open subsets U , V ⊂ Hm maps any pointp ∈ U ∩∂ Hm to ∂ Hm (for instance, look at the local homology H∗(U , U \p; R)). Therefore weequivalently have

∂M = x ∈ M |(∀α)(x ∈ Uα⇒ φα(x) ∈ ∂ Hm).

From this it is easy to see that ∂M is a closed subset of M , and that the maps φα|Uα∩∂M : Uα ∩

∂M → ∂ Hm ∼= Rm−1 comprise an atlas which makes ∂M into a smooth (m − 1) manifold(without boundary). The subset int(M) = M \∂M is open in M and is an m-manifold (withoutboundary).

Smooth maps, diffeomorphisms, and tangent bundles of manifolds with boundary are definedin the obvious way, straightforwardly generalizing the case of manifolds without boundary.

We quote the following standard result:

Theorem 8.16 (Collar neighborhood theorem). [Lee, Theorem 9.25] If M is a smooth manifold

with boundary then there is a diffeomorphism Φ : (−1, 0] × ∂M → U where U ⊂ M is an open

neighborhood of ∂M, such that Φ(0, x) = x for all x ∈ ∂M.

There is another noteworthy manifold without boundary associated to a manifold with bound-ary M , called the double of M and denoted DM . To construct it, choose a collar neighborhoodΦ : (−1, 0] × ∂M → U as described above, and also let M+ denote an additional copy of M ,with the general point x ∈ M corresponding to the point x+ ∈ M+. Now define

DM =int(M)

∐int(M+)

∐(−1, 1)× ∂M

For t ∈ (−1, 0), x ∈ ∂M , (t, x)∼ Φ(t, x)

For t ∈ (0, 1), x ∈ ∂M , (t, x)∼ Φ(−t, x)+

Since int(M), int(M+), and (−1, 1)×∂M are all smooth manifolds and the equivalence relationglues open sets of these smooth manifolds together by diffeomorphisms, DM is a smooth mani-fold. M and M+ both appear as closed subsets of DM , with DM = M ∪M+ and M ∩M+ = ∂M .

One can think of DM visually as follows: imagine that there is a mirror along the boundary∂M of M ; then DM is the union of M together with its image M+ as seen in the mirror.

If M is an oriented manifold with boundary, then one naturally obtains an orientation on ∂M

by requiring a collar neighborhood Φ : (−1, 0]× ∂M → U ⊂ M to be orientation-preserving—this is called the “outer normal first” convention, since it means that ∂M is oriented in such away that, at a point x ∈ ∂M , an oriented basis for Tx M is given by e1, . . . , em where e1 is atangent vector in Tx M that points outward with respect to ∂M and e2, . . . , em is an orientedbasis for Tx∂M .

Still assuming M oriented, we can orient the double DM = M∪M+ by having the orientationon M coincide with the given orientation, while having the orientation on the other copy M+ ofM be opposite to the given orientation on M (corresponding to the fact that the map (t, x) 7→Φ(−t, x)+ by which M+ is glued to (−1, 1) × ∂M is orientation-reversing with respect to thegiven orientation on M).

Now let R= Z/2Z or Z and assume that our smooth m-manifold with boundary M is compact,and if R= Z also assume that M is oriented. Then DM is a compact smooth m-manifold (withoutboundary), and ∂M is a compact smooth (m− 1)-manifold (without boundary), and if R = Z

these manifolds are oriented as well. Hence Theorem 8.7 gives fundamental classes

[DM] := oDM ,DM ∈ Hm(DM ; R) [∂M] := o∂M ,∂M ∈ Hm−1(∂M ; R)

Since a manifold without boundary is not a manifold, Theorem 8.7 does not give us a funda-mental class for M itself, but we remedy that with the following:


Definition 8.17. Let M be a compact smooth m-manifold with boundary, and let R = Z/2Z or,if M is oriented, let R = Z. Then the fundamental class of M , denoted [M], is the element ofHm(M ,∂M ;R) which is given as the image of [DM] ∈ Hm(DM ; R) under the following compo-sition of maps:

Hm(DM ;R) // Hm(DM , M+; R) // Hm(M ,∂M ; R)

where the first map is the usual projection-induced map in the long exact sequence of the pair(DM , M+), and the second map is the inverse of the excision isomorphism.

If q ∈ int(M), then we have an excision isomorphism H∗(int(M), int(M)\q; R)∼= H∗(M , M\q;R), so we can define oM ,q to be the image of oint(M),q under this isomorphism. To partly jus-

tify Definition 8.17, we prove:

Proposition 8.18. If M is as in Definition 8.17 and q ∈ int(M) then where iq : (M ,∂M) →(M , M \ q) is the inclusion we have iq

∗[M] = oM ,q.

(In fact, just like with fundamental classes of compact manifolds without boundary, [M] isthe unique element of Hm(M ,∂M ;R) that has this property for all q ∈ int(M), but we willneither prove nor use this.)

Proof. We have a composition of excision isomorphisms

H∗(int(M), int(M) \ q;R) // H∗(M , M \ q; R) // H∗(DM , DM \ q; R) .

As observed in the second paragraph of the proof of Claim 8.11, this composition sends oint(M),q

to oDM ,q. So the definition of oM ,q is such that oM ,q maps to oDM ,q under the inclusion-induced

map H∗(M , M \ q;R)→ H∗(DM , DM \ q; R) (and is the unique element with this property,since the map is an isomorphism).

Now there is a commutative diagram of inclusion-induced maps

H∗(DM ,∅;R) // H∗(DM , M+; R) // H∗(DM , DM \ q; R)

H∗(M ,∂M ; R)

∼=

OO

iq∗ // H∗(M , M \ q; R)

∼=

OO

where the vertical maps are isomorphisms by excision.Now the composition on the top line sends [DM] to oDM ,q by the defining property of a

fundamental class. Meanwhile [M] ∈ H∗(M ,∂M ; R) is by definition the image of [DM] underthe composition of the first map on the top line and the inverse of the first vertical map. So [M]is sent to oDM ,q by the maps in the commutative square. So the fact that the square commutes

implies that iq∗[M] ∈ H∗(M , M \q;R) is sent by the second vertical map to oM ,q. So by the first

paragraph of the proof we must have iq∗[M] = oM ,q.

The following will have significant consequences for the relationship between characteristicclasses and manifolds with boundary.

Theorem 8.19. In the long exact sequence of the pair (M ,∂M):

Hm(M ;R) // Hm(M ,∂M ; R)∂ // Hm−1(∂M ; R)

j∗ // Hm−1(M ; R)

we have

∂ [M] = [∂M]

Consequently j∗[∂M] = 0 where j : ∂M → M is the inclusion.


(Actually in the case that R = Z we will just prove that ∂ [M] = εm[∂M] for some undeter-mined εm ∈ −1, 1 that depends on the dimension m but not otherwise on M ; this suffices formost applications. The reader can check on the model case of M = Dm where Dm is the closedm-dimensional unit disk that indeed εm = 1.)

Proof. The fact that j∗[M] = 0 clearly follows from the statement that ∂ [M] = [∂M] using theexactness of the long exact sequence of the pair. So we must show that ∂ [M] obeys the propertythat uniquely characterizes the fundamental class [∂M], namely that for all p ∈ ∂M we haveip∗ (∂ [M]) = o∂M ,p where ip

∗ : H∗(∂M ;R) → H∗(∂M ,∂M \ p; R) is induced by the inclusion

(∂M ,∅)→ (∂M ,∂M \ p).Let B be a closed coordinate ball around p in the smooth manifold ∂M . Then the collar

neighborhood theorem identifies the closure C0 of a neighborhood of p in M with the product

I × B where I = [− 1

2, 0] (and where C0 ∩ ∂M is identified with 0 × B). We can slightly

“smooth the corners” of the cylinder C0 to obtain a smooth manifold with boundary C ⊂ M ,diffeomorphic to the closed m-ball, such that ∂ C∩∂M = B. Note that, along B, the orientationsof ∂ C and ∂M are the same. Indeed we may choose an orientation-preserving diffeomorphism

F : Sm−1 → C which maps the north pole N ∈ Sm−1 to p and which maps the open northern

hemisphere UN ⊂ Sm−1 to B. There is then a commutative diagram

Hm−1(UN , UN \ N;R)F∗ //

ι∗

Hm−1(int(B), int(B) \ p; R)

jBC∗

Hm−1(S

m−1, Sm−1 \ N; R)F∗ // Hm−1(∂ C ,∂ C \ p; R)

where the vertical maps are induced by inclusions; in view of the second paragraph of the proof

of Claim 8.11 we have ι∗oUN ,N = oSm−1,N and jBC∗ oint(B),p = o∂ C ,p, and since F is orientation-

preserving we have F∗oUN ,N = oint(B),p and F∗oSm−1,N = o∂ C ,p. Moreover where Dm denotes

the m-dimensional closed unit ball (so ∂ Dm = Sm−1), the map F : Sm−1 → ∂ C extends to an

orientation-preserving diffeomorphism F : Dm → C . If x ∈ int(Dm) with F(x) = q ∈ int(C)

then F∗oDm,x = oC ,q.

If q ∈ int(C) one easily sees that the inclusion-induced map i int(C),q∗ : H∗(M , M \ int(C); R)→

H∗(M , M \ q;R) is an isomorphism (for instance by using excision to reduce to the case thatM = Rm and then using arguments similar to those at the start of the proof of Claim 8.13). Sosince, again for q ∈ int(C), we have iq

∗[M] = oM ,q by Proposition 8.18, it follows by splitting

iq : (M ,∂M)→ (M , M \ q) as a composition

(M ,∂M)i int(C)

// (M , M \ int(C))i int(C),q

// (M , M \ q)

that i int(C),p∗ (i int(C)

∗ [M]) = oM ,q, and in particular that i int(C)∗ [M] generates Hm(M ,\int(C); R)

(which is isomorphic to R). Moreover if x ∈ Dm is such that F(x) = q then in the commutativediagram

H∗(Dm, Sm−1;R)

F∗ //

H∗(M , M \ int(C); R)

H∗(D

m, Dm \ x)F∗ // H∗(M , M \ q; R)


we see that all maps are isomorphisms (by excision in the case of the horizontal maps), and

that F∗oDm,x = oM ,q. Therefore the class z ∈ H∗(Dm, Sm−1; R) which is sent under the inclusion-

induced map H∗(Dm, Sm−1;R)→ H∗(D

m, Dm\x; R) to oM ,x has the property that F∗z = i int(C)∗ [M] ∈

H∗(M , M \ int(C);R). Note that this class z is independent of the manifold M (it also does notdepend on the point x ∈ int(Dm), as can be seen by an argument as in the proof of Claim 8.13).

Now consider the following commutative diagram, where the horizontal maps on the leftare all inclusion-induced and those on the right are induced by the composition of an inclusion

with F : Dm ∼= C or F : Sm−1 ∼= ∂ C , and the first row of vertical maps consists of connectinghomomorphisms in various long exact sequences of pairs (and where we delete the coefficientring R from the notation):

Hm(M ,∂M)i int(C)∗ //

∂

Hm(M , M \ int(C))

∂

Hm(Dm, Sm−1)

∼=oo

∂

Hm−1(∂M) //

ip∗

Hm−1(M \ int(C))

Hm−1(Sm−1)oo

Hm−1(∂M ,∂M \ p)

∼=

α// Hm−1(M , M \ (int(C)∪ p)) Hm−1(S

m−1, Sm−1 \ N)∼=

βoo

(The indicated isomorphisms are either excision isomorphisms or compositions of excision

isomorphisms with the isomorphisms induced by the homeomorphisms F : Sm−1 → ∂ C or

F : Dm→ C .)Since the fixed class z ∈ Hm(D

m, Sm−1) maps to i int(C)∗ [M] ∈ Hm(M , M \ int(C)), the com-

mutativity of the diagram shows that z ∈ Hm(Dm, Sm−1) and [M] ∈ Hm(M ,∂M) have the same

image in the lower center group Hm−1(M , M \ (int(C) ∪ p)). Now both of the vertical mapson the right are isomorphisms if m > 1, and if m = 1 then their composition is still an isomor-

phism. Thus in any case the full composition Hm(Dm, Sm−1) → Hm(M , M \ (int(C) ∪ p))

is an isomorphism. So since z generates Hm(Dm, Sm−1) ∼= R (and hence the image of z in

Hm(M , M \ (int(C) ∪ p)) generates Hm(M , M \ (int(C) ∪ p)) ∼= R), since z and [M] havethe same image in Hm(M , M \ (int(C)∪ p)), and since α is an isomorphism, this proves thatip∗ (∂ [M]) generates Hm−1(∂M ,∂M \ p).

In the case that R= Z/2Z this completes the proof, since then o∂M ,p is the unique generator

of Hm−1(∂M ,∂M \ p), and so we have shown that ip∗ (∂ [M]) is equal to it.

If R= Z, let y denote the image of z in Hm−1(Sm−1, Sm−1 \ N), so y has the same image as

does ip∗ (∂ [M]) in Hm−1(M , M \ (int(C)∪ p)). Since the composition of the two right vertical

maps of the above large commutative diagram is an isomorphism, so that y is a generator, wecan write oSm−1,N = εm y for some εm ∈ −1, 1 depending only on m and not on M . (In factεm = 1 as the reader can check based on our conventions, but we’ll omit the proof.) We have acommutative diagram

Hm−1(∂M ,∂M \ p)

α

++VVVVV

VVVV

VVVV

VVVV

V

Hm−1(UN , UN \ N)

44iiiiiiiiiiiiiiii

**UUUUU

UUUU

UUUU

UUU

Hm−1(M , M \ (int(C)∪ p))

Hm−1(Sm−1, Sm−1 \ p)

β

44hhhhhhhhhhhhhhhhhh


where each map is an isomorphism (given by excision together with a map induced by the

orientation-preserving diffeomorphism F : (Sm−1, UN ) → (∂ C , B)). Since the left maps aboveare obtained by composing inclusions with orientation-preserving diffeomorphisms, they respectthe orientation classes: oUN ,N is mapped respectively to o∂M ,p and oSm−1,N , and hence since the

diagram commutes we have α(o∂M ,p) = β(oSm−1,N ).

Meanwhile we saw earlier that α(ip∗ (∂ [M])) = β(y) = εmβ(oSm−1,N ). So since α is injective

it follows that ip∗ (∂ [M]) = εmo∂M ,p where again εm ∈ −1, 1 depends neither on p nor on M .

So by the uniqueness of [∂M] this proves that εm∂ [M] = [∂M].

9. CHARACTERISTIC CLASSES AND COBORDISM

We now bring the discussion back to vector bundles and Stiefel–Whitney classes. From thecollar neighborhood theorem one can see that if M is a smooth manifold with boundary thenthe restriction T M |∂ M splits as a direct sum:

T M |∂ M∼= R⊕ T (∂M)

where the trivial line bundle R comes from the tangent bundle to the (−1, 0] factor in (−1, 0]×∂M . So where j : ∂M → M is the inclusion (so T M |∂ M = j∗T M), the total Stiefel–Whitneyclasses obey

j∗w(T M) = w( j∗T M) = w(R)∪ w(T (∂M)) = w(T (∂M))

Definition 9.1. Let Y be a compact smooth n-manifold and let I = (i1, . . . , ik) ∈ Nk be a tuple of

natural numbers with i1+ · · ·+ ik = n. Then the Ith Stiefel–Whitney number of Y is the elementof Z/2Z given by

wI(Y ) = ⟨wi1(T Y )∪ · · · ∪ wik

(T Y ), [Y ]⟩

Given a smooth manifold Y , one might ask whether there is a smooth manifold with boundaryM such that ∂M = Y . As stated the answer is obviously yes: let M = [0, 1) × Y . But if Y iscompact and we require M to also be compact then the story is more complicated.

Theorem 9.2. Let Y be a compact smooth manifold such that there is a compact smooth manifold

with boundary M such that ∂M = Y . Then all Stiefel–Whitney numbers wI(Y ) are equal to zero.

Proof. We have seen that for all i we have wi(T Y ) = j∗wi(T M) where j : Y = ∂M → M is theinclusion. So for any I = (i1, . . . , ik) with i1 + · · ·+ ik = T Y we have

⟨wI(Y ) = ⟨wi1(T Y )∪ · · · ∪ wik

(T Y ), [Y ]⟩

= ⟨( j∗wi1(T M))∪ · · · ∪ ( j∗wik

(T M)), [Y ]⟩

= ⟨ j∗(wi1(T M)∪ · · · ∪ wik

(T M)), [Y ]⟩

= ⟨wi1(T M)∪ · · · ∪ wik

(T M), j∗[Y ]⟩

= 0

since by Theorem 8.19 j∗[Y ] = 0.

Example 9.3. For a somewhat silly example, let Y be a compact zero-manifold, i.e. a finitediscrete collection of points. The only Steifel–Whitney number to speak of is w(0)(Y ), and the

fundamental class [Y ] is, as one can easily see, represented by the chain given by the sum of thevarious points of Y . So since T Y , like any vector bundle, has w0(T Y ) = 1, the Stiefel–Whitneynumber w(0)(Y ) is equal to the number of points of Y , reduced modulo 2.


Consistently with this, it is easy to see that a compact zero-manifold occurs as the boundaryof a compact 1-manifold with boundary if and only if the zero-manifold has an even number ofpoints.

Example 9.4. Let Y = RPn. By Proposition 7.10 we have w(T Y ) = (1+e)n+1 where H∗(Y ;Z/2Z) =Z/2Z[e]

⟨en+1⟩. Meanwhile [Y ] is the unique nonzero element of Hn(RPn;Z/2Z), so ⟨en, [Y ]⟩ = 1. So

for I = (i1, . . . , ik) we have

wI(Y ) =

n+ 1

i1

· · ·

n+ 1

ik

mod 2

In particular if n is even, so that

n+1

n

= n+ 1 is odd, then w(n)(RPn) = 1, proving that RPn

does not arise as the boundary of a compact manifold when n is even.On the other hand if n is odd then for any I = (i1, . . . , ik) with i1 + · · · + ik = n, some i j

must be odd. But then since n + 1 is even

n+1

i j

is also even (writing n + 1 = 2r, one has,

modulo 2, (1+ x)n+1 = (1+ x2)r , which has no odd powers of x in its expansion). So for n oddall Stiefel–Whitney numbers of RPn are zero. There are indeed various ways of constructingmanifolds with boundary equal to RPn for any odd n. For instance, writing n = 2m − 1, the

map p : RP2m−1→ CPm−1 that sends a line in R2m to the complex line in Cm which it spans is a

fiber bundle with fiber RP1 ∼= S1, and this bundle can be extended to a D2-bundle E → CPm−1

such that ∂ E = RP2m−1.

Consistent with the above two examples is the following celebrated result of Thom, whichgives a converse to Theorem 9.2.

Theorem 9.5. [T, Théorème IV.10] If Y is a compact smooth manifold such that all Stiefel–

Whitney numbers wI(Y ) are zero then there is a compact smooth manifold with boundary M such

that ∂M = Y .

The proof of this (which together with related work is a large part of the reason that Thomwon a Fields medal in 1958) is beyond the scope of the course. For a partial exposition see [MS,Chapter 18].

Definition 9.6. • If M0 and M1 are two compact smooth m-dimensional manifolds, acobordism between M0 and M1 is a compact smooth (m+1)-dimensional manifold withboundary W such that ∂W is diffeomorphic to the disjoint union M0

∐M1.

• Two compact smooth m-manifolds M0 and M1 are said to be cobordant if there exists acobordism between them (in which case we write M0 ∼ M1).

It should not be difficult for you to check that ∼ is an equivalence relation (the Collar Neigh-borhood Theorem helps with transitivity). If we denote by Nm the set of equivalence classes,then Nm forms an abelian group under the operation of disjoint union: indeed, associativity andcommutativity are obvious, the empty manifold serves as an identity, and since [0, 1]×M canbe seen as giving a cobordism between M

∐M and the empty manifold every element of Nm is

its own inverse. Better still, N∗ = ⊕mNm is a commutative ring under Cartesian product, withthe one-point manifold serving as multiplicative identity.

Corollary 9.7. Two compact m-manifolds M0 and M1 are cobordant if and only if all of their

Stiefel–Whitney numbers are equal.

Proof. This follows directly from Theorems 9.2 and 9.5 together with the general identity

(33) wI

M0

∐M1

= wI(M0) + wI(M1)


Indeed, since wI takes values in Z/2Z, if wI(M0) = wI(M1) then wI(M0

∐M1) = 0, so by

Theorem 9.5 there is a cobordism W between M0

∐M1 and ∅, which can be equivalently seen

as a cobordism between M0 and M1. Conversely if M0 ∼ M1 then Theorem 9.2 shows thatwI(M0

∐M1) = 0 for all I and so wI(M0) = wI(M1) by (33).

To prove (33), let M = M0

∐M1 and for j = 0, 1 let i j : M j → M be the inclusion. Evidently

we have

i∗jT M = T M j .

If k : (M ,∅)→ (M , M\M j) is the inclusion we see that k∗(i j)∗[M j] = k∗(i0)∗[M0] + (i1)∗[M1]

,

and the defining property of M j implies that for p ∈ M j the class k∗(i j)∗[M j] is sent to oM ,p under

the inclusion (M , M \M j)→ (M , M \ p). Since this holds for both j = 0 and j = 1 and since

k∗(i j)∗[M j] = k∗(i0)∗[M0] + (i1)∗[M1]

it follows that (i0)∗[M0]+(i1)∗[M1] obeys the defining

property of the fundamental class. So

(i0)∗[M0] + (i1)∗[M1] = [M].

Thus

wI(M) = ⟨wi1(T M)∪ · · · ∪ wik

(T M), (i0)∗[M0] + (i1)∗[M1]⟩

=i∗0

wi1(T M)∪ · · · ∪ wik

(T M)

, [M0]+i∗0

wi1(T M)∪ · · · ∪ wik

(T M)

, [M1]

= ⟨wi1(T M0)∪ · · · ∪ wik

(T M0), [M0]⟩+ ⟨wi1(T M1)∪ · · · ∪ wik

(T M1), [M1]⟩

= wI(M0) + wI(M1)

Using this together with an easily-proven result describing the behavior of Stiefel–Whitneynumbers under Cartesian products (which we leave to the reader to formulate) and some moresophisticated results about properties of Stiefel–Whitney classes, one can in fact precisely de-termine the whole unoriented cobordism ring N∗, see [MS, Problem 16-F]. N∗ turns out to befreely generated as a (Z/2Z)-algebra by the cobordism classes of certain specific n-manifolds Yn

where n varies through natural numbers such that n+ 1 is not a power of two. Thus any mani-fold at all is cobordant to a disjoint union of products of these standard manifolds. As a specialcase of this, one can show that any compact 3-manifold has all of its Stiefel–Whitney numbersequal to zero—hence every compact 3-manifold is the boundary of some compact 4-manifoldwith boundary.

9.1. Pontryagin classes and oriented cobordism. In the case that M is a compact orientedsmooth m-manifold, we have a fundamental class [M] ∈ Hm(M ;Z), and so if we can associatesome characteristic class a(T M) ∈ Hm(M ;Z) to the tangent bundle T M then we would obtain acharacteristic number ⟨a(T M), [M]⟩ ∈ Z similar to the Stiefel–Whitney numbers but now livingin Z rather than in Z/2Z.

At the moment the only characteristic classes at our disposal that live in integer-coefficientcohomology are the Chern classes, but these are defined only for complex vector bundles. So ifwe happen to have the structure of a rank-n complex vector bundle on the tangent bundle T M

(in which case M is 2n-dimensional as a real manifold), then for a tuple of natural numbersI = (i1, . . . , ik) with i1 + · · ·+ ik = n we can form the Chern number

cI(T M) := ⟨ci1(T M)∪ · · · ∪ cik

(T M), [M]⟩ ∈ Z

If M is a complex manifold (i.e. if M has an atlas of coordinate charts whose transition func-tions are holomorphic diffeomorphisms between open subsets of Cn) then T M naturally has thestructrure of a complex vector bundle; it is also true that a symplectic manifold has a natural


isomorphism class of complex bundle structures on its tangent bundle (namely the ones thatare “compatible” with the symplectic form in a standard sense). So for a complex or symplecticmanifold we have Chern classes and Chern numbers; these in principle depend on the complexor symplectic structure and not just on the smooth structure (or the oriented smooth structure)of the underlying manifold.

But the typical oriented smooth manifold does not have any natural complex structure on itstangent bundle, and so does not have Chern numbers in the above sense. There is however asimple trick that allows us to move into the world of complex vector bundles: we can tensor thetangent bundle with C.

More specifically, for any vector space V over R we may form a vector space over C by taking

VC = V ⊗R C = v1 + iv2|v1, v2 ∈ V

Of course scalar multiplication in this complex vector space is given by

(a+ i b)(v1 + iv2) = (av1 − bv2) + i(bv1 + av2)

(So evidently as a special case we have (Rn)C = Cn.) We will find it somewhat more notationallyconvenient to regard VC as the real vector space V ⊕ V together with the complex scalar multi-plication operation given by combining the given real scalar multiplication on V with followingrule for multiplication by i:

i(v1, v2) = (−v2, v1)

Now if π: E → M is a real rank-n vector bundle we obtain a new complex rank-n vectorbundle EC (the “complexification” of E): the underlying topological space of EC is the Whitneysum E ⊕ E, and on each fiber EC

m= Em ⊕ Em we put the complex vector space structure of the

previous paragraph: i(e1, e2) = (−e2, e1). To each local trivialization

E|U → U ×Rn

e 7→ (π(e),φ(e))

we associate a local complex-vector bundle trivialization for EC given by

EC|U → U ×Cn

(e1, e2) 7→ (π(e1),φ(e1) + iφ(e2))

These trivializations are complex-linear on each fiber, and so EC does indeed have the structureof a rank-n complex vector bundle.

The Pontryagin classes of E will (modulo a sign convention) be certain Chern classes of EC—these exist for any vector bundle over a paracompact Hausdorff space. Before making a formaldefinition we will make some observations about the Chern classes of EC.

To set this up, suppose that π: W → M is a rank-n complex vector bundle, with local trivial-izations of the shape

w 7→ (π(w),φ(w)).

To W we can associate the conjugate bundle W , defined as follows. As a real vector bundle, W

is equal to W . However for w ∈ W =W , the effect of multiplying w by i is opposite: if iW w andiW w denote the result of multiplying w by i when we consider w as an element of W and W

respectively, then iW w = −iW w. Associated to the above local trivializations w 7→ (π(w),φ(w))for W we have local complex vector bundle trivializations

w 7→ (π(w),φ(w))

for W , where of course φ(w) denotes the complex conjugate of φ(w) ∈ Cn


Relatedly, if F : W → CN is a complex pre-classifying map for W , then the complex conjugateF : W → CN is a complex pre-classifying map for W .

Proposition 9.8. If π: W → M is a finite-type rank-n complex vector bundle over a paracompact

Hausdorff space then ci(W ) = (−1)ici(W ) for all i ∈ N.

Proof. As topological spaces, the complex projective bundles PC(W ) and PC(W ) are identical,as are their bundle projections π: PC(W ) = PC(W ) → M . Let F : W → CN be a complex

pre-classifying map for W , so that F : W → CN is a complex pre-classifying map for W . Then

where zW = −PC(F)∗e(γ1(CN )) and zW = −PC(F)

∗e(γ1(CN )), the Chern classes of W and W

determined by the relations c0(W ) = c0(W ) = 1 and

n∑

i=0

π∗ci(W )∪ zn−iW= 0

n∑

i=0

π∗ci(W )∪ zn−i

W= 0

Define b : CPN−1 → CPN−1 by b([x0 : · · · : xN−1]) = [ x0 : · · · : xN−1]. Consider the restric-

tions of b and e(γ1(CN )) to the standard copy of S2 ∼= CP1 ⊂ CPN−1 (namely CP1 = [x0 :

x1 : 0 : · · · : 0]). We see that b restricts to this copy of S2 as a reflection and hence acts with

degree −1. So since the restriction H2(CPN−1;Z)→ H2(CP1;Z) is an isomorphism with image

generated by the restriction of e(γ1(CN )), it follows that b∗e(γ1(CN )) = −e(γ1(CN )).Now we have PC(F) = b PC(F), so it follows from the definitions of zW and zW that zW =

−zW . So the defining relation for the Chern classes of W gives

0=

n∑

i=0

(−1)n−ici(W )∪ zn−iW= (−1)n

n∑

i=0

(−1)ici(W )∪ zn−iW

But these relations are also satisfied with ci(W ) in place of (−1)ici(W ), so it follows that indeed

ci(W ) = (−1)ici(W ).

In particular this shows that typically it is not the case that a complex vector bundle is isomor-phic to its conjugate bundle (rather, one can use a Hermitian metric on the bundle to show thatthe conjugate bundle is isomorphic to the dual Hom(γ,C) of the original vector bundle). How-ever those complex vector bundles that arise as complexifications of real bundles are exceptionsto this:

Proposition 9.9. For any real vector bundle π: E→ M there is an isomorphism of complex vector

bundles EC ∼= EC.

Proof. Viewing the underlying real vector bundle of EC as E⊕E define G : EC→ EC by G(e1, e2) =

(e1,−e2). This is obviously an isomorphism of real vector bundles, so to see that it is an isomor-phism of complex vector bundles we have to see that it intertwines the operations of multipli-cation by i in the domain and range. We have

G(iEC(e1, e2)) = G(−e2, e1) = (−e2,−e1)

and similarly

iEC

G(e1, e2) = iEC(e1,−e2) = −iEC(e1,−e2) = −(e2, e1),

so indeed G(iEC(e1, e2)) = iEC

G(e1, e2).

Corollary 9.10. For any finite-type real vector bundle E over a paracompact Hausdorff space we

have 2ci(EC) = 0 for all odd i.

Proof. For odd i we have ci(EC) = −ci(E

C) = −ci(EC), where the first identity follows from

Proposition 9.8 and the second from Proposition 9.9


So there is not much interesting information to be obtained from the odd Chern classes of

EC (in principle there might be a little bit, since they can be nontrivial elements of H2i(M ;Z)with order two). Accordingly we ignore them and define:

Definition 9.11. If π: E→ M is any finite-type real vector bundle over a paracompact Hausdorffspace M , the Pontryagin classes of E, denoted pi(E), are the classes

pi(E) = (−1)ic2i(EC) ∈ H4i(M ;Z)

The total Pontryagin class is p(E) =∑

i pi(E).

Nothing significant should be read into the factor of (−1)i; it is just a normalization factorwhich makes some other formulas work better.

It is straightforward to see that an isomorphism of vector bundles induces a complex iso-morphism of their complexifications, and that the complexification of the pullback of a vectorbundle is isomorphic as a complex vector bundle to the pullback of the complexification, inview of which the naturality of the Chern classes immediately implies that Pontryagin classesenjoy the naturality property f ∗pi(E) = pi( f

∗E). Obviously p0(E) = 1, and if E has real rank n

(so that EC has complex rank n) we have pi(E) = 0 for i > n/2. There is also something thatserves well enough as a Whitney sum formula, though we have to adjust for the two-torsioninformation that is neglected by the Pontryagin classes:

Proposition 9.12. For two vector bundles E0, E1→ M we have 2p(E0 ⊕ E1) = 2p(E0)∪ p(E1)

Proof. For any k, using the Whitney sum formula for Chern classes and Corollary 9.10, as wellas the obvious isomorphism (E0 ⊕ E1)

C ∼= EC0⊕ EC

1,

pk(E0 ⊕ E1) = (−1)kc2k(EC0⊕ EC

1)

= (−1)k∑

l+m=2k

cl(EC0)∪ cm(E

C1)

= (−1)k∑

i+ j=k

c2i(EC0)∪ c2 j(E

C1) + (elements of order 2)

= (−1)k∑

i+ j=k

(−1)i+ j pi(E0)∪ p j(E1) + (elements of order 2)

and so

2pk(E0 ⊕ E1) = 2∑

i+ j=k

pi(E0)∪ p j(E1).

Since this holds for all k the result follows.

Definition 9.13. Let M be a compact oriented smooth 4n-dimensional manifold and let I =

(i1, . . . , ik) be a tuple of numbers with i1 + · · ·+ ik = n. Then the Ith Pontryagin number of M isthe integer

pI (M) = ⟨pi1(T M)∪ · · · ∪ pik

(T M), [M]⟩

For any oriented manifold M let −M denote the same smooth manifold with the oppositeorientation. It is not hard to check that [−M] = −[M], so since reversing the orientation of M

does not affect the Pontryagin numbers of T M it follows that pI(−M) = −pI (M). In particularif there is a diffeomorphism φ : M → M which reverses the orientation of M then all Pontryaginnumbers of M vanish.

Here is another situation in which the Pontryagin numbers vanish:


Theorem 9.14. Suppose that N is a compact smooth oriented (4n+ 1)-manifold, with ∂ N = M.

Then pI (M) = 0 for all I .

Proof. The proof is of course similar to that of the corresponding statement for Stiefel–Whitneynumbers. The assumption implies that where j : M → N is the inclusion we have an isomor-phism of oriented bundles j∗T N ∼= R⊕T M . So since p(R) = 1, it follows from Proposition 9.12that, for all i, 2pi(T M) = 2 j∗pi(T N). So for I = (i1, . . . , ik),

2kpI(T M) = ⟨(2pi1(T M))∪ · · · ∪ (2pik

(T M)), [M]⟩

= 2k⟨ j∗(pi1(T N)∪ · · · ∪ pik

(T N)), [M]⟩

= 2k⟨pi1(T N)∪ · · · ∪ pik

(T N), j∗[M]⟩

But j∗[M] = 0 by Theorem 8.19, so 2kpI (M) = 0, i.e. (since we are working in Z, in which 2k

is not a zero-divisor) pI (M) = 0.

Two oriented compact smooth manifolds M0 and M1 are said to be oriented-cobordant if thereis an oriented compact smooth manifold W with boundary such that ∂W is diffeomorphic as anoriented manifold to (−M0)

∐M1. This is an equivalence relation, and just as in the unoriented

case the equivalence classes of oriented n-manifolds form an abelian group Ωn under disjointunion. Theorem 9.14 and the same argument as in the proof of Corollary 9.7 show that twooriented-cobordant manifolds necessarily have the same Pontryagin numbers. Thom proved anear-converse to Theorem 9.14 ([T, Corollaire IV.16]), to the effect that if M0 and M1 have thesame Pontryagin numbers then there is a positive integer k such that the disjoint union of k

copies of M0 is cobordant to the disjoint union of k copies of M1.To compute Pontryagin classes pi(E) for some particular bundles E we will first see how they

can be expressed in terms of the Chern classes of E in the special case that E is a already acomplex vector bundle. First we observe:

Proposition 9.15. If E is a complex vector bundle then EC is isomorphic as a complex vector bundle

to E ⊕ E.

Proof. Identifying EC as a real vector bundle with E ⊕ E with the complex structure given asusual by i(e1, e2) = (−e2, e1), we may define G : EC → E ⊕ E by G(e1, e2) = (e1 + ie2, e1 − ie2)

(where i denotes multiplication by i in e, which of course is opposite to multiplication by i inE). This is obviously an isomorphism of real vector bundles, and is easily seen to be complexlinear on each fiber.

Now let us compute the Pontryagin classes of the tangent bundle of the complex manifoldCPn. Arguing in the same way as in the case of RPn one checks that there is an isomorphism

of complex vector bundles TCPn ⊕C ∼= Hom(γ,C)n+1 where γ = γ1(Cn+1) is the tautological

bundle, with c(γ) = 1+ e where H∗(CPn;Z) =Z[e]

⟨en+1⟩. Now a Hermitian metric on γ yields an

isomorphism Hom(γ,C) ∼= γ. So TCPn ⊕C ∼= γ⊕(n+1), whence c(TCPn) = (1− e)n+1. Likewise

TCPn ⊕C ∼= γ⊕(n+1), so c(TCPn) = (1+ e)n+1. Thus by Proposition 9.15,

c(TCPn)C

= c(TCPn)∪ c(TCPn) = (1− e2)n+1.

So

pi(TCPn) = (−1)ic2i

(TCPn)C

=

n+ 1

i

e2i .

Equivalently, the total Pontryagin class of TCPn is p(TCPn) = (1+ e2)n+1.


Corollary 9.16. For each nonnegative integer k, CP2k is not the boundary of any compact oriented

smooth manifold.

Proof. It follows from what we have just done that the top Pontryagin class pk(TCP2k) is equal to

(2k+1)e2k. Since e2k generates H4k(CP2k;Z), it evaluates as ±1 on [CP2k]. Thus p(k)(CP2k) =

±(2k+ 1) 6= 0 and the conclusion follows from Theorem 9.14.

Aside from applications to cobordism, Pontryagin classes have applications to other topolog-ical questions. Here is one:

Theorem 9.17. For all positive integers k the tangent bundle TS4k is not isomorphic to any complex

vector bundle. In particular, S4k is not diffeomorphic to a complex manifold.

Remark 9.18. Of course S2 is diffeomorphic to the complex manifold CP1. It turns out that theonly values of m for which TSm is isomorphic to a complex vector bundle are m = 2 and 6. It

is still an open question whether S6 is diffeomorphic to a complex manifold.

Proof. Write E = TS4k. The Pontryagin numbers of S4k vanish because it is the boundary of the(4k + 1)-ball (or alternatively because it admits an orientation-reversing diffeomorphism). In

particular p(k)(S4k) = 0, and hence pk(E) = 0, i.e. c2k(E

C) = 0. Of course c j(EC) = 0 for all

j > 0 other than j = 2k also, since H2 j(S4k;Z) = 0 for such j. Thus EC has total Chern classc(EC) = 1.

If E were (isomorphic to) a complex vector bundle then we would have EC ∼= E ⊕ E byProposition 9.15 and hence

1= c(E ⊕ E) = c(E)∪ c(E) = (1+ c2k(E))∪ (1+ c2k(E))

= (1+ c2k(E))∪ (1+ c2k(E)) = 1+ 2c2k(E)

where we have used Proposition 9.8 as well as the fact that H j(S4k;Z) = 0 for j /∈ 0, 4k.Thus we would have c2k(E) = 0. But c2k(E) is the top Chern class of E, so by Theorem 6.25 it

is equal to the Euler class of E = TS4k. So if TS4k were isomorphic to a complex vector bundlethen its Euler class would be zero. The proof of the Theorem will thus be complete if we show

that TS4k has nonzero Euler class.It is a general fact that for a compact oriented smooth manifold M the quantity ⟨e(T M), [M]⟩

is equal to the Euler characteristic of M , from which the conclusion follows since S4k has Euler

characteristic 2. But let us prove that e(TS4k) 6= 0 (and more generally that e(TSn) 6= 0 for n

even) independently of this general result.

Again write E = TSn, so we can identify E with (~x , ~v) ∈ Sn×Rn+1|~x ·~v = 0. By stereographicprojection of the second factor onto the sphere, we can identify E with (~x , ~y) ∈ Sn×Sn|~y 6= −~x.Under this identification the complement of the zero-section is given by

E0 = (~x , ~y)|~x 6= ~y 6= −~x.

In other words, where

∆= ~x , ~x) ⊂ Sn × Sn Z = (~x ,−~x) ⊂ Sn × Sn,

we have

(E, E0)∼= (Sn × Sn \ Z , Sn × Sn \ (∆∪ Z))


We have a commutative diagram

Hn(E, E0;Z)π∗ // Hn(E;Z)

Hn(Sn × Sn, Sn × Sn \∆;Z)i∗ //

j∗

OO

Hn(Sn × Sn;Z)

k∗

OO

// Hn(Sn × Sn \∆;Z)

where j∗ is an isomorphism by excision and the bottom row is exact. The desired Euler class

e(E) is the image of the Thom class τ ∈ Hn(E, E0;Z) ∼= Z (which is a generator of that group)under the composition of π∗ with the isomorphism H∗(E;Z) → H∗(Sn;Z) given by restrictionto the zero-section. In particular e(E) 6= 0 if and only if k∗ i∗ 6= 0, which by the exactness ofthe bottom row holds if and only if the restriction of k∗ to the kernel of the inclusion-inducedmap H∗(Sn × Sn;Z)→ H∗(Sn × Sn \∆;Z) is nonzero.

Now k : E → Sn × Sn embeds E as Sn × Sn \ Z . Moreover Sn × Sn \ Z deformation retractsto ∆, while Sn × Sn \∆ deformation retracts to Z . So after composing with the cohomologyisomorphisms induced by these deformation retractions, we see that e(E) 6= 0 if and only ifthere is a class α ∈ H∗(Sn × Sn;Z) such that α|Z = 0 but α|∆ 6= 0. There is indeed such a classwhen n is even: Let π1,π2 : Sn × Sn → Sn be the projections onto the first and second factorsand let ω be a generator for Hn(Sn;Z). Now put α = π∗

1ω + π∗

2ω. Where f : Sn → Z is the

embedding ~x 7→ (~x ,−~x) and A: Sn → Sn is the antipodal map, we see that π1 f = 1Sn whileπ2 f = A, so that f ∗α = ω+ A∗ω = 0 when n is even since in this case A has degree −1. Sosince f is a diffeomorphism to Z this proves that α|Z = 0. Meanwhile where g : Sn→ ∆ is theembedding ~x 7→ (~x , ~x) we see that g∗α= 2ω 6= 0, so that α|∆ 6= 0, completing the proof.

REFERENCES

[BT] R. Bott and L. Tu. Differential forms in algebraic topology. Graduate Texts in Mathematics, 82. Springer-Verlag,

New York, 1982.

[FP] R. Fritsch and R. Piccinini. Cellular structures in topology Cambridge University Press, 1990.

[H] A. Hatcher. Algebraic topology

[Hi] M. Hirsch. Differential topology Graduate Texts in Mathematics, 61. Springer-Verlag, New York, 1976.

[La] S. Lang. Algebra, revised 3rd edition. Graduate Texts in Mathematics, 211. Springer-Verlag, New York, 2002.

[Lee] J. Lee. Introduction to smooth manifolds. Graduate Texts in Mathematics, 218. Springer-Verlag, New York, 2003.

[MS] J. Milnor and J. Stasheff. Characteristic Classes. Annals of Mathematics Studies, Princeton University Press, 1974.

[T] R. Thom. Quelques propriétés globales des variétés différentiables. Comment. Math. Helv. 28, (1954), 17–86.

[U1] M. Usher. Math 8210, Fall 2011 Lecture Notes, part 1.

[U2] M. Usher. Math 8210, Fall 2011 Lecture Notes, part 2.

http://www.math.cornell.edu/~hatcher/AT/ATpage.html

http://www.math.uga.edu/~usher/8210-notes.pdf

http://www.math.uga.edu/~usher/8210-notes2.pdf

fall 2012 math 8230 (vector bundles) lecture notes

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