Introduction to higher homotopy groups and

obstruction theory

Michael Hutchings

February 17, 2011

Abstract

These are some notes to accompany the beginning of a second-semester algebraic topology course. The goal is to introduce homotopygroups and their uses, and at the same time to prepare a bit for thestudy of characteristic classes which will come next. These notes arenot intended to be a comprehensive reference (most of this material iscovered in much greater depth and generality in a number of standardtexts), but rather to give an elementary introduction to selected basicideas.

Contents

0 Prerequisites 2

1 Higher homotopy groups 3

2 The Hurewicz isomorphism theorem 5

3 Dependence of ⇡k on the base point 8

4 Fiber bundles 11

5 Homotopy properties of fiber bundles 14

6 The long exact sequence of a fibration 17

1

7 First example of obstruction theory 21

8 Eilenberg-MacLane spaces 22

9 Whitehead’s theorem 26

10 Orientations of sphere bundles 28

11 The Euler class of an oriented sphere bundle 30

12 Homology with twisted coe�cients 35

0 Prerequisites

The prerequisites for this course are the fundamental group, covering spaces,singular homology, cohomology, cup product, CW complexes, manifolds, andPoincare duality.

Regarding covering spaces, recall the Lifting Criterion: if eX

p! X is a

covering space, if f : (Y, y0) ! (X, x0), and if p(ex0) = x0, then f lifts to a

map ef : (Y, y0) ! ( e

X, ex0) with p �

ef = f if and only if

f⇤⇡1(Y, y0) ⇢ p⇤⇡1( eX, ex0).

Regarding homology, there are many ways to define it. All definitions ofhomology that satisfy the Eilenberg-Steenrod axioms give the same answerfor any reasonable space, e.g. any space homotopy equivalent to a CW com-plex. We will often use cubical singular homology, because this is convenientfor some technical arguments. To define the homology of a space X, weconsider “singular cubes”

� : I

k! X

where I = [0, 1]. The cube I

k has 2k codimension 1 faces

F

0i = {(t1, . . . , tk) | ti = 0},

F

1i = {(t1, . . . , tk) | ti = 1},

for 1 i k, each of which is homeomorphic to I

k�1 in an obvious manner.We now define a boundary operator on formal linear combinations of singular

2

cubes by

@� :=kX

i=1

(�1)i⇣�|F 0

i

� �|F 1i

⌘. (0.1)

This satisfies @2 = 0. We could try to define Ck(X) to be the free Z-modulegenerated by singular k-cubes and Hk(X) = Ker(@)/ Im(@). However this isnot quite right because then the homology of a point would be Z in everynonnegative degree. To fix the definition, we mod out by the subcomplex ofdegenerate cubes that are independent of one of the coordinates on I

k, andthen it satisfies the Eilenberg-Steenrod axioms.

1 Higher homotopy groups

Let X be a topological space with a distinguished point x0. The fundamentalgroup ⇡1(X, x0) has a generalization to homotopy groups ⇡k(X, x0), definedfor each positive integer k.

The definition of ⇡k is very simple. An element of ⇡k(X, x0) is a homotopyclass of maps

f : (Ik, @I

k) �! (X, x0).

Here I = [0, 1]. Equivalently, an element of ⇡k(X, x0) is a homotopy class ofmaps (Sk

, p) ! (X, x0) for some distinguished point p 2 S

k.The group operation is to stack two cubes together and then shrink:

fg(t1, . . . , tk) :=

⇢f(2t1, t2, . . . , tk), t1 1/2,

g(2t1 � 1, t2, . . . , tk), t1 � 1/2.

[Draw picture.] It is an exercise to check that ⇡k(X, x0) is a group, with theidentity element given by the constant map sending I

k to x0.A map � : (X, x0) ! (Y, y0) defines an obvious map

�⇤ : ⇡k(X, x0) �! ⇡k(Y, y0),

f 7�! f � �.

This makes ⇡k a functor from pointed topological spaces to groups. Moreover,�⇤ is clearly invariant under based homotopy of �, so if � is a based homotopyequivalence then �⇤ is an isomorphism. (We will see in §3 how to remove the“based” condition.)

3

We also define ⇡0(X) to be the set of path components of X, althoughthis has no natural group structure.

Here are two nice properties of the higher homotopy groups.

Proposition 1.1. (a) ⇡k(X ⇥ Y, (x0, y0)) = ⇡k(X, x0)⇥ ⇡k(Y, y0).

(b) If k > 1, then ⇡k(X, x0) is abelian.

Proof. (a) Exercise. (b) [Draw picture]

Example 1.2. If k > 1, then ⇡k(S1, p) = 0.

Proof. Recall that there is a covering space R ! S

1. Any map f : S

k! S

1

lifts to a map ef : S

k! R, by the Lifting Criterion, since S

k is simply con-nected. Since R is contractible, e

f is homotopic to a constant map. Projectingthis homotopy to S

1 defines a homotopy of f to a constant map.

Example 1.3. More generally, the same argument shows that if the universalcover of X is contractible, then ⇡k(X, x0) = 0 for all k > 1. For example,this holds if X is a Riemann surface of positive genus. This argument is aspecial case of the long exact sequence in homotopy groups of a fibration,which we will learn about later.

Example 1.4. We will prove shortly that1

⇡k(Sn, x0) '

⇢0, 1 k < n,

Z, k = n.

(1.1)

The higher homotopy groups ⇡k(Sn) for k > n are very complicated,

and despite extensive study are not completely understood, even for n = 2.This is the bad news about higher homotopy groups: despite their simpledefinition, they are generally hard to compute2. However, for any given

1This can also be seen using some di↵erential topology. Choose a point x1 2 Sn withx0 6= x1. Let f : (Sk, p) ! (Sn, x0). By a homotopy we can arrange that f is smooth.Moreover if k < n then we can arrange that x1 /2 f(Sk). Then f maps to Sn

\ {x1} ' Rn,which is contractible, so f is homotopic to a constant map. If k = n then we can arrangethat f is transverse to x1 so that x1 has finitely many inverse images, to each of which isassociated a sign. It can then be shown that the signed count #f�1(x1) 2 Z determinesthe homotopy class of f in ⇡k(Sk, x0). See e.g. Guillemin and Pollack.

2For example, recall that the Seifert-Van Kampen theorem gives an algorithm for com-puting, or at least finding a presentation of, ⇡1(X,x0) whenever X can be cut into simplepieces. The idea is that a loop in X can be split into paths which live in the various pieces.This fails for higher homotopy groups because if one attemps to cut a sphere into pieces,then these pieces might be much more complicated objects.

4

space we can usually compute at least the first few homotopy groups. Andhomotopy groups have important applications, for example to obstructiontheory as we will see below.

2 The Hurewicz isomorphism theorem

To compute some higher homotopy groups, we begin by studying the relationbetween higher homotopy groups and homology. The key ingredient is theHurewicz homomorphism

� : ⇡k(X, x0) �! Hk(X),

defined as follows. Recall that the standard orientation of S

k determines acanonical isomorphism

Hk(Sk) ' Z.

The generator is the fundamental class [Sk] 2 Hk(Sk). If f : (Sk

, p) !(X, x0) represents [f ] 2 ⇡k(X, x0), we define

�[f ] := f⇤[Sk] 2 Hk(X).

Alternatively, if we use cubical singular homology, then a map f : (Ik, @I

k) !(X, x0), regarded as a singular cube, defines a cycle in the homology class�[f ]. By the homotopy invariance of homology, �[f ] is well-defined, i.e.depends only on the homotopy class of f . It is an exercise to check that �is a homomorphism.

Also, it follows immediately from the definition that the Hurewicz map �is natural, in the following sense: If : (X, x0) ! (Y, y0), then the diagram

⇡k(X, x0)�

���! Hk(X)??y ⇤

??y ⇤

⇡k(Y, y0)�

���! Hk(Y )

(2.1)

commutes. That is, � is a natural transformation of functors from ⇡k to Hk.Recall now that if X is path connected, then � induces an isomorphism

⇡1(X, x0)ab ' H1(X).

5

This fact has the following generalization, asserting that if X is also simplyconnected, then the first nontrivial higher homotopy group is isomorphic tothe first nontrivial reduced homology group, and implying equation (1.1) forthe first nontrivial homotopy groups of spheres.

Theorem 2.1 (Hurewicz isomorphism theorem). Let k � 2. Suppose thatX is path connected and that ⇡i(X, x0) = 0 for all i < k. Then the Hurewiczmap induces an isomorphism

⇡k(X, x0) ' Hk(X).

To prove this theorem, we will need the following useful facts about ho-motopy groups. Below, FjI

k denotes the set of j-dimensional faces of thek-cube.

Lemma 2.2. Let (X, x0) be a pointed space.

(a) For any k � 1, if f : (Sk, p) ! (X, x0) is homotopic, without fixing the

base points, to a constant map, then [f ] = 0 2 ⇡k(X, x0).

(b) For any k � 2, let f : @Ik+1! X be a map sending every (k � 1)-

dimensional face to x0. Then

[f ] =X

�2Fk

(Ik+1)

[f |�] 2 ⇡k(X, x0).

Proof. We will just sketch the argument and leave the details as an exercise.(a) Given a homotopy {ft} with f0 = f and f1 constant, one can use the

trajectory of p, namely the path � : I ! X sending t 7! ft(p), to modify thisto a homotopy sending p to x0 at all times. The endpoint of the homotopywill then be the composition of � with a map (Sk

, p) ! (I, 0). Since I iscontractible, we can further homotope this rel basepoints to a constant map.(Warning: More generally, if f, g : (Sk

, p) ! (X, x0) are homotopic withoutfixing the base points, it does not follow that [f ] = [g] 2 ⇡k(X, x0). See §3.)

(b) We use a homotopy to shrink the restrictions of f to the k-dimensionalfaces of I

k+1, so that most of @Ik+1 is mapped to x0. Identifying @Ik+1' S

k,and using the fact that k � 3, we can then move around these k-cubes untilthey are lined up as in the definition of composition in ⇡k.

6

Proof of the Hurewicz isomorphism theorem.3 Using singular homology withcubes, we define a map

: Ck(X) �! ⇡k(X, x0)

as follows. The idea is that a generator of Ck(X) is a cube whose boundarymay map anywhere in X, and we have to modify it, via a chain homotopy,to obtain a cube whose boundary maps to x0. To do so, we define a map

K : Ci(X) �! Ci+1(X) (2.2)

for 0 i k as follows.Since X is path connected, for each 0-cube p 2 X we can choose a path

K(p) from x0 to p. For each 1-cube � : I ! X, there is a map @I

2! X

sending the four faces to x0, �, K(�(0)), and K(�(1)). Since X is simplyconnected, this can be extended to a map K(�) : I

2! X such that

@K(�) = � �K(@�). (2.3)

Continuing by induction on i, if 1 i < k, then for each i-cube � : I

i! X,

we can choose4 an (i+1)-cube K(�) : I

i+1! X which sends the faces to x0,

�, and the faces of K(@�), and therefore satisfies equation (2.3). Finally, if� : I

k! X is a k-cube, then there is a map @Ik+1

! X sending the faces tox0, �, and the faces of K(@�). Let F denote the face sent to x0. Identifying(@Ik+1

/F, F ) ' (Ik, @I

k), this gives an element (�) 2 ⇡k(X, x0). Moreover,it is easy to see that the sum of the faces other than F is homologous to (�) regarded as a cube, i.e. there is a cube K(�) with

@K(�) = � �K@� � �( (�)). (2.4)

While may depend on the above choices, we claim that induces a mapon homology which is inverse to �. To start, we claim that if � : I

k+1! X

is a (k + 1)-cube, then (@�) = 0. (2.5)

3Later in the course we will see a shorter and slicker proof using spectral sequences,but which I think ultimately has the same underlying geometric content.

4It is not really necessary to make infinitely many choices in this proof, but the ar-gument is less awkward this way. Incidentally, when � is degenerate, we can and shouldchoose K(�) to be degenerate as well, so that we have a well-defined map (2.2) on thecubical singular chain complex.

7

To see this, first note that by Lemma 2.2(b), we have (@�) = [f ], where f :@I

k+1! X sends each k-dimensional face of I

k+1 to of the correspondingface of �. By cancelling stu↵ along adjacent faces, f is homotopic, withoutfixing base points, to �|@Ik+1 . This is homotopic to a constant map since itextends over I

k+1. So by Lemma 2.2(a) we conclude that [f ] = 0. It followsfrom (2.5) that induces a map ⇤ : Hk(X) ! ⇡k(X, x0).

It follows immediately from (2.4) that � � = idHk

(X).Also, we can make the choices in the definition of K so that:

(*) If i < k and if � : I

i! X is a constant map to x0, then K(�) is also a

constant map to x0.

It is then easy to see that ⇤ � � = id⇡k

(X,x0).

3 Dependence of ⇡k

on the base point

We will now show that if X is path connected, then the homotopy groupsof X for di↵erent choices of base point are isomorphic, although not alwaysequal. This is worth understanding properly, since analogous structures ariselater in the course and also are common elsewhere in mathematics.

If � : [0, 1] ! X is a path from x0 to x1, we define a map

�� : ⇡k(X, x1) �! ⇡k(X, x0)

as follows. Let us temporarily reparametrize the k-cube as I

k = [�1, 1]k. Iff : [�1, 1]k ! X and t = (t1, . . . , tk) 2 I

k, let m = max{|ti|} and define

��(f)(t) :=

⇢f(2t), m 1/2,

�(2(1�m)), m � 1/2.

[Draw picture.] Clearly this gives a well-defined function on homotopy groups.It is an exercise to show that �� is a group homomorphism.

The following additional facts are easy to see:

(i) If � is homotopic to �0 (rel endpoints), then

�� = ��0 : ⇡k(X, x1) �! ⇡k(X, x0).

(ii) Suppose that �1 is a path from x0 to x1 and �2 is a path from x1 to x2,and let �2�1 denote the composite path. Then

��2�1 = ��2��1 : ⇡k(X, x2) �! ⇡k(X, x0).

8

(iii) If � is the constant path at x0, then

�� = id⇡k

(X,x0) .

The above three properties5 imply:

Proposition 3.1. If X is path connected, then

⇡k(X, x0) ' ⇡k(X, x1) (3.1)

for any two points x0, x1 2 X. Moreover, if X is simply connected, then thisisomorphism is canonical, and so ⇡k(X) is a well-defined group without thechoice of a base point.

If X is not simply connected, then the isomorphism (3.1) might not becanonical. In particular, for a noncontractible loop � based at x0, the iso-morphism �� of ⇡k(X, x0) might not be the identity. In general, by (i)–(iii)above, these isomorphisms define an action of ⇡1(X, x0) on ⇡k(X, x0). Whenk = 1, this action is just conjugation in ⇡1(X, x0), which of course is nontriv-ial whenever ⇡1(X, x0) is not abelian. If the action of ⇡1(X, x0) on ⇡k(X, x0)is nontrivial, then ⇡k(X) is not a well-defined group without the choice ofa base point. All one can say in this case is that ⇡k(X) is a well-definedisomorphism class of groups. See also Exercise 3.5 below.

We now construct, for arbitrary k > 1, examples of spaces in which theaction of ⇡1 on ⇡k is nontrivial. Let X be any space and let f : X ! X bea homeomorphism. The mapping torus of f is the quotient space

Yf :=X ⇥ [0, 1]

(x, 1) ⇠ (f(x), 0).

For example, if X = [�1, 1] and f is multiplication by �1, then Yf is theMobius band. If f is the identity, then Yf = X ⇥ S

1.There is a natural inclusion ı : X ! Yf sending x 7! (x, 0). Pick a base

point x0 2 X, and let y0 = ı(x0).

Lemma 3.2. For k � 2, the inclusion induces an isomorphism

ı⇤ : ⇡k(X, x0)'�! ⇡k(Yf , y0).

5A fancy way of describing these three properties is that ⇡k(X, ·) is a functor fromthe fundamental groupoid of X to groups. Alternatively, ⇡k(X·) is a “twisted coe�cientsystem” on X, see §12.

9

Proof. The mapping torus can equivalently be defined as

Yf =X ⇥ R

(x, t + 1) ⇠ (f(x), t).

Hence X⇥R is a covering space of Yf . By the Lifting Criterion, since k � 2,a map (Sk

, p) ! (Yf , y0) lifts to a map (Sk, p) ! (X ⇥ R, (x0, 0)). Since R

is contractible, we can find a homotopy from this map to a map (Sk, p) !

(X⇥{0}, (x0, 0). Projecting this homotopy back down to Yf shows that ı⇤ issurjective. Applying the same argument to homotopies S

k⇥ I ! Yf shows

that ı⇤ is injective. (Compare Example 1.2, and the long exact sequence ofa fibration in §6.)

Now suppose that x0 is a fixed point of f . This defines a loop � in Yf basedat x0, sending t 7! (x0, t). It is an exercise to show that the correspondingisomorphism of ⇡k(Yf , y0) is given by (the inverse of) f⇤, i.e.:

Proposition 3.3. The following diagram commutes:

⇡k(X, x0)f�1⇤

���! ⇡k(X, x0)

ı⇤

??y ı⇤

??y

⇡k(Y, y0)�

�

���! ⇡k(Y, y0).

Example 3.4. To get an explicit example where �� is nontrivial, let X = S

k,and let f : S

k! S

k be a degree �1 homomorphism with a fixed point x0. Bythe naturality of the Hurewicz isomorphism (2.1), f⇤ = �1 on ⇡k(S

k) = Z.

Exercise 3.5. If X is path connected, then there is a canonical bijection

[Sk, X] = ⇡k(X, x0)/⇡1(X, x0).

Here the left hand side denotes the space of free (no basepoints) homotopyclasses of maps from S

k to X, and the right hand side denotes the quotientof ⇡k(X, x0) by the action of ⇡1(X, x0).

Exercise 3.6. Show that if � : X ! Y is a homotopy equivalence then�⇤ : ⇡k(X, x0) ! ⇡k(Y,�(x0)) is an isomorphism. Hint: First show that if{�t}t2[0,1] is a homotopy of maps from X to Y , and if � : [0, 1] ! Y is thepath defined by �(t) = �t(x0), then

(�0)⇤ = �� � (�1)⇤ : ⇡k(X, x0) ! ⇡k(Y,�0(x0)).

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4 Fiber bundles

A fiber bundle is a kind of “family” of topological spaces. These are impor-tant objects of study in topology and in this course, and also will help uscompute homotopy groups.

Definition 4.1. A fiber bundle is a map of topological spaces ⇡ : E ! B

such that there exists a topological space F with the following property:

(Local triviality) For all x 2 B, there is a neighborhood U of x in B, and ahomeomorphism ⇡

�1(U) ' U ⇥ F , such that the diagram

⇡

�1(U)'

���! U ⇥ F

⇡

??y??y

U

=���! U

commutes, where the map U ⇥ F ! U is projection on to the firstfactor.

In particular, for each x 2 B, we have ⇡�1(x) ' F . Thus the fiber bundlecan be thought of as a family of topological spaces, each homeomorphic toF , and parametrized by B. One calls F the fiber , E the total space, and B

the base space. A fiber bundle with fiber F and base B is an “F -bundle overB”. Also, ⇡�1(x) is the fiber over x, denoted by Ex. We denote the fiberbundle by the diagram

F ���! E

??y

B,

or sometimes just by E.

Definition 4.2. If ⇡1 : E1 ! B and ⇡2 : E2 ! B are fiber bundles over thesame base space, then a morphism6 of fiber bundles from ⇡1 to ⇡2 is a mapf : E1 ! E2 such that ⇡2 � f = ⇡1.

Example 4.3. For any B and F , we have the trivial bundle E = B ⇥ F ,with ⇡(b, f) = b.

6Di↵erent definitions are possible. A less restrictive definition allows the base spacesto be di↵erent. A more restrictive definition requires the map to be a homeomorphism oneach fiber.

11

The local triviality condition says that any fiber is locally isomorphicto a trivial bundle. Also, we will see in Corollary 5.8 below that any fiberbundle over a contractible CW complex is trivial. However, when B has somenontrivial topology, fiber bundles over B can have some global “twisting”.

Example 4.4. A covering space is a fiber bundle in which the fiber F hasthe discrete topology. In this case an isomorphism of fiber bundles is thesame as an isomorphism of covering spaces.

Example 4.5. Let f : X ! X be a homeomorphism. Then the mappingtorus Yf is the total space of a fiber bundle

X ���! Yf??y⇡

S

1

where ⇡(x, t) = t mod 1. Exercise: every fiber bundle over S

1 arises in thisway.

The previous example has the following generalization, called the “clutch-ing construction”. Let Homeo(F ) denote the space of homeomorphisms7 ofF , let k > 1, and let � : S

k�1! Homeo(F ). Identify

S

k = D

k[Sk�1 D

k,

and define a fiber bundle over S

k by

E = (Dk⇥ F ) [Sk�1⇥F (Dk

⇥ F ),

where the two copies of D

k⇥ F are glued together along S

k�1⇥ F by the

map

S

k�1⇥ F �! S

k�1⇥ F,

(x, f) 7�! (x,�(x)(f)).

Note that the fiber bundle depends only on the homotopy class of �.Conversely, it follows from Corollary 5.8, applied to B = D

k, that everyfiber bundle over S

k arises this way.

7We topologize the space Maps(X,Y ) of continuous maps from X to Y using thecompact-open topology. For details on this see e.g. Hatcher. A key property of thistopology is that if Y is locally compact, then a map X ! Maps(Y,Z) is continuous i↵ thecorresponding map X ⇥ Y ! Z is continuous.

12

Example 4.6. S

1-bundles over S

2 are classified by nonnegative integers8.To see this, first observe that Homeo(S1) deformation retracts9onto O(2).Now given an S

1-bundle over S

2, one can arrange that � maps to SO(2), bychanging the identification of one of the halves ⇡�1(D2) with D

2⇥ S

1, viaidD2 cross an orientation-reversing homeomorphism of S

1. The nonnegativeinteger associated to the bundle is then | deg(�)|; we will temporarily callthis the “rotation number”10. Note that switching the identification of bothhalves will change the sign of deg(�).

Exercise 4.7. More generally, show that the clutching construction definesa bijection

{F -bundles over S

k}

isomorphism=⇡k�1(Homeo(F ), idF )

Homeo(F )

where Homeo(F ) acts on ⇡k�1(Homeo(F ), idF ) by conjugation in Homeo(F ).

Example 4.8. The Hopf fibration

S

1���! S

3

??y

S

2

is given by the map

S

3! C2

\ {0}! CP

1' S

2.

Exercise 4.9. The Hopf fibration has “rotation number” 1. The unit tangentbundle of S

2 has “rotation number” 2.

Definition 4.10. A section of a fiber bundle ⇡ : E ! B is a map s : B ! E

such that ⇡ � s = idB.

Intuitively, a section is a continuous choice of a point in each fiber.

8It is more common to consider oriented S1-bundles over S2, which are classified byintegers. We will discuss these later.

9This is a good exercise. More generally, one can ask whether Homeo(Sk) deformationretracts to O(k + 1). This is known to be true for k = 2 and k = 3, the latter result beinga consequence of the “Smale conjecture” proved by Hatcher (asserting that the inclusionSO(4) ! Di↵(S3) is a homotopy equivalence) and a result of Cerf. It is false for all k � 4.

10This is really the absolute value of the Euler class for a choice of orientation of thebundle; we will discuss this later.

13

Example 4.11. A section of a trivial bundle E = B ⇥ F is just a maps : B ! F .

Nontrivial bundles may or may not posess sections. We will study thisquestion systematically a little later. For now you can try to convince yourselfthat an S

1-bundle over S

2 with nonzero rotation number does not have asection.

5 Homotopy properties of fiber bundles

We will now show, roughly, that the fiber bundles over a CW complex B

depend only on the homotopy type of B. We will use the following termi-nology: if E is a fiber bundle over B, and if U ⇢ B, then a “trivializationof E over U” is a homeomorphism � : ⇡�1(U) ' U ⇥ F commuting with theprojections.

Lemma 5.1. Any fiber bundle E over B = I

k is trivial.

Proof. By the local triviality condition and compactness of I

k, we can sub-divide I

k into N

k subcubes of size 1/N , such that E is trivial over eachsubcube. We now construct a trivialization of E over the whole cube, onesubcube at a time. By an induction argument which we leave as an exercise,it is enough to prove the following lemma.

Lemma 5.2. Let X = I

k and let A ⇢ I

k be the union of all but one of the(k� 1)-dimensional faces of I

k. Suppose that E is a trivial fiber bundle overX. Then a trivialization of E over A extends to a trivialization of E overX.

Proof. We can regard E = X⇥F . The trivialization of E over A can then beregarded as a map � : A ! Homeo(F ), and we have to extend this to a map� : X ! Homeo(F ). Such an extension exists because there is a retractionr : X ! A, so that we can define � = � � r.

Before continuing, we need to introduce the notion of pullback of a fiberbundle ⇡ : E ! B by a map f : B

0! B. The pullback f

⇤E is a fiber bundle

14

over B

0 defined by11

f

⇤E := {(x, y) | x 2 B

0, y 2 E, f(x) = ⇡(y)} ⇢ B

0⇥ E.

The map f

⇤E ! B

0 is induced by the projection B

0⇥E ! B

0. The definitionimplies that the fiber of f

⇤E over x 2 B

0 is equal to the fiber of E overf(x) 2 B.

Example 5.3. In the special case when B

0 is a subset of B and f is theinclusion map, f

⇤E = ⇡

�1(B0) is called the restriction of E to B

0 and isdenoted by E|B0 . In particular, if f = idB, then f

⇤E = E.

Example 5.4. If f is a constant map to x 2 B, then f

⇤E is the trivial

bundlef

⇤E = B

0⇥ Ex.

Example 5.5. If E is the mapping torus of � : X ! X, and if f : S

1! S

1

has degree d, then f

⇤E is the mapping torus of �d.

Similarly, we will see that if E is an S

1-bundle over S

2 with rotationnumber r, and if f : S

2! S

2 has degree d, then f

⇤E is an S

1-bundle overS

2 with rotation number |dr|.

Lemma 5.6. Let B be a CW complex and let E be a fiber bundle overB ⇥ [0, 1]. Then

E|B⇥{0} ' E|B⇥{1},

where both sides are regarded as fiber bundles over B.

Proof. Let f : B ⇥ [0, 1] ! B be the projection. We will construct anisomorphism

f

⇤(E|B⇥{0}) ' E. (5.1)

Restricting this isomorphism to B ⇥ {1} then implies the lemma.We define the isomorphism (5.1) to be the identity over B ⇥ {0}. We

now extend this isomorphism over B ⇥ [0, 1], one cell (in B) at a time, by

11That is, f⇤E is the fiber product of f and ⇡ over B0. There is a commutative diagram

f⇤E ����! E??y

??y⇡

B0 f����! B.

15

induction on the dimension. (This is legitimate because a map from a CWcomplex is continuous if and only if it is continuous on each closed cell.) Thismeans that without loss of generality, B = D

k, and the isomorphism (5.1) hasalready been defined over D

k⇥ {0} and @Dk

⇥ [0, 1]. By Lemma 5.1, E andf

⇤(E|B⇥{0}) are trivial when B = D

k⇥[0, 1], so after choosing a trivialization,

the desired isomorphism (5.1) is equivalent to a map D

k⇥[0, 1] ! Homeo(F ).

As in the proof of Lemma 5.2, such a map from D

k⇥ {0} [ @Dk

⇥ [0, 1] canbe extended over D

k⇥ [0, 1].

Proposition 5.7. Let B

0 be a CW complex, let E be a fiber bundle over B,and let f0, f1 : B

0! B be homotopic maps. Then

f

⇤0 E ' f

⇤1 E

as fiber bundles over B

0.

Proof. The homotopy can be described by map

f : B

0⇥ [0, 1] ! B

with f |B⇥{i} = fi. Then

f

⇤0 E = (f ⇤E)|B⇥{0} ' (f ⇤E)|B⇥{1} = f

⇤1 E,

where the middle isomorphism holds by Lemma 5.6.

Corollary 5.8. Any fiber bundle over a contractible CW complex is trivial.

Proof. Let B be a contractible CW complex. Contractibility means thatthere is a point x 2 B and a homotopy between idB and the constant mapf : B ! {x}. So by Proposition 5.7, if E is a fiber bundle over B, then

E = id⇤B E ' f

⇤E = B ⇥ Ex,

which is a trivial bundle.

More generally, a homotopy equivalence between B1 and B2 induces abijection between isomorphism classes of F -bundles over B1 and B2. (Wewill see a nicer way to understand this when we discuss classifying spaceslater in the course.)

16

Remark 5.9. Usually one considers fiber bundles with some additionalstructure. For example, a vector bundle is a fiber bundle in which eachfiber has the structure of a vector space (and the local trivializations can bechosen to be fiberwise vector space isomorphisms). A key example is the tan-gent bundle of a smooth n-dimensional manifold M ; this is a vector bundleover M whose fiber over each point is an n-dimensional real vector space.Vector bundles will be a major object of study later in the course.

6 The long exact sequence of a fibration

The following is a useful tool for computing some homotopy groups.

Theorem 6.1. Let F ! E ! B be a fiber bundle. Pick y0 2 E, let x0 =⇡(y0) 2 B, and identify F = ⇡

�1(x0). Then there is a long exact sequence

· · ·! ⇡k+1(B, x0) ! ⇡k(F, y0) ! ⇡k(E, y0) ! ⇡k(B, x0) ! ⇡k�1(F, y0) ! · · ·

(6.1)

The exact sequence terminates12 at ⇡1(B, x0). The arrows ⇡k(F, y0) !⇡k(E, y0) and ⇡k(E, y0) ! ⇡k(B, x0) are induced by the maps F ! E andE ! B in the fiber bundle. The definition of the connecting homomorphism⇡k(B, x0) ! ⇡k�1(F, y0) and the proof of exactness will be given later. Letus first do some computations with this exact sequence.

Example 6.2. If E ! B is a covering space, then it follows from the exactsequence that ⇡k(E, y0) ' ⇡k(B, x0) for all k � 2. This generalizes Exam-ple 1.3.

Example 6.3. Applying the exact sequence to the Hopf fibration S

1!

S

3! S

2, we get an exact sequence

⇡3(S1) ! ⇡3(S

3) ! ⇡3(S2) ! ⇡2(S

1).

We know that this is 0 ! Z ! ⇡3(S2) ! 0, and therefore

⇡3(S2) ' Z. (6.2)

Furthermore, the Hopf fibration S

3! S

2 is a generator of this group.

12One can also continue the sequence with maps of sets ⇡1(B, x0) ! ⇡0(F ) ! ⇡0(E)such that the “kernel” of each map, interpreted as the inverse image of the componentcontaining y0, is the image of the previous map.

17

Remark 6.4. There is a nice explicit description of the isomorphism (6.2).Any homotopy class in ⇡3(S

2) can be represented by a smooth map f : S

3!

S

2. By Sard’s theorem, for a generic point p 2 S

2, the map f is transverse top, so that the inverse image f

�1(p) is a smooth link in S

2. This link inheritsan orientation from the orientations of S

3 and S

2. Let q be another genericpoint in S

2, and define the Hopf invariant of f to be the linking number13

of f

�1(p) and f

�1(q):

H(f) := `(f�1(p), f�1(q)) 2 Z.

The integer H is a homotopy invariant of f and defines the isomorphism(6.2). This is a special case of the Thom-Pontrjagin construction, which wemay discuss later in the course.

The long exact sequence can be used to compute a few homotopy groupsof Lie groups, by the use of suitable fiber bundles. For example, let us tryto compute some homotopy groups of U(n). There is a map U(n) ! S

2n�1

which sends A 7! Av, where v is some fixed unit vector.

Exercise 6.5. This gives a fiber bundle U(n� 1) ! U(n) ! S

2n�1.

We know that U(1) ' S

1. The n = 2 case is then a fiber bundle

S

1! U(2) ! S

3.

So by the long exact sequence, ⇡1U(2) = Z, ⇡2U(2) = 0, and ⇡3U(2) = Z. Ingeneral, while the long exact sequence of this fiber bundle does not allow usto compute ⇡k(U(n)) in all cases (since we do not know all of the homotopygroups of spheres), it does show that ⇡k(U(n)) is independent of n when n

is su�ciently large with respect to k. This group is denoted by ⇡k(U). Infact, Bott Periodicity, which we may discuss later in the course, asserts that⇡k(U) is isomorphic to Z when k is odd and 0 when k is even.

We now construct the long exact sequence. It works in a more generalcontext than that of fiber bundles, which we now define.

13Recall that the linking number of two disjoint oriented links L1 and L2 in S3 isthe intersection number of L1 with a Seifert surface for L2. Equivalently, if L1 has kcomponents, then H1(S3

\L1) ' Zk, with a natural basis given by little circles around eachof the components of L1 with appropriate orientations; and the linking number is obtainedby expressing [L2] 2 H1(S3

\ L1) in this basis and taking the sum of the coe�cients. Thelinking number is symmetric, and can be computed from a link diagram by counting thecrossings of L1 with L2 with appropriate signs and dividing by 2.

18

Definition 6.6. A map ⇡ : E ! B (not necessarily a fiber bundle) hasthe homotopy lifting property with respect to a pair (X, A) if given mapsf : X ⇥ I ! B and g : X ⇥ {0}[A⇥ I ! E with ⇡ � g = f |X⇥{0}[A⇥I , thereexists h : X ⇥ I ! E extending g such that ⇡ � h = f .

Example 6.7. Recall that if ⇡ : E ! B is a covering space, then any pathin B lifts to a path in E, given a lift of its initial endpoint. This means that⇡ : E ! B has the homotopy lifting property with respect to the pair (pt, ;).

Definition 6.8. A map ⇡ : E ! B is a Serre fibration if it has the homotopylifting property with respect to the pair (Ik

, @I

k) for all k.

Lemma 6.9. Every fiber bundle is a Serre fibration.

Proof. Let (X, A) = (Ik⇥ I, I

k⇥ {0} [ @Ik

⇥ I), let ⇡ : E ! B be a fiberbundle, and let f : X ! B. We are given a lift g to E of the restrictionof f to A, and we must extend this to a lift of f over all of X. Now g isequivalent to a section of f

⇤E|A, and the problem is to extend this section

over X. We know that f

⇤E is trivial, so after choosing a trivialization, a

section is equivalent to a map to F . But a map A ! F extends over X

because there is a retraction r : X ! A.

Exercise 6.10. Let E = {(x, y) 2 R2| 0 y x 1}, let B = I, and

let ⇡ : E ! B send (x, y) 7! x. Then ⇡ is a Serre fibration, but not a fiberbundle.

Exercise 6.11. A map ⇡ : E ! B is a fibration if it has the homotopylifting property with respect to (X,�) for all spaces X.

(a) Show that a fibration is a Serre fibration. Hint: The key fact is that(Ik

⇥ I, I

k⇥ {0} [ @Ik

⇥ I) is homeomorphic to (Ik⇥ I, I

k⇥ {0}).

(b) Show that all the fibers of a fibration over a path connected space B arehomotopy equivalent. Hint: Use the homotopy lifting property whereX is a fiber.

(c) Show that all the fibers of a Serre fibration over a path connected spaceB have isomorphic homotopy and homology groups.

By Lemma 6.9, Theorem 6.1 follows from:

19

Theorem 6.12. If ⇡ : E ! B is a Serre fibration, then there is a long exactsequence in homotopy groups (6.1).

Proof. For k � 1 we need to define a connecting homorphism

� : ⇡k+1(B, x0) �! ⇡k(F, y0).

Consider an element of ⇡k+1(B, x0) represented by a map f : I

k+1! B

sending @Ik+1 to x0. We can lift f to E over I

k⇥ {0}[ @Ik

⇥ I, by mappingthat set to y0. By the homotopy lifting property, this lift extends to a lift h :I

k⇥ I ! E of f . Then h|Ik⇥{1}, regarded as a map on I

k, sends (Ik, @I

k) !(F, y0). We now define

�[f ] := [h|Ik⇥{1}].

One can check that this is well-defined on homotopy classes, and that this isa homomorphism. With this definition, the proof of exactness is a more orless straightforward exercise.

Exercise 6.13. Let E be an S

1-bundle over S

2 with rotation number r.Then the connecting homomorphism

� : Z ' ⇡2(S2) ! ⇡1(S

1) ' Z

is multiplication by ±r.

Example 6.14. Let X be a topological space and x0 2 X. The based loopspace ⌦X = {� : [0, 1] ! X | �(0) = �(1) = x0}. Define the space of pathsstarting at x0 to be PX = {� : [0, 1] ! X | �(0) = x0}. This fits into thepath fibration

⌦X ���! PX

??y⇡

X

where ⇡(f) = f(1). It is an easy exercise to check that this is a fibration.Now PX is contractible, so the long exact sequence gives

⇡k(⌦X, x0) ' ⇡k+1(X, x0)

for k � 1. This can also be seen directly from the definition of homotopygroups. But one can get a lot more information out of the path fibration aswe will see later.

Although we have only taken some very first steps in computing homotopygroups, let us now turn to some of their applications.

20

7 First example of obstruction theory

The rough idea of obstruction theory is simple. Suppose we want to constructsome kind of function on a CW complex X. We do this by induction: if thefunction is defined on the k-skeleton X

k, we try to extend it over the (k +1)-skeleton X

k+1. The obstruction to extending over a (k+1)-cell is an elementof ⇡k of something. These obstructions fit together to give a cellular cochain o

on X with coe�cients in this ⇡k. In fact this cochain is a cocycle, so it definesan “obstruction class” in H

k+1(X; ⇡k(something)). If this cohomology classis zero, i.e. if there is a cellular k-cochain ⌘ with o = �⌘, then ⌘ prescribes away to modify our map over the k-skeleton so that it can be extended overthe (k + 1)-skeleton.

Let us now make this idea more precise by proving some theorems. Fornow, if X is a CW complex, C⇤(X) will denote the cellular chain complex.To simplify notation, we assume that each cell has an orientation chosen, sothat Ck(Z) is the free Z-module generated by the k-cells. If e is a k-cell andif e

0 is a (k � 1)-cell, we let h@e, e0i 2 Z denote the coe�cient of e

0 in @e.

Theorem 7.1. Let X be a CW complex. Then

[X, S

1] = H

1(X; Z).

Proof. Let ↵ be the preferred generator of H

1(S1; Z) = Z. Define a map

� : [X, S

1] �! H

1(X; Z),

[f : X ! S

1] 7�! f

⇤↵.

We want to show that � is a bijection.Proof that � is surjective: let ⇠ 2 C

1(X; Z) with �⇠ = 0. We need to findf : X ! S

1 with f

⇤↵ = [⇠]. We construct f on the k skeleton by induction

on k. First, we send the 0-skeleton to a base point p 2 S

1. Next, if � is aone-cell, we extend f over � so that f |� has winding number ⇠(�) around S

1.If e : D

2! X is a 2-cell, then since ⇠ is a cocycle,

X

�

h@e,�i⇠(�) = 0.

It is not hard to see (and we will prove a more general statement in Lemma 8.3(a)below) that the left hand side of the above equation is the winding numberof f � e|@D2 around S

1. Hence, we can extend f over the 2-cell.

21

Now assume that f has been defined over the k-skeleton X

k for some k �

2. We can then extend f over the (k + 1)-skeleton, because the obstructionto extending over any (k + 1)-cell lives in ⇡k(S

1) = 0.Proof that � is injective: Let f0, f1 : X ! S

1, and suppose that f

⇤0↵ =

f

⇤1↵. We can homotope14

f0 and f1 so that they send the 0-skeleton of X tothe base point p 2 S

1. Then f

⇤i ↵ is represented by the cellular cochain �i that

sends each 1-cell � to the winding number of fi|� around S

1. The assumptionf

⇤0↵ = f

⇤1↵ then means that there is a cellular 0-cochain ⌘ 2 C

0(X; Z) with�0 � �1 = �⌘. That is, if � is a 1-cell with vertices �(0) and �(1), then

�0(�)� �1(�) = ⌘(�(1))� ⌘(�(0)). (7.1)

We now regard f0 and f1 as defining a map X⇥{0, 1}! S

1, and we wantto extend this to a homotopy X⇥[0, 1] ! S

1. We extend over X

0⇥[0, 1] such

that if x is a 0-cell, then the restriction to {x} ⇥ [0, 1] has winding number�⌘(x) around S

1. Then equation (7.1) implies that there is no obstruction toextending the homotopy over X

1⇥ [0, 1]. Finally, for k � 2, if the homotopy

has been extended over X

k�1⇥[0, 1], then there is no obstruction to extending

the homotopy over X

k⇥ [0, 1], since ⇡k(S

1) = 0.

Remark 7.2. The group structure on S

1 induces a group structure on[X, S

1], and it is easy to see that this agrees with the group structure onH

1(X; Z) under the above bijection.

8 Eilenberg-MacLane spaces

Theorem 7.1 can be regarded as giving a homotopy-theoretic interpretationof H

1(·; Z) for CW complexes. We now introduce an analogous interpretationof more general cohomology groups.

Consider a path connected space Y with only one nontrivial homotopygroup, i.e.

⇡k(Y ) '

⇢G, k = n,

0, k 6= n.

(8.1)

14We can do this because in general, any CW pair (X,A) has the homotopy extensionproperty. That is, any map (X ⇥ {0} [ (A⇥ I) ! Y extends to a map X ⇥ I ! Y . Oneconstructs the extension cell-by-cell, similarly to the proof of Lemma 5.6. In the presentcase, A is the 0-skeleton of X. A related argument proves the cellular approximation

theorem, which asserts that any continuous map between CW complexes is homotopic toa cellular map.

22

Such a space is called an Eilenberg-MacLane space. We also say that “Y isa K(G, n)”.

For example, S

1 is a K(Z, 1). Another example is that CP

1 is a K(Z, 2).Here one defines CP

1 = limk!1 CP

k with the direct limit topology, wherethe inclusion CP

k! CP

k+1 is induced by the inclusion Ck+1\ {0} !

Ck+2\ {0}. This is a CW complex with one cell in each even dimen-

sion. One can prove that CP

1 is a K(Z, 2) by using the cellular approx-imation theorem, as in the proof of Proposition 8.1 below, to show that⇡n(CP

1) = ⇡n(CP

k) whenever 2k � n+1. One can then use the long exactsequence in homotopy groups associated to the fibration S

1! S

2k�1! CP

k

to show that ⇡n(CP

k) = 0 if n 6= 2 and k > 1 and 2k � 1 > n. Similarly,RP

1 is a K(Z/2, 1).Explicit constructions of K(G, n)’s can be di�cult in general, but there

is still the following general existence result:

Proposition 8.1. For any positive integer n and any group G (abelian ifn > 1):

(a) There exists a CW complex Y satisfying (8.1).

(b) The CW complex Y in (a) is unique up to homotopy equivalence.

Proof. (a) Consider a presentation of G by generators and relations. Let Y

n

be a wedge of n-spheres, one for each generator. For each relation, attachan (n + 1)-cell whose boundary is the sum of the n-spheres in the relation.Now we have a CW complex Y

n+1 with ⇡n(Y n+1) ' G, and ⇡i(Yn+1) = 0 for

i < n. (When n = 1 this follows from the Seifert-Van Kampen theorem, andwhen n > 1 this follows from the Hurewicz theorem.) However ⇡n+1(Y

n+1)might not be zero. One can attach (n+2)-cells to kill ⇡n+1(Y

n+1), i.e. one canrepresent each element of ⇡n+1(Y

n+1) by a map S

n+1! Y

n+1 and use this asthe attaching map for an (n+2)-cell. Let Y

n+2 denote the resulting complex.Then ⇡n+1(Y

n+2) = 0, because the cellular approximation theorem impliesthat an element of ⇡n+1(Y

n+2) can be represented by a map S

n+1! Y

n+1.The cellular approximation similarly implies that attaching (n+2)-cells doesnot a↵ect ⇡i for i n. One then attaches (n + 3)-cells to kill ⇡n+2, and soon. (The resulting CW complex might be very complicated.)

(b) When G is abelian, this will be proved in Corollary 8.4 below. Thecase when n = 1 and G is nonabelian follows from a related argument whichwe omit.

23

Theorem 8.2. If G is abelian, and if Y is a K(G, n), then an identificationof ⇡k(Y ) with G determines, for any CW complex X, an isomorphism

H

n(X; G) ' [X, Y ].

Proof. This is similar to the proof of Theorem 7.1. We begin by defining amap

� : [X, Y ] �! H

n(X; G)

as follows. Fix an identification of ⇡n(Y ) with G. (Note that our assumptionsimply that ⇡n(Y ) does not depend on the choice of base point.) By theHurewicz theorem and the universal coe�cient theorem, we then obtain anidentification

H

n(Y, G) = Hom(G, G).

So there is a canonical element idG 2 H

n(Y, G). Now given f : X ! Y , wedefine

�[f ] := f

⇤(idG) 2 H

n(X; G).

We will prove that � is a bijection.To do so, we need the following lemma. Part (a) of the lemma is a useful

“homotopy addition lemma” which generalizes Lemma 2.2(b). Part (b) ofthe lemma gives a more concrete description of the map � when f sends the(n � 1)-skeleton X

n�1 to a base point y0 2 Y . (There is no obstruction tohomotoping f to have this property.)

Lemma 8.3. Suppose f : X ! Y sends X

n�1 to y0. Then:

(a) For any (n + 1)-cell e : D

n+1! X,

X

�

h@e,�i[f � �] = [f � e|Sn ] 2 ⇡n(Y, y0).

(b) �[f ] 2 H

n(X; G) is represented by a cellular cocycle � 2 C

ncell(X; G)

which sends an n-cell � : D

n! X to

�(�) = [f � �] 2 ⇡n(Y, y0) = G. (8.2)

Proof. Since f factors through the projection X ! X/X

n�1, for both (a)and (b) we may assume without loss of generality that X

n�1 is a point x0.

24

Proof of (a): By the Hurewicz theorem,

⇡n(Xn, x0) = Hn(Xn) = C

celln (X). (8.3)

By equation (8.3) and the naturality of the Hurewicz isomorphism under themap e|Sn : S

n! X

n, we have

[e|Sn ] =X

�

h@e,�i[�] 2 ⇡n(Xn, x0).

Applying f⇤ to this equation completes the proof.Proof of (b): Note that � is a cellular cocycle by part (a), and so it

determines a cohomology class in H

n(X; G). To show that this cohomologyclass is f

⇤(idG) as desired, since the pullback on H

n(·; G) by the inclusionX

n! X is injective, we may assume without loss of generality that X = X

n

(in addition to X

n�1 = {x0}). Then H

n(X; G) = C

ncell(X; G), so we just have

to check that �(f)(�) = [f � �] for � : D

n! X an n-cell. This follows from

naturality of the Hurewicz isomorphism under the map f � � : S

n! Y .

This completes the proof of Lemma 8.3.

Continuing the proof of Theorem 8.2, we now show that � is a bijection.To prove that � is surjective, let ↵ 2 H

n(X; G), and let � 2 C

ncell(X; G) be

a cellular cocycle representing ↵. Define f : X ! Y cell-by-cell as follows.First define f |Xn�1 = y0. Extend f over X

n so that (8.2) holds. Since� is a cocycle, it follows from Lemma 8.3(a) that there is no obstructionto extending f over X

n+1. There is then no obstruction to extending overhigher skeleta since ⇡i(Y ) = 0 for i > n. By Lemma 8.3(b), �(f) = ↵.

The proof that � is injective is completely analogous to the proof for thecase n = 1, G = Z that we have already seen.

Corollary 8.4. If G is abelian, then a K(G, n) that is a CW complex isunique up to homotopy equivalence.

Proof. Suppose that Y and Y

0 are both CW complexes and K(G, n)’s. Fixidentifications ⇡n(Y ) = ⇡n(Y 0) = G. By Theorem 8.2,

[Y, Y ] = [Y, Y

0] = [Y 0, Y ] = [Y 0

, Y

0] = Hom(G, G).

Furthermore, it follows from the definition of this correspondence that thecomposition of two maps between Y and/or Y

0 corresponds to the compo-sition of the corresponding homomorphisms from G to G, and idY and idY 0

25

correspond to idG. Now let f : Y ! Y

0 and g : Y

0! Y correspond to idG.

Then g � f also corresponds to idG, so g � f is homotopic to idY . Likewisef � g is homotopic to idY 0 .

9 Whitehead’s theorem

We now use another, simpler cell-by-cell construction to prove the followingtheorem, which shows that homotopy groups give a criterion for a map to bea homotopy equivalence.

Theorem 9.1 (Whitehead’s theorem). Let X and Y be path connected CWcomplexes and let f : X ! Y be a continuous map. Suppose that f inducesisomorphisms on all homotopy groups. Then f is a homotopy equivalence.

Proof. We first reduce to a special case. Define the mapping cylinder

Cf :=(X ⇥ I) [ Y

(x, 1) ⇠ f(x).

Then f is the composition of two maps X ! Cf ! Y , where the first mapis an inclusion sending x 7! (x, 0), and the second map sends (x, t) 7! f(x)and y 7! y. Furthermore, the map Cf ! Y is a homotopy equivalence sinceit comes from a deformation retraction of Cf onto Y . So it is enough toshow that the inclusion X ! Cf is a homotopy equivalence. By the cellularapproximation theorem, to prove Whitehead’s theorem we may assume thatf is cellular. Then Cf is a CW complex and X ⇥ {0} is a subcomplex.In conclusion, to prove Whitehead’s theorem, we may assume that X is asubcomplex of Y and f is the inclusion.

So assume that X is a subcomplex of Y and that the inclusion inducesisomorphisms on all homotopy groups. We now construct a (strong) defor-mation retraction of Y onto X. This consists of a homotopy F : Y ⇥ I ! Y

such that F (y, 0) = y for all y 2 Y ; F (x, t) = x for all x 2 X and t 2 I; andF (y, 1) 2 X for all y 2 Y . We construct F one cell at a time. This reducesto the following problem: given

F : (Dk⇥ {0}) [ (@Dk

⇥ I) �! Y, (9.1)

@D

k⇥ {1} 7�! X,

extend F to a map D

k⇥ I ! Y sending D

k⇥ {1}! X.

26

The restriction of F to @Dk⇥ {1} defines an element ↵ 2 ⇡k�1(X) (mod

the action of ⇡1(X)). Since F extends to a map (9.1), it follows that ↵maps to 0 in ⇡k�1(Y ). Since the inclusion induces an injection on ⇡k�1, itfollows that we can extend ↵ to a map D

k⇥ {1} ! X. Now F is defined

on @(Dk⇥ I) ' S

k, and we need to extend F over all of D

k⇥ I. This may

not be possible. The map we have so far represents an element � 2 ⇡k(Y )(mod the action of ⇡1(Y )), and we need this element to be zero. If � 6= 0,then since the map ⇡k(X) ! ⇡k(Y ) is surjective, we can change our choiceof extension of F over D

k⇥ {1} to arrange that � = 0. (Note that these

choices can be made for all k-cells independently, so there is no homologicalissue as in our previous obstruction theory arguments.)

Remark 9.2. If all we know is that X and Y have isomorphic homotopygroups, then X and Y need not be homotopy equivalent (which by White-head’s theorem means that the isomorphisms on homotopy groups might notbe induced by a map f : X ! Y ).

In general, Whitehead’s theorem can be hard to apply, because it maybe hard to check that a map induces isomorphisms on all homotopy groups.Fortunately, there is an alternate version of Whitehead’s theorem which canbe more practical:

Theorem 9.3. Let X and Y be simply connected CW complexes and letf : X ! Y be a continuous map. Suppose that f induces isomorphisms onall homology groups. Then f is a homotopy equivalence.

This can be proved using a relative version of the Hurewicz theorem, seee.g. Hatcher.

Corollary 9.4. Let X be a closed, oriented, simply connected n-dimensionalmanifold with Hi(X) = 0 for 0 < i < n. Then X is homotopy equivalent toS

n.

Proof. We can find an embedding of the closed ball D

n into X. Define amap f : X ! S

n by sending the interior of D

n homeomorphically to thecomplement of the north pole, and the rest of X to the north pole. ThenTheorem 9.3 applies to show that f is a homotopy equivalence.

27

10 Orientations of sphere bundles

We now use the ideas of obstruction theory to begin to analyze fiber bundles.

Definition 10.1. Let E be an S

k-bundle over B with k � 0. An orientationof E is a choice of generator of the reduced15 homology e

Hk(Ex) ' Z for eachx 2 B. This should depend continuously on x in the following sense. If E istrivial over a subset U ⇢ B, then a trivialization E|U ' U ⇥ S

k determines

identifications Ex ' S

k, and hence isomorphisms eHk(E|x) ' e

Hk(Sk), for

each x 2 U . We require that if U is connected, then the chosen generatorsof e

Hk(E|x) all correspond to the same generator of eHk(S

k). Note that thiscontinuity condition does not depend on the choice of trivialization over U .

The bundle E is orientable if it possesses an orientation. It is oriented ifmoreover an orientation has been chosen.

Example 10.2. If B is a smooth n-dimensional manifold, then choosinga metric on B gives rise to an S

n�1-bundle STB ! B consisting of unitvectors in the tangent bundle, and an orientation of STB is equivalent to anorientation of B in the usual sense.

Example 10.3. The mapping torus of a homeomorphism f : S

k! S

k,regarded as an S

k-bundle over S

1, is orientable if and only if f is orientation-preserving.

Exercise 10.4. Oriented S

1-bundles over S

2, up to orientation-preservingisomorphism, are classified by Z. (We will prove a generalization of this inTheorem 11.8 below.)

An orientation is equivalent to a section of the orientation bundle O(E),a double cover of B defined as follows. An element of O(E) is a pair (x, ox)

where x 2 B and ox is a generator of eHk(Ek). We topologize this as follows.

If U is an open subset of B and o is an orientation of E|U , let V (U, o) ⇢ O(E)

denote the set of pairs (x, ox) where x 2 B and ox is the generator of eHk(Ex)

determined by o. The sets V (U, o) are a basis for a topology on O(E). Withthis topology, O(E) ! B is a 2:1 covering space, and an orientation of E isequivalent to a section of O(E).

We now want to give a criterion for orientability of a sphere bundle. Bycovering space theory, a path � : [0, 1] ! B induces a bijection

�� : O(E)�(0) �! O(E)�(1).

15Of course the word “reduced” here only makes a di↵erence when k = 0.

28

Moreover, if � is homotopic to �0 rel endpoints, then �� = ��0 . So if x0 2 B

is a base point, we obtain a “monodromy” homomorphism

� : ⇡1(B, x0) ! Aut(O(E)x0) = Z/2. (10.1)

Since Z/2 is abelian, this homomorphism descends to the abelianization of⇡1(B, x0), so combining these homomorphisms for all path components ofB gives a map H1(B) ! Z/2. By the universal coe�cient theorem, this isequivalent to an element of H

1(B; Z/2), which we denote by

w1(E) 2 H

1(B; Z/2).

Proposition 10.5. Let E ! B be an S

k-bundle with k � 1. Assume thatthe path components of B are connected (e.g. B is a CW complex). Then:

(a) E is orientable if and only if w1(E) = 0 2 H

1(B; Z/2).

(b) If E is orientable, then the set of orientations of E is an a�ne space16

over H

0(B; Z/2).

Proof. (a) Without loss of generality, B is path connected. Since O(E) ! B

is a 2:1 covering space, it has a section if and only if it is trivial. And thecovering space is trivial if and only if the monodromy (10.1) is trivial.

(b) H

0(B; Z/2), regarded as the set of maps B ! Z/2 that are con-stant on each path component, acts on the set of orientations of E in anobvious manner. This action is clearly free, and the continuity condition fororientations implies that it is transitive.

When B is a CW complex, we can understand (a) in terms of obstruc-tion theory as follows. We can arbitrarily choose an orientation over the0-skeleton. The obstruction to extending this over the 1-skeleton is a 1-cocycle ↵ 2 C

1(B; Z/2). This 1-cycle represents the class w1(E). If ↵ = d�,then � tells us how to switch the orientations over the 0-skeleton so that theyextend over the 1-skeleton. There is then no further obstruction to extendingover the higher skeleta.

16An a�ne space over an abelian group G is a set X with a free and transitive G-action.For x, y 2 X, we can define the di↵erence x � y 2 G to be the unique g 2 G such thatg · y = x. The choice of an “origin” x0 2 X determines a bijection G ! X via g 7! g · x0.However this identification depends on the choice of x0. For example, if A is an m ⇥ nreal matrix and b 2 Rm, then the set {x 2 Rn

| Ax = b}, if nonempty, is naturally ana�ne space over Ker(A).

29

Exercise 10.6. w1 is natural, in that if f : B

0! B, then

w1(f⇤E) = f

⇤w1(E) 2 H

1(B; Z/2). (10.2)

Equation (10.2) implies that w1 is a “characteristic class” of sphere bun-dles. Specifically, it is equivalent to the first Stiefel-Whitney class (usuallydefined for real vector bundles rather than sphere bundles). We will studycharacteristic classes more systematically later in the course. Our next ex-ample of a characteristic class is the Euler class.

11 The Euler class of an oriented sphere bun-dle

Let E ! B be an oriented S

k bundle with k � 1. Assume temporarily thatB is a CW complex. We now define the Euler class

e(E) 2 H

k+1(B; Z). (11.1)

This is the “primary obstruction” to the existence of a section of E, in thesense of Proposition 11.4 below.

First, we choose a section s0 of E over the 0-skeleton B

0 (just pick anypoint in the fiber over each 0-cell). Now if we have a section si�1 over the(i� 1)-skeleton, and if i k, then we can extend to a section si over the i-skeleton. The reason is that if e : D

i! B is an i-cell, then we know that the

pullback17 bundle e

⇤E over D

i is trivial, so we can identify e

⇤E ' D

i⇥ S

k.The section si�1 induces a map @Di = S

i�1! S

k, and this extends overDi because ⇡i�1(S

k) = 0. By induction we obtain a section sk over thek-skeleton.

Now let us try to extend sk over the (k + 1)-skeleton. Let e : D

k+1! B

be a (k + 1)-cell. We choose an orientation-preserving trivialization e

⇤E '

D

k+1⇥ S

k. Then sk defines a map @Dk+1! S

k, which we can identify withan element of ⇡k(S

k) = Z. This assignment of an integer to each (k + 1)-celldefines a cochain

o(sk) 2 C

k+1(B; Z).

Lemma 11.1. o(sk) is a cocycle: �o(sk) = 0.

17The reason that we use pullback bundles here instead of restriction is that the map emight not be injective on @Di.

30

Proof. Consider a (k + 2)-cell ⇠ : D

k+2! B. We need to show that

X

�

h@⇠,�io(sk)(�) = 0. (11.2)

To understand the following proof of (11.2), it helps to draw pictures in thecase k = 1.

For each (k + 1)-cell �, let p� denote the center point of �. Let P denotethe set of points p�. Using smooth approximation and Sard’s theorem, wecan homotope18 the attaching map ⇠|Sk+1 so that:

(i) The inverse image of p� is a finite set of points x 2 S

k+1, to each ofwhich is associated a sign ✏(x).

(ii) The inverse image of a small (k + 1)-disk D� around p� consists of one(k+1)-disk Dx ⇢ S

k+1 for each x 2 ⇠

�1(p�), such that the restriction of⇠ to Dx is a homeomorphism which is orientation-preserving i↵ ✏(x) =+1.

Given (i) and (ii), it follows from the definition of the cellular boundary mapthat

h@⇠,�i =X

x2⇠�1(p�

)

✏(x). (11.3)

Next, we can extend the section sk over B

k+1\ P . Also, choose a triv-

ialization ⇠

⇤E ' D

k+2⇥ S

k. The extended section sk then defines a mapf : S

k+1\ ⇠

�1(P ) ! S

k. By (ii) above, for each x 2 ⇠

�1(p�), we have

[f |@Dx

] = ✏(x)o(sk)(�) 2 ⇡k(Sk).

Combining this with (11.3) gives

X

�

h@⇠,�io(sk)(�) =X

x2⇠�1(P )

[f |@Dx

] 2 ⇡k(Sk).

Now let X denote the complement in S

k+1 of the interiors of the balls Dx; thisis a compact manifold with boundary. By naturality of the Hurewicz isomor-phism, the right hand side of the above equation corresponds to �f⇤[@X] 2Hk(S

k). But this is zero, since f extends over X.

18This homotopy may change B, but it does not change the left side of the equation(11.2) that we want to prove.

31

Definition 11.2. The Euler class (11.1) is the cohomology class of the co-cycle o(sk).

Lemma 11.3. The cohomology class e(E) is well-defined.

Proof. Let sk and tk be two sections over the k-skeleton. We will constructa cellular cochain ⌘ 2 C

k(B; Z) with

�⌘ = o(sk)� o(tk). (11.4)

The idea is to try to find a homotopy from sk to tk. Consider the pullbackof E to B ⇥ I with the product CW structure; we can then regard sk andtk as sections defined over B

k⇥ {0} and B

k⇥ {1} respectively. There is no

obstruction to extending this section over B

k�1⇥ I, so that we now have a

section uk over the k-skeleton of B ⇥ I. By Lemma 11.1,

�o(uk) = 0 2 C

k+2(B ⇥ I; Z). (11.5)

Now define the required cellular cochain ⌘ 2 C

k(B; Z) as follows: If ⇢ isa k-cell in B, then

⌘(⇢) := o(uk)(⇢⇥ I) 2 Z.

(That is, ⌘(⇢) is the obstruction to obstruction to extending the homotopyover ⇢.) Then equation (11.4), evaluated on a (k + 1)-cell � in B, followsfrom equation (11.5) evaluated on the (k + 2)-cell � ⇥ I in B ⇥ I.

To put this all together, we have:

Proposition 11.4. Let E be an oriented S

k-bundle over a CW complex B.Then there exists a section of E over the (k + 1)-skeleton B

k if and only ife(E) = 0 2 H

k(B; Z).

Proof. ()) If there exists a section sk over the k-skeleton that extends overthe (k + 1)-skeleton, then by construction the cocycle o(sk) = 0.

(() Suppose that e(E) = 0. Let sk be a section over the k-skeleton; thenwe know that o(sk) = �⌘ for some ⌘ 2 C

k(B; Z). Keeping sk fixed over the(k� 1)-skeleton, we can modify it to a new section tk over the k-skeleton, sothat over each k-cell, sk and tk di↵er by ⌘; see Exercise 11.5 below. That is,equation (11.4) holds. Then o(tk) = 0, which means that tk extends over the(k + 1)-skeleton.

32

Exercise 11.5. Let X be a path connected space such that ⇡k(X) is abelianand the action of ⇡1 on ⇡k is trivial (so that ⇡k(X) is just the set of homotopyclasses of maps S

k! X with no base point). Let f : S

k�1! X. Then

the set of homotopy classes of extensions of f to a map D

k! X is an

a�ne space over ⇡k(X). The di↵erence between two extensions g+ and g�is obtained by regarding g+ and g� as maps defined on the northern andsouthern hemispheres respectively of S

k, and gluing them together along theequator (where they are both equal to f) to obtain a map S

k! X.

One can show, using the cellular approximation theorem, that the Eulerclass as defined above is natural and does not depend on the CW structureon B. We omit the details. In fact, later in the course we will see alternatedefinitions of the Euler class which do not use a CW structure at all.

Example 11.6. For an oriented S

1-bundle over S

2, the Euler class agreeswith the rotation number. This follows from the definitions if we use acell decomposition of S

2 with two 2-cells corresponding to the northern andsouthern hemispheres.

Remark 11.7. The Euler class is called the “primary obstruction” to theexistence of a section. There can also be “secondary obstructions” and higherobstructions, involving higher homotopy groups of spheres. That is, whenk > 1 and dim(B) > k + 1, it is possible that e(E) = 0 and yet no sectionover all of B exists. An example of this (which requires a bit of proof) isgiven by an S

2-bundle over S

4 in which a nonzero element of ⇡3(SO(3)) = Zis used in the clutching construction.

On the other hand, when k = 1 the Euler class is the only obstruction tothe existence of a section, because the higher homotopy groups of S

1 vanish.More generally, we have the following very satisfying result, which can beregarded as a higher-dimensional analogue of Theorem 7.1.

Theorem 11.8. Let B be a CW complex. Then the Euler class definesa bijection from the set of oriented S

1 bundles over B, up to orientation-preserving isomorphism, to H

2(B; Z).

Proof. To prove that the Euler class is injective, we use the fact that thespace of orientation-preserving homeomorphisms of S

1 deformation retractsonto S

1. Given two oriented S

1 bundles over B, there is then no obstructionto finding an isomorphism between them over the 1-skeleton. It is an exercise

33

to show that the obstruction to finding an isomorphism over the 2-skeletonis the di↵erence between the Euler classes. If this obstruction vanishes, thenthere is no obstruction to extending the isomorphism over higher skeleta sincethe higher homotopy groups of S

1 are trivial.Now let us prove that the Euler class is surjective. Let o 2 C

2(B; Z) bea cellular cocycle. We will construct an oriented S

1 bundle E ! B and asection s1 of E over the 1-skeleton such that o is the obstruction to extendings1 over the 2-skeleton.

Over the 1-skeleton, we define E|B1 = B

1⇥ S

1, and let s1 be a constantsection.

We now inductively extend E over the k-skeleton by gluing in one copyof D

k⇥ S

1 for each k-cell � : D

k! E. To do so we need a gluing map

S

k�1⇥ S

1! E|Bk�1 which projects to the attaching map �|Sk�1 : S

k�1!

B

k�1 and which restricts to an orientation-preserving homeomorphism oneach fiber. That is, we need to specify an orientation-preserving bundleisomorphism

S

k�1⇥ S

1�! (�|Sk�1)⇤(E|Bk�1). (11.6)

When k = 2, the map (11.6) is an orientation-preserving bundle iso-morphism S

1⇥ S

1! S

1⇥ S

1, i.e. a map S

1! Homeo+(S1), where the

superscript ‘+’ indicates orientation-preserving. We choose this to be a mapS

1! S

1 with degree �o(�) 2 Z. This ensures that o is the obstruction toextending s1 over the 2-skeleton.

When k � 3, we can choose any bundle isomorphism (11.6); we just needto show check that such an isomorphism exists, i.e. that (�|Sk�1)⇤(E|Bk�1) istrivial. By the injectivity of the Euler class, it is enough to check that

e((�|Sk�1)⇤(E|Bk�1)) = 0 2 H

2(Sk�1; Z).

When k = 3, this follows from the cocycle condition �o = 0, similarly to theproof of Lemma 11.1. When k > 3, this is automatic.

Remark 11.9. In §8 we saw another geometric interpretation of H

2(B; Z),namely as [B; CP

1]. The resulting bijection from [B; CP

1] to the set ofisomorphism classes of oriented S

1-bundles over B can be described directlyas follows. There is a “universal” oriented S

1-bundle E ! CP

1, namely thedirect limit of the bundles S

1! S

2k+1! CP

k (which are defined as in theHopf fibration). One then associates, to f : B ! CP

1, the pullback bundlef

⇤E ! B.

34

To describe the situation that oriented S

1 bundles are classified by homo-topy classes of maps to CP

1, one says that CP

1 is a “classifying space” fororiented S

1-bundles. We will discuss the more general theory of classifyingspaces later in the course.

12 Homology with twisted coe�cients

A natural context in which to generalize some of the previous discussion isprovided by homology with “twisted” or “local” coe�cients.

Definition 12.1. A local coe�cient system on a space X consists of thefollowing:

(a) For each x 2 X, an abelian group Gx,

(b) for each path � : [0, 1] ! X, a homomorphism �� : G�(0) �! G�(1),such that:

(i) �� depends only on the homotopy class of � rel endpoints,

(ii) If �1 and �2 are composable paths then ��1�2 = ��1��2 ,

(iii) If � is a constant path then �� is the identity.

Note that properties (i)–(iii) imply that �� is in fact an isomorphism.

We sometimes denote a local coe�cient system by G or by {Gx}, leaving theisomorphisms �� implicit.

Example 12.2. A constant local coe�cient system is obtained by settingGx = G for some fixed group G, and �� = idG for all �.

Example 12.3. If n > 1, or n = 1 and ⇡1(X, x0) is abelian for all x0 2 X,then {⇡n(X, x)} is a local coe�cient system on X.

Example 12.4. If E ! B is a Serre fibration, then {H⇤(Ex)} is a localcoe�cient system on B.

Example 12.5. A bundle of groups is a covering space eX ! X such that

each fiber has the structure of an abelian group, which depends continuouslyon x 2 X (in the sense that there are local trivializations which are groupisomorphisms on each fiber). Any bundle of groups gives rise to a localcoe�cient system on X. If X is a “reasonable” space, then the converse istrue.

35

Remark 12.6. If X is path connected and x0 2 X is a base point, then alocal coe�cient system on X determines a monodromy homomorphism

⇡1(X, x0) �! Aut(Gx0).

Conversely, if G is any abelian group, then any homomorphism ⇡1(X, x0) !Aut(G) is the monodromy of a local coe�cient system on X with Gx = G.

Now let G = {Gx} be a local coe�cient system on X. We want to definethe homology with local coe�cients H⇤(X, G ), as well as the cohomologyH

⇤(X, G ). When G is a constant local coe�cient system with Gx = G, thiswill agree with the usual singular (co)homology with coe�cients in G.

Let � : I

k! X be a singular cube. For every path � : I ! I

k, theisomorphism ���� defines an isomorphism G�(�(0)) ' G�(�(1)). Since I

k iscontractible, these isomorphisms are canonical, so that the groups G�(t) fort 2 I

k are all isomorphic to a single group, which we denote by G�. Also, if�

0 is a face of �, then G�0 = G�.Now define C⇤(X, {Gx}) to be the free Z-module generated by pairs (�, g)

where g 2 G�, modulo degenerate cubes as usual, and modulo the relation

(�, g1) + (�, g2) = (�, g1 + g2).

(One can also describe this as finite linear combinations of distinct nonde-generate cubes, where the coe�cient of each cube � is an element of G�.)Define the di↵erential @(�, g) by the usual formula (0.1), with g inserted ineach term. We have @

2 = 0 as usual. The homology of this complex isthen H⇤(X, {Gx}). The cohomology H

⇤(X, {Gx}) is the homology of thechain complex consisting of functions that assign to each singular cube � anelement of the group G�, with the di↵erential given by the dual of (0.1).

If X is a CW complex, then we can analogously define cellular homologyand cohomology with local coe�cients.

Example 12.7. A local coe�cient system G on S

1 is specified by a groupG and a monodromy map � : G ! G. If we choose a cell structure withone 0-cell e0 and one 1-cell e1, then under appropriate identifications, thedi↵erential in the cellular chain complex C⇤(S

1; G ) is given by

@(e1, g) = (e0, g � �g).

Hence H0(S1; G ) = G/ Im(1 � �) and H1(S

1; G ) = Ker(1 � �). AlsoH

k(S1; G ) = H1�k(S1; G ).

36

Example 12.8. Let B be a CW complex, let k � 1, and let E ! B bean S

k-bundle without an orientation (and possibly nonorientable). Then ourprevious construction of the Euler class generalizes to give a cohomologyclass

e(E) 2 H

k+1(B; {Hk(Ex)}),

which vanishes if and only if E has a section over the (k + 1)-skeleton. Notehere that the local coe�cient system {Hk(Ex)} is isomorphic to the constantlocal coe�cient system Z exactly when E is orientable.

Example 12.9. Let E be a fiber bundle over a CW-complex B with fiber F .Suppose that F is path connected. Let k be the smallest positive integer suchthat ⇡k(F ) 6= 0; suppose that ⇡k(F ) is abelian and that ⇡1(F ) acts triviallyon ⇡k(F ). Then the previous example generalizes to give a cohomology class

↵ 2 H

k+1(B; {⇡k(Ex)})

which vanishes if and only if E has a section over the (k + 1)-skeleton.

Example 12.10. One can use local coe�cients to generalize Poincare dualityto manifolds which are not necessarily oriented or even orientable. Any n-dimensional manifold X has a local coe�cient system O, such that Ox =Hn(X, X\{x}) ' Z. An orientation of X, if one exists, is a section of O whichrestricts to a generator of each fiber. In any case, regardless of whether ornot an orientation exists, if X is a compact n-dimensional manifold withoutboundary, and if G is any local coe�cient system on X, then there is acanonical isomorphism

H

k(X; G ) = Hn�k(X; G ⌦ O). (12.1)

This isomorphism is given by cap product with a canonical fundamental class[X] 2 Hn(X; O). (If X has a triangulation, e.g. if X is smooth, then thisfundamental class is represented in the cellular chain complex by the sum ofall the n-simplices. This sum is well-defined without any orientation choicesif one uses coe�cients in O.) To prove this, take your favorite proof ofPoincare duality19 and insert local coe�cients everywhere.

An instructive example is X = CP

2. Consider the usual CW structurewhich has one i-cell ei for i = 0, 1, 2. The usual cellular chain complex

19My favorite proof, for smooth manifolds, is the one using Morse homology. We willsee this later in the course.

37

C

cell⇤ (CP

2) is then by @e2 = ±2e1 and @e1 = 0, where the sign depends onchoices of orientations of the cells. Now fix an orientation of CP

2 at thecenter point of each of the cells. This allows us to also regard C

cell⇤ (CP

2; O)as generated by e0, e1, e2. However since the orientation of CP

2 gets reversedif one goes from one end of e1 to the other or from one side of e2 to theother, one now has @e2 = 0 and @e1 = ±2e0, where the sign depends onthe orientation choices. It is then easy to check that Poincare duality (12.1)holds in this example.

38