contentsstolz/2016s_math60440/homework_solutions/... · contents 1. homework assignment # 1 1 2....

SOLUTIONS TO HOMEWORK PROBLEMS

Contents

1. Homework Assignment # 1 12. Homework Assignment # 2 63. Homework Assignment # 3 84. Homework Assignment # 4 125. Homework Assignment # 5 166. Homework Assignment # 6 227. Homework Assignment # 7 258. Homework Assignment # 8 299. Homework Assignment # 9 3510. Homework Assignment # 10 3911. Homework Assignment # 11 44

1. Homework Assignment # 1

1. Let p : X → X be covering map which sends x0 ∈ X to x0 ∈ X. Show that the inducedmap

p∗ : πn(X, x0) −→ πn(X, x0)

is injective for n ≥ 1. Hint: use the homotopy lifting property for the covering map p.

Proof. Since the induced map p∗ is a homomorphism, it suffices to show that the kernel of p∗ is

trivial. So let f : (In, ∂In)→ (X, x0) be a map of pairs representing an element in the kernel

of p∗. In other words, there is a homotopy H between the map p ◦ f : (In, ∂In) → (X, x0)and the constant map x0 : (In, ∂In)→ (X, x0). More explicitly this means that we have thatH is a map H : In × I → X with

H(s, 0) = p ◦ f(s) for s ∈ In

H(s, t) = x0 for s ∈ ∂In or t = 1.

In particular, the outer square of the diagram

In × 0� _

��

f // X

p��

In × I

H

77

H // X

commutes. By the homotopy lifting property for the covering space X → X, there is a

map H making the whole diagram commutative. We claim that H is the desired homotopy1

2 SOLUTIONS TO HOMEWORK PROBLEMS

between f : (In, ∂In)→ (X, x0) and the constant map x0 : (In, ∂In)→ (X, x0). So we needto show that

• H(s, t) = x0 for all s ∈ ∂In and t ∈ I.

• H(s, 1) = x0 for all s ∈ In and

To prove the first property suppose that F ⊂ ∂In is one of the (n − 1)-dimensional facesthat form the boundary of In. The map H maps all of F × I to the base point x0 ∈ X, and

hence the lift H maps F × I to the fiber p−1(x0) ⊂ X. Since F × I is connected, H mapsall of F × I to a connected component of p−1(x0), that is, to one point of the discrete space

p−1(x0). Since H maps (s, 0) to x0, it maps every point of p−1(x0) to x0. This proves thefirst property above.

The second property follows by the same argument applied to HIn×{1}: since H is a lift

of H and H maps In × {1} to the basepoint x0 ∈ X, H maps the connected space In × {1}to one point of the fiber p−1(x0). This point must be x0, since H maps the boundary ofIn × {1} to x0 by the first property. �

2. Show that for closed 2-manifolds Σ1, Σ2, the Euler characteristic of the connected sumΣ1#Σ2 is given by

χ(Σ1#Σ2) = χ(Σ1) + χ(Σ2)− 2

Proof. Cover both surfaces by patterns of polygons, and let us write Vi (resp. Ei resp. Fi)for the number of vertices (resp. edges resp. faces) on Xi. Let us assume that we chosethese pattern such that both contain a face which is a polygon with k vertices and edges forsome k. The point of this is that removing that face from both surfaces and identifying theresulting surfaces along the polygon results in the connected sum X1#X2 and the patternson both surfaces fit together to give a pattern of polygons on X1#X2. Let us determine V(= number of vertices), E (= number of edges) and F (= number of faces) of this patternon X1#X2:

faces: F = F1 + F2 − 2 (each face on X1 or X2 gives a face on X1#X2 except the twopolygons that we removed in order to form the connected sum).

edges: E = E1 +E2− k (each edge on X1 or X2 gives an edge on X1#X2 except thereare k pairs of edges that get identified with each other when we form the connectedsum).

vertices: V = V1 + V2 − k (each vertex on X1 or X2 gives a vertex on X1#X2 exceptthere are k pairs of vertices that get identified with each other when we form theconnected sum).

This implies

χ(X1#X2) = V − E + F

= (V1 + V2 − k)− (E1 + E2 − k) + (F1 + F2 − 2)

= V1 − E1 + F1 + V2 − E2 + F2 − 2

= χ(X1) + χ(X2)− 2

�

3. a) Calculate the homology groups of the connected sum RP2# . . .#RP2 of k copies of thereal projective plane (we’ve calculated the homology groups of the surface of genus g ≥ 0

SOLUTIONS TO HOMEWORK PROBLEMS 3

in class; after doing this homework problem, by the Classification Theorem for surfaces, wehave then calculated the homology groups of all compact connected surfaces). Hint: Use therepresentation of RP2# . . .#RP2 as the quotient space of a polygon obtained by identifyingpairs of edges. For calculating the homology of the associated chain complex it will be usefulto work with a convenient basis for the free Z-module C1, which is not the obvious basisgiven by the edges of polygonal pattern we use for the calculation.

b) Can the Euler characteristic of a compact connected surface be expressed in terms of itshomology groups?

Remark: Later we will define the Euler characteristic of more general topological spaces interms of their homology groups. We will generalize the description of the Euler characteristicas an alternating sum of the number of vertices, edges and faces to an important class ofspaces known as “CW complexes”.

Proof. Part a). We recall that the connected sum RP2# . . .#RP2 of k copies of the pro-jective plane RP2 can be described as the quotient of the 2k-gon with edge identificationsgiven by the picture

(1)

RP2# . . .#RP2︸︷︷︸k

≈ak

ak

a1

a1

a2

a2

. . .

f

The resulting pattern given by this polygon on the surface P# . . .#P has one vertex v,k edges a1, . . . , ak and one face f . Hence the chain complex associated to this pattern lookslike

Zv Za1 ⊕ · · · ⊕ Zak∂1oo Zf∂2oo

Since there is only one vertex involved, we have ∂1(ai) = v − v = 0 for any edge ai andhence ∂1 ≡ 0. To determine ∂2(f) we notice that every edge label occurs exactly twice, witharrows pointing in the same direction as the arrow for f . Hence ∂2(f) = 2a1 + 2a2 + · · ·+ak.It follows that H0 = Z, H2 = 0, and

H1 = Za1 ⊕ · · · ⊕ Zak/Z2(a1 + · · ·+ ak).


To identify this quotient group, we choose a different basis of the free abelian group C1,namely a1, . . . , ak−1, c, with c = a1 + · · ·+ ak. Then we see

H1 = Za1 ⊕ · · · ⊕ Zak−1 ⊕ Zc/Z2c ∼= Z⊕ · · · ⊕ Z︸︷︷︸k−1

⊕Z/2.

Part b). Can the Euler characteristic of a compact connected surface be expressed in termsof its homology groups?

Inspection of the Euler characteristic calculation (resp. homology group calculation) forcompact connected surfaces X shows that

χ(X) = rkH0(X)− rkH1(X) + rkH2(X),

where rkHq(X) is the rank of the abelian group Hq(X). �

4. Let Σ be a compact surface, which is not necessarily connected. Then Σ has finitely manyconnected components Σ1, . . . ,Σk each of which is a closed connected surface.

a) How can the homology groupHq(Σ) be expressed in terms of the homology groupsHq(Σi)?b) What is the group H0(Σ)?

Proof. Part a). We recall that the homology group Hk(Σ) is defined as the k-th homologygroup of the chain complex (C∗(Σ,Γ), ∂). Here Γ is a pattern of polygons on the surface Σi,equipped with an orientation for all edges and faces. The group Ck(Σi,Γi) is the free abeliangroup generated by the vertices (for k = 0) resp. edges (for k = 1) resp. faces (for k = 2).For an oriented edge e ∈ C1(Σ) the boundary ∂1(e) ∈ C0(Σ,Γ) is tip(e) − tail(e). For anoriented face f ∈ C2(Σ) the boundary ∂2(f) is

∑±e, where the sum runs over all edges e

which are edges of the polygon f ; the sign is positive if the orientations of e and f agree,and negative otherwise.

Given a pattern of polygons with orientations Γi on each surface Σi, the union of all thesevertices, oriented edges and faces can be viewed as providing us with a pattern of polygonsΓ on Σ (with orientations on edges and faces). Denoting by V i the set of vertices of Γi, andby V the corresponding sets for the pattern of polygons Γ on Σ whose restriction to Σi ⊂ Σis Γi. Then the set V is the disjoint union V 1 q · · · q Vk of the sets V i and it follows that

C0(Σ,Γ) = Z[V ] = Z[V 1 q · · · q V k] ∼=k⊕i=1

Z[V i] =k⊕i=1

C0(Σi,Γi),

where the middle isomorphism comes from the map

Z[V 1 q · · · q V k]∼=−→

k⊕i=1

Z[Vi]

which sends the generator v ∈ V j ⊂ V 1 q · · · q V k ⊂ Z[V 1 q · · · q V k] to the element(c1, . . . , ck) where ci ∈ Z[V i] is the trivial element for i 6= j and ci = v ∈ V i ⊂ Z[V i] fori = j. Similarly, we have

C1(Σ,Γ) ∼=k⊕i=1

C1(Σi,Γi) and C2(Σ,Γ) ∼=

k⊕i=1

C2(Σi,Γi)

for the 1-chains and 2-chains consisting of linear combinations of edges and faces.


Concerning the boundary homomorphisms ∂n : Cn(Σ,Γ)→ Cn−1(Σ,Γ) for n = 1, 2, if e isan edge or f is a face for the pattern Γ, then e is an edge or f is a face belonging to thepattern Γi on Σi for some i, and hence

∂1(e) = ∂i1(e) and ∂2(f) = ∂i2(f)

where ∂in : Cn(Σi,Γi) → Cn−1(Σi,Γi) is the boundary homomorphism for the pattern Γi onΣi. It follows that the diagram

Cn(Σ,Γ)∂k //

∼=��

Cn−1(Σ,Γ)

∼=��⊕k

i=1Cn(Σi,Γi)

⊕∂in //

⊕ki=1Cn−1(Σi,Γ

i)

is commutative. We summarize this discussion by saying that the chain complex (C∗(Σ,Γ), ∂∗)is isomorphic to the direct sum chain complex

⊕i∈I Cn−1(Σi,Γ

i)oo⊕

i∈I Cn(Σi,Γi)

⊕∂inoo

⊕i∈I Cn+1(Σi,Γ

i)

⊕∂in+1oo oo

Lemma 1. Let (Ci∗, ∂

i∗), i ∈ I be a collection of chain complexes parametrized by some index

set I (not necessarily finite). Then the n-th homology group of the direct sum chain complex

⊕i∈I C

in−1

⊕∂in−1oo

⊕i∈I C

in

⊕∂inoo

⊕i∈I C

in+1

⊕∂in+1oo

⊕∂in+1oo

is isomorphic to⊕

i∈I Hn(Ci∗, ∂

i∗).

Proof. Let us write Zin for the n-cycles of the chain complex Ci

∗, and Zn for the n-cyclesof the direct sum chain complex. Similarly, let Bi

n, Bn be the n-boundaries of the chaincomplex Ci

∗ resp. the direct sum chain complex. Then

Zn = ker

(⊕i∈I

∂in :⊕i∈I

Cin −→ Ci

n−1

)∼=⊕i∈I

ker ∂in =⊕i∈I

Zin

Similarly, Bn∼=⊕

i∈I Bin and hence

Hn(sum chain complex) =ZnBn

∼=⊕

Zin⊕

Bin

∼=⊕ Zi

n

Bin

=⊕

Hn(Ci∗)

�

This finishes the proof of part (a).

Part b). By the Classification Theorem for surfaces, any connected closed surface is home-omorphic to either the sphere, or a connected sum of tori or real projective planes. Ourcalculations show that for all these surfaces the homology group H0(Σ) is isomorphic to Z.Part (a) implies that H0(Σ) for a not necessarily connected surface is the direct sum of Z’swith each copy of Z corresponding to a connected component of Σ. �



1. Show that the abelian groups Z/mn and Z/m⊕Z/n are isomorphic if and only if m andn are relatively prime.

Proof. We claim that the homomorphism Z/mn→ Z/m⊕Z/n which sends [k] to ([k], [k]) isan isomorphism if m, n are relatively prime. The condition gcd(m,n) = 1 implies that thereare integers a, b such that am+ bn = 1. Consider the homomorphism Z/m⊕Z/n→ Z/mngiven by ([k], [`]) 7→ [bnk + am`]. This is well-defined since it sends ([m], [0]) to [bnm] =[0] ∈ Z/mn and ([0], [n]) to [amn] = [0] ∈ Z/mn. It is easy to check that this is an inverse.

If m, n are not relatively prime, then the least common multiple of m and n is strictlysmaller than mn. Hence multiplication by the least common multiple gives zero for everyelement in Z/m⊕ Z/n, but not Z/mn showing that these groups can’t be isomorphic. �

2. Show that the singular chain complex of a topological space X is in fact a chain complex;i.e., that ∂q ◦ ∂q+1 = 0, where ∂q : Cq(X)→ Cq−1(X) is the boundary map.

Proof. It will suffice to show that ∂q ◦ ∂q+1 = 0 on the generators of Cq+1 since ∂q and ∂q+1

are both homomorphisms. Consider σ : ∆q+1 → X in Cq+1(X).

∂q ◦ ∂q+1(σ)

=∂q

(q∑j=0

(−1)j∂q+1(σ) ◦ [e0, ..., ej, ..., eq+1]

)

=

q∑i=0

(−1)i

(q+1∑j=0

(−1)jσ ◦ [e0, ..., ej, ..., eq+1] ◦ [e0, ..., ei, ..., eq]

)=

∑0≤i<j≤q+1

(−1)i+jσ ◦ [e0, ..., ei, ..., ej, ..., eq+1]

+∑

0≤j<i≤q+1

(−1)i+j+1σ ◦ [e0, ..., ej, ..., ei, ..., eq+1]

=∑

0≤i<j≤q+1

(−1)i+jσ ◦ [e0, ..., ei, ..., ej, ..., eq+1]

−∑

0≤j<i≤q+1

(−1)i+jσ ◦ [e0, ..., ej, ..., ei, ..., eq+1]

=0.

Hence, since ∂q ◦∂q+1 = 0 on the generators of Cq+1(X), ∂q ◦∂q+1 = 0, and the singular chaincomplex of a topological space X is in fact a chain complex. �

3. Let X be a topological space with path components Xα, α ∈ A. Show that Hq(X) isisomorphic to

⊕α∈AHq(Xα).

Proof. Since the image of a singular simplex q-simplex σ : ∆q → X is path connected, itsimage is contained in exactly one of the path connected components Xα. It follows that


Sq(X), the set of singular q-simplices in X, is given by

Sq(X) =∐α∈A

Sq(Xα),

the disjoint union of the sets Sq(Xα). As in problem 4 in assignment # 1, this leads to anisomorphism⊕

α∈A

Cq(Xα) =⊕α∈A

Z[Sq(Xα)] ∼= Z[∐α∈A

Sq(Xα)] = Z[Sq(X)] = Cq(X).

From the construction it is clear that this isomorphism restricted to Cq(Xα) is the map(iα)∗ : Cq(Xα) → Cq(X) induced by the inclusion iα : Xα → X. In particular, this map iscompatible with boundary maps and hence the map⊕

α∈A

(iα)∗ :⊕α∈A

Cq(Xα) −→ C∗(X)

is an isomorphism of chain complexes. By the lemma we proved for problem 4, the homologyof the direct sum of chain complexes is isomorphic to the direct sum of the homology groups.This proves the desired isomorphism. �

4. Show that the Hurewicz map h : π1(X, x0) → H1(X) given by [γ] → [[γ]] is a homomor-phism.

Proof. Let [γ], [γ′] ∈ π1(X, x0), and let γγ′ denote the concatenation of paths γ and γ′. Toshow that h is a homomorphism, we need to verify that h([γ][γ′]) = h([γ]) + h([γ′]); i.e.,[[γγ′]] = [[γ]] + [[γ′]]. We will do so by showing that γ + γ′ − γγ′ ∈ B1(X) which will thenimply that [[γγ′]] = [[γ + γ′]] = [[γ]] + [[γ′]], as desired. Consider the singular 2-simplex,σ : ∆2 → X, defined as the composition

∆2[e0,

12

(e0+e1),e1]//∆1 γγ′ //X

Notice that ∂2(σ) = γ′ − γγ′ + γ, so γ′ − γγ′ + γ ∈ B1(X) as desired. Thus, h is indeed ahomomorphism. �

5. The goal of this problem is to show that the homomorphism

h : πab1 (X, x0)→ H1(X)

induced by the Hurewicz homomorphism h is in fact an isomorphism for path connected spaceX; here πab1 (X, x0) is the abelianized fundamental group of X. The idea is to construct aninverse to h as follows. Choose for every point x ∈ X a path λx from x0 to x. In class weshowed that we have a well-defined map

Ψ: C1(X)/B1(X) −→ πab1 (X, x0) given by [[γ]] 7→ [λγ(0) ∗ γ ∗ λγ(1)]

for any singular 1-simplex γ, also known as path γ : I → X. Here λγ(0) ∗ γ ∗ λγ(1) is theconcatenation of the path λγ(0) (from x0 to γ(1)), the path γ (from γ(0) to γ(1)) and thepath λγ(1) (from γ(1) to x0, obtained by running the path λγ(1) from x0 to γ(1) backwards).Show that the restriction of Ψ to H1(X) ⊂ C1(X)/B1(X) provides an inverse to the map h.

Proof. We need to show:

(a) Ψ|H1(X) ◦ h = idπab1 (X,x0);


(b) h ◦Ψ|H1(X) = idH1(X).

To prove (a), let [γ : (I, ∂I)→ (X, x0)] ∈ πab1 (X, x0). Then

Ψ|H1(X) ◦ h([γ]) = Ψ|H1(X)([[γ]])

= [λγ(0)γλγ(1)] = [λx0γλx0 ]

= [λx0 ][γ][λx0 ] = [λx0 ][λx0 ][γ] = [γ],

which shows Ψ|H1(X) ◦ h = idπab1 (X,x0).

To prove (b), we will use the following statements:

(i) If γ, γ′ : I → X are paths with γ(1) = γ′(0), and γγ′ : I → X their concatenation, thenγ′ − γγ′ + γ ∈ B1(X).

(ii) γ + γ ∈ B1(X).

Statement (i) we proved in problem (4) (note that the argument used there does not requireγ and γ′ to be based loops; it is only necessary that γ(1) = γ′(0) in order to be able to havethe concatenated path γγ′). To deduce the second claim, we apply part (i) for γ′ = γ whichimplies γ − γγ + γ ∈ B1(X). We note that γγ is a loop based at γ(0) which is homotopicto the constant loop. Hence its image under the Hurewicz map [[γγ]] ∈ H1(X) is trivial. Inother words, γγ ∈ B1(X), which proves (ii).

With these preliminaries, we can calculate for any path γ:

(2)

h ◦Ψ([[γ]]) = h([λγ(0)γλγ(1)] = [[λγ(0)γλγ(1)]]

= [[λγ(0)]] + [[γ]] + [[λγ(1)]] = [[λγ(0)]] + [[γ]]− [[λγ(1)]]

= [[γ]] + [[λ(∂(γ))]],

where λ : C0(X) → C1(X) is given on generators by x 7→ λx (and hence λ maps ∂γ =γ(1) − γ(0) to λγ(0) − λγ(1)). Since formula (2) holds for the generators γ ∈ C1(X), it holdfor every element z ∈ C1(X); in particular, if z is a cycle, we have h ◦Ψ([[z]]) = [[z]], whichis what we wanted to prove. �


1. Calculate the reduced homology groups of the subspace X ⊂ R3 which is the union of thesphere S2 and the x-axis.

Proof. The natural idea is to apply the Mayer-Vietoris sequence to X = U∪V , where U = S2,and V is the x-axis. Unfortunately, this doesn’t work since these aren’t open subsets of Xas required by the Mayer-Vietoris Theorem. The way around this problem is to work withlarger open subsets U ′, V ′ ⊂ X that contain U resp. V as deformation retracts. This meansthat there is a map

rU : U ′ → U such that rU ◦ iU = idU and iU ◦ rU ∼ idU ′ ,

where iU : U → U ′ is the inclusion map, and similarly for V . There are many possible choicesfor U ′, V ′, for example we can take

U ′ := S2 ∪ x-axis without the origin V ′ := x-axis ∪ {(x, y, z) ∈ S2 | x 6= 0}.The maps rU , rV , the homotopy between iU ◦rU and idU ′ , and the homotopy between iV ◦rVand idV ′ can be written down explicitly in formulas, but it might be more helpful to describe


them in a geometric way as follows: the subset {±1} ⊂ R\{0} is a deformation retract (takethe linear homotopy between the identity on R \ {0} and the map x 7→ x/|x|). It followsthat U = S2 is a deformation retract of U ′ := S2 ∪ x-axis without the origin.

Similarly, V ′ is the union of the x-axis and the two open disks given by the left resp.right hemisphere of S2. The center of each of these disks is a deformation retract of thedisk. These deformation retractions and homotopies fit together to show that V = x-axisis a deformation retract of V ′. Finally, using both retractions (for the x-axis without 0 andthe two disks), we obtain the set U ∩ V = {(−1, 0, 0), (1, 0, 0)} as a deformation retract ofU ′ ∩ V ′.

These homotopy equivalences allow us in particular to calculate the reduced homologygroups of U ′, V ′ and U ′ ∩ V ′ as

Hk(U′) ∼= Hk(U) ∼=

{Z k = 2

0 k 6= 2

Hk(V′) ∼= Hk(V ) ∼= Hk(pt) = 0

Hk(U′ ∩ V ′) ∼= Hk(U ∩ V ) ∼=

{Z k = 0

0 k 6= 0

These calculations show that in the Mayer-Vietoris sequence for X = U ′∪V ′, there are onlytwo non-trivial reduced homology groups of U ′, V ′ or U ′ ∩ V ′ that contribute. Here are therelevant portions of the Mayer-Vietoris sequence:

0 = H2(U ′ ∩ V ′) // H2(U ′)⊕H2(V ′) // H2(X)∂ // H1(U ′ ∩ V ′) = 0

The exactness of the sequence implies that the middle map is an isomorphism, and hence

H2(X) ∼= H2(U ′)⊕H2(V ′) ∼= Z.

0 = H1(U ∩ V ) // H1(X)∂ // H0(U ′ ∩ V ′) // H0(U ′)⊕H0(V ′) = 0

The exactness of the sequence implies that ∂ is an isomorphism, and hence H1(X) ∼= H0(U ′∩V ′) ∼= Z. For k 6= 1, 2 the exactness of the Mayer-Vietoris sequence implies that Hk(X) = 0.Summarizing our result, we find

Hk(X) ∼=

{Z k = 1, 2

0 k 6= 1, 2

�

2. Let x1, . . . , xl be distinct points of Rn. Calculate the reduced homology groups of thespace Rn \ {x1, . . . , xl}. Hint: Compare the homology groups of Rn \ {x1, . . . , xl} with thoseof Rn via the Mayer-Vietoris sequence.

Proof. Let D1, . . . , Dl ⊂ Rn be disjoint open disks with center x1, . . . , xl. Let

U := D1 ∪ · · · ∪Dl and V := Rn \ {x1, . . . , xl}.

Hk(U ∩ V ) = Hk(qli=1(Di \ {xi}) ∼=l⊕

i=1

Hk(Di \ {xi}) ∼=l⊕

i=1

Hk(Sn−1) ∼=

{Zl k = 0, n− 1

0 k 6= 0, n− 1


Hk(U) = Hk(qli=1(Di) ∼=l⊕

i=1

Hk(Di) ∼=

{Zl k = 0

0 k 6= 0

We note that that the inclusion map iU : U ∩V → U is a bijection on connected components,and hence it induces an isomorphism on H0. Next we consider the Mayer-Vietoris sequencefor the decomposition of Rn as the union of the open subsets U and V :

// Hk+1(Rn)∂ // Hk(U ∩ V )

iU∗ ⊕iV∗ // Hk(U)⊕ Hk(V ) // Hk(Rn) //

Since Rn is contractible, its the reduced homology groups Hk(Rn) vanish, and hence the mapiU∗ ⊕ iV∗ is an isomorphism by exactness of the Mayer-Vietoris sequence. For k = 0 the map

iU∗ is an isomorphism on the homology group Hk as noted above, and hence also on Hk. It

follows that H0(V ) = 0. For k > 0, the vanishing of Hk(U) implies

Hk(V ) ∼= Hk(U ∩ V ) ∼=

{Zl k = 0, n− 1

0 k 6= 0, n− 1

So, summarizing, we have

Hk(Rn \ {x1, . . . , xl}) ∼=

{Zl k = n− 1

0 k 6= n− 1

�

3. Let X be a topological space and let ΣX be the suspension of X which is defined as thequotient space X × [0, 1]/ ∼, where the equivalence relation is generated by (x, 0) ∼ (x′, 0)

and (x, 1) ∼ (x′, 1) for all x, x′ ∈ X. Show that Hk(X) ∼= Hk+1(ΣX) (this is called thesuspension isomorphism).

Hint: note that the suspension of Sn is homeomorphic to Sn+1, and think of the suspension

isomorphism as a generalization of the isomorphism Hq(Sn) ∼= Hq+1(Sn+1) proved in class.

That proof used the decomposition of Sn+1 as a union of U = Sn+1 \ north pole and V =Sn+1 \ south pole. In this more general situation, the subspaces U = {[x, t] ∈ Σ | 0 ≤ t < 1}and V = {[x, t] ∈ Σ | 0 < t ≤ 1} of ΣX play an analogous roll.

Proof. We will apply the Mayer-Vietoris sequence for the decomposition of the suspensionΣX as the union of the two open subspaces U and V . We note that the spaces U and V(can be pictured as “cones”) are both contractible, i.e., homotopy equivalent to the 1-pointspace pt. To see this, let i : pt → ΣX be the inclusion map that sends pt to the “conepoint” [x, 0] ∈ U , and let r : ΣX → pt be the unique projection map. Then r ◦ i = idpt, andi ◦ r : U → U sends every point to the cone point, and it remains to show that this map ishomotopic to the constant map. A homotopy

H : ΣX × I −→ ΣX is given by ([x, t], s) 7→ [x, st]

The argument for V is analogous. We claim that the inclusion map i : X → U ∩ V , x 7→[x, 1/2], is a homotopy equivalence. To see this, let r : U ∩ V → X be the projection mapgiven by [x, t] 7→ x. Then r ◦ i = idX , and i ◦ r is homotopic to idU∩V via the homotopy

H : (U ∩ V )× I −→ U ∩ V H([x, t], s) = [x, (1− s)1

2+ st]


We remark that (1− s)12

+ st is the linear path from 1/2 (for s = 0) to t (for s = 1). Writingdown the Mayer-Vietoris sequence for reduced homology groups we have

// Hk(U)⊕ Hk(V ) // Hk(ΣX)∂ // Hk−1(U ∩ V ) //

∼= r∗��

Hk−1(U)⊕ Hk−1(V ) //

0 Hk−1(X) 0

The exactness of the sequence implies that ∂ is an isomorphism, and hence the composition

r∗ ◦ ∂ : Hk(ΣX)→ Hk−1(X) is the desired isomorphism. �

4. (a) Let

Af //

α��

Bg //

β��

C

γ��

A′f ′// B′

g′// C ′

Af // B

g // C

A′

α′

OO

f ′// B′

g′//

β′

OO

C ′

γ′

OO

be a commutative diagram of abelian groups and homomorphisms whose rows are exact atB and B′. Moreover, α ◦ α′ = idA′ , β ◦ β′ = idB′ , and γ ◦ γ′ = idC′ . Show that the sequence

kerα −→ ker β −→ ker γ

is exact (the maps are the restrictions of f resp. g).(b) Show that there is a reduced version of the Mayer-Vietoris long exact sequence, where

all homology groups are replaced by the corresponding reduced groups. We recall that the

reduced homology group Hk(X) of a topological space X is by definition the kernel of themap p∗ : Hk(X) → Hk(pt) induced by the projection map p : X → pt. Hint: compare theMayer-Vietoris sequence for X = U ∪ V with the Mayer-Vietoris sequence for the spaceX ′ = pt considered as the union of the open subsets U ′ = V ′ = pt via the projection mapp : X → pt. Then apply part (a).

Proof. Part (a). For a ∈ kerα, its image under g◦f in ker γ is trivial, since the compositiong◦f is zero by exactness of the top row. Let b ∈ ker β and assume that g(b) = 0 ∈ ker γ ⊂ C.Then by exactness of the top row, there is some a ∈ A with f(a) = b. In general, α(a) willbe non-zero, and so we need to modify a in order to produce an element in the kernel of α.We note that a := a− α′α(a) is in the kernel of α, since

α(a− α′α(a)) = α(a)− αα′α(a) = α(a)− α(a) = 0.

Here the second equality holds since αα′ = idA′ . We check that modifying a by subtractingα′α(a) doesn’t change its image under f :

f(a) = f(a− α′α(a)) = f(a)− f(α′α(a)) = b− β′f ′α(a) = b− β′βf(a) = b− β′βb = b

Here the third (resp. fourth) equation holds due to the commutativity of the left square ofthe first (resp. second) diagram. The last equality follows from the assumption b ∈ ker β.


Part (b). Consider the following commutative diagram

∂ // Hk(U ∩ V ) //

��

Hk(U)⊕Hk(V ) //

��

Hk(X) //

��∂ // Hk(pt) // Hk(pt)⊕Hk(pt) // Hk(pt) //

where the vertical maps are induced by the projection maps to pt. The rows are exactsequences: the top row is the Mayer-Vietoris sequence for X = U ∪V , while the bottom rowis the Mayer-Vietoris sequence for X ′ = U ′ ∪ V ′, where X ′ = U ′ = V ′ = pt. We note thatthe choice of a point x ∈ U ∪ V can be interpreted as a map X ′ → X compatible with thedecompositions X = U ∪ V , X ′ = U ′ ∪ V ′, i.e., we have maps U ′ → U , V ′ → V , U ′ ∩ V ′ →U ∩ V , compatible with the inclusion maps and hence these maps induce homomorphismfrom the bottom Mayer-Vietoris sequence to the top one which gives a commutative diagram(like the second diagram in part (a)). Finally, the composition

ptx // U ∩ V p // pt

of the inclusion map and the projection map is the identity map of pt, and hence the analo-gous statement holds for the homomorphisms they induce in homology. The correspondingstatement holds for U ∩ V replaced by U , V or X, and hence we can apply part (a) toargue that the sequence given by the kernels of the vertical maps is exact. The kernel of thevertical maps are by definition the reduced homology groups. �


1. Let 0 → A∗f∗−→ B∗

g∗−→ C∗ → 0 be a short exact sequence of chain complexes. Showthat the sequence

... −→ Hq(A)f∗−→ Hq(B)

g∗−→ Hq(C)∂−→ Hq−1(A) −→ ...

is exact at Hq(C) and Hq(B).

Proof. Exactness at Hq(C). We need to show im g∗ = ker ∂. Recall that for q ∈ Z, thechain map g∗ induces a map g∗ : Hq(B) −→ Hq(C) where g∗([b]) = [g(b)]. Let g∗([b]) ∈im g∗ ⊂ Hq(C), where [b] ∈ Hq(B) (so b ∈ Zq(B)), then ∂g∗([b]) = ∂([g(b)]). Recall theconstruction of ∂: ∂ assigns c, which is g(b) in our case, to an a ∈ Aq−1 for which f(a) = ∂(b′),with b′ chosen so that c = g(b) = g(b′) for b ∈ Bq. Since we showed that the choice of such ab′ is irrelevant, we can then choose b′ to be b. Hence, f(a) = ∂(b) = 0 since b ∈ Zq(B). Themap fq is injective (by the exactness of the given sequence), so we have that a = 0, whence∂g∗([b]) = ∂([g(b)]) = [0]. Therefore, im g∗ ⊂ ker ∂ at Hq(C).

To show im g∗ ⊃ ker ∂, let [c] ∈ ker ∂ ⊂ Hq(C). Using the notation from the constructionof ∂, there exists a b ∈ Bq such that g(b) = c and an a ∈ Aq−1 for which f(a) = ∂(b), andsince [c] ∈ ker ∂, ∂a′ = a for some a′ ∈ Aq. We notice that the element b − f(a′) ∈ Bq is aq-cycle since

∂(b− f(a′)) = ∂b− ∂f(a′) = f(a)− f∂a′ = 0,

whence [b− f(a′)] ∈ Hq(B). Furthermore, by the exactness of the original sequence, gf = 0,so


g(b− f(a′)) = g(b)− gf(a′) = g(b)− 0 = g(b) = c.

Thus, [c] = [g(b− f(a′))] = g∗([b− f(a′)]) ∈ im g∗, and im g∗ ⊃ ker ∂.

Exactness at Hq(B). To show im f∗ ⊂ ker g∗, let f∗([a]) = [f(a)] ∈ im f∗ ⊂ Hq(B),where a ∈ Zq(A). Then, by the exactness of the original sequence, gf = 0, so g∗([f(a)]) =[gf(a)] = [0]. Hence, im f∗ ⊂ ker g∗.

To show im f∗ ⊃ ker g∗, let [b] ∈ Hq(B) such that g∗([b]) = [0]; i.e., g(b) ∈ im(∂ : Cq+1 →Cq). Therefore, ∃c ∈ Cq+1 such that ∂(c) = g(b). Since g is onto (by the exactness of thegiven sequence), ∃b′ ∈ Bq+1 such that g(b′) = c. The commutativity of the diagram (of thechain complexes, chain maps, and boundary maps) yields:

g∂(b′) = ∂g(b′) = ∂(c) = g(b).

Since each g is a homomorphism, we then have g(b− ∂(b′)) = 0, so b− ∂(b′) ∈ ker g = im f .Thus, ∃a ∈ Aq such that f(a) = b − ∂(b′). Again, by the commutativity of the diagram,f(∂a) = ∂f(a) = ∂b − ∂∂b′ = ∂b = 0, so a = 0 since f is injective (ker f = 0). Thus,a ∈ Zq(A) and [a] ∈ Hq(A). Moreover, since b−f(a) = ∂(b′), f∗([a]) = [f(a)] = [b] ∈ Hq(B).Thus, im f∗ ⊃ ker g∗. To recap, we now have that:

... −→ Hq(A)f∗−→ Hq(B)

g∗−→ Hq(C)∂−→ Hq−1(A) −→ ...

is exact at Hq(C) and Hq(B). �

2. Suppose the following diagram of abelian groups and group homomorphisms is commu-tative with exact rows:

−−−→ Cq+1∂q+1−−−→ Aq

fq−−−→ Bqgq−−−→ Cq

∂q−−−→ Aq−1 −−−→cq+1

y aq

y bq

y cq

y aq−1

y−−−→ C ′q+1

∂′q+1−−−→ A′qf ′q−−−→ B′q

g′q−−−→ C ′q∂′q−−−→ A′q−1 −−−→

Assuming in addition that the maps cq are isomorphisms show that there is a long exactsequence of the form

// Aq // A′q ⊕Bq// B′q // Aq−1

// A′q−1 ⊕Bq−1// B′q−1

//

First define carefully the homomorphisms in the above sequence. Then prove exactness ateach location.

Proof. We define the maps in the above sequence as follows:

αq : Aq −→ A′q ⊕Bq a 7→ (aq(a), fq(a))

βq : A′q ⊕Bq −→ B′q (a′, b) 7→ f ′q(a′)− bq(b)

γq : B′q −→ Aq−1 b′ 7→ ∂qc−1q g′q(b

′)

Exactness at B′q. First we show γqβq = 0. For (a′, b) ∈ A′q ⊕Bq we have

γqβq(a′, b) = ∂qc

−1q g′q(f

′qa′ − bqb) = −∂qc−1

q g′qbqb = ∂qgqb = 0

Here the second equality holds due to g′qf′q = 0, the third follows from the commutativity of

the third square, and the last is due to ∂qgq = 0.


To show ker γq ⊂ imβq let b′ ∈ B′q with γqb′ = ∂qc

−1q g′qb

′ = 0. By exactness at Cq there is

an element b ∈ Bq such that gqb = c−1q g′qb

′ or equivalently

g′qb′ = cqgqb = g′qbqb,

where the second equality follows from commutativity of the third square. It follows thatb′ − bqb is in the kernel of g′q and hence by exactness at B′q, there is an element a′ ∈ A′q withf ′qa′ = b′ − bqb. This implies

βq(a′,−b) = f ′qa

′ + bqb = b′

which shows that b′ is in the image of βq.Exactness at A′q ⊕Bq. First we show βqαq = 0. For a ∈ Aq we have

βqαqa = βq(aqa, fqa) = f ′qaqa− bqfqa = 0

due to the commutativity of the second square.To show ker βq ⊂ imαq, let (a′, b) ∈ A′q ⊕Bq with

βq(a′, b) = f ′qa

′ − bqb = 0.

Then we havecqgqb = g′qbqb = g′qf

′qa′ = 0,

where the first equality is due to the commutativity of the third square, and the last is dueto exactness at B′q. Since cq is an isomorphism, this implies gqb = 0 and hence by exactnessat Bq, there is an element a ∈ Aq with fqa = b. If we could show aqa = a′, we would bedone. However we can only say the following:

f ′q(aqa− a′) = f ′qaqa− f ′qa′ = bqfqa− bqb = 0

where the second equality follows from the commutativity of the second square and ourassumption f ′qa

′ = bqb. Since f ′q is not necessarily injective, we can’t conclude that aqa = a′,but thanks to exactness at A′q, it implies that there is an element c′ ∈ C ′q+1 with ∂q+1c

′ =aqa − a′. Moreover, since cq+1 is an isomorphism, there is a c ∈ Cq+1 with cq+1c = c′. Nowwe modify the element a ∈ Aq by defining a := a− ∂q+1c. We calculate

fqa = fq(a− ∂q+1c) = fqa = b

aqa = aq(a− ∂q+1c) = aqa− ∂′q+1cq+1c = aqa− (aqa− a′) = a′

This shows that αq(a) = (a′, b) as desired.Exactness at Aq. First let us show αq ◦ γq+1 = 0. For b′ ∈ B′q+1 we have

αqγq+1b′ = αq(∂q+1c

−1q+1g

′q+1b

′)

= (aq∂q+1c−1q+1g

′q+1b

′, fq∂q+1c−1q+1g

′q+1b

′)

= (∂′q+1g′q+1b

′, 0) = (0, 0)

since the compositions fq∂q+1 and ∂′q+1g′q+1 are zero due to the exactness at Aq resp. A′q.

To show kerαq ⊂ imγq+1, let a ∈ Aq with αqa = (aqa, fqa) = (0, 0). By exactness at Aqthere is an element c ∈ Cq+1 with ∂q+1c = a. Then

∂′q+1cq+1c = aq∂q+1c = aqa = 0

and hence by exactness at C ′q+1, there is an element b′ ∈ B′q+1 with g′q+1b′ = cq+1c. This

impliesγq+1b

′ = ∂q+1c−1q+1g

′q+1b

′ = ∂q+1c = a,


which shows that a is in the image of γq+1. �

3. Prove the Brouwer Fixed Point Theorem: Any continuous map f : Dn → Dn from theclosed n-disk to itself has a fixed point, that is, there is some point x0 ∈ Dn such thatf(x0) = x0.

Hint: Prove by contradiction. More precisely, assuming that f has no fixed point, constructa retraction r : Dn → Sn−1 (i.e., a continuous map whose restriction to Sn−1 is the identity),contradicting the fact proved in class that Sn−1 is no retract of Dn.

Proof. Assuming that f has no fixed point, the line through x and f(x) intersects the sphereSn−1 in exactly two points. Let r(x) ∈ Sn−1 be the intersection point closer to x than tof(x). If x is a point on the sphere, then x is clearly that intersection point; in other words,the map r : Dn → Sn−1 restricted to Sn−1 is the identity on Sn−1. It remains to show thatr is continuous.

The idea is to write down a formula for r(x), and to argue that r is a composition of basicfunctions that we know are continuous from calculus. To derive the formula for r(x), wenote that every point of the line through f(x) and x is of the form

x+ t(x− f(x)) for t ∈ R.

Moreover, the point r(x) = x+t(x−f(x)) is characterized by the two conditions ||r(x)||2 = 1and t ≥ 0. The first condition is a quadratic equation for t. Explicitly, setting v = v(x) =x− f(x) we have

||r(x)||2 = 〈r(x), r(x)〉 = 〈tv + x, tv + x〉 = ||v||2t2 + 2〈v, x〉t+ ||x||2

and hence the quadratic formula gives us the following non-negative solution for t:

(3) t =−b+

√b2 − 4ac

2awhere a = ||v(x)||2, b = 2〈v(x), x〉, c = ||x||2 − 1.

We note that the functions a(x), b(x), c(x) are continuous maps Dn → R, since they areexpressed as compositions of functions that are well-known to be continuous. This impliesthat the function t = t(x) defined by equation (3) is a continuous function: the only thing tomake sure is that the denominator function a(x) = ||v(x)||2 is nowhere 0 (this is guaranteedby the assumption that v(x) = x− f(x) 6= 0 for all x ∈ Dn), and that the expression underthe square root is positive (this follows geometrically since we have always two intersectionpoints of the line through x and f(x) and the sphere; this is equivalent to having two solutionsof the quadratic equation; this in turn is equivalent to a positive expression under the squareroot in the quadratic formula). Putting everything together, we conclude that the map

r : Dn −→ Sn−1 given by r(x) = x+ t(x)(x− f(x))

is continuous since t(x) and f(x) are continuous maps. �

4. The open cone CX of a topological space X is the quotient space

CX := (X × [0, 1))/(X × {0}).

(a) Show that the cone of Sn−1 is homeomorphic to the open n-disk Dn. In particular, thecone of Sn−1 is a manifold of dimension n.


(b) Show that the cone of X is not a manifold of dimension n if the homology groups of Xare not isomorphic to those of Sn−1. Hint: Relate the local homology groups of CX atthe cone point (the image of X × {0} under the projection map X × [0, 1) → CX) tothe homology groups of X.

Proof. Part (a). Consider the map

f : CSn−1 −→ Dn given by [x, t] 7→ xt.

The map f is a continuous, since its composition with the projection map Sn−1 × [0, t) →CSn−1 is given by (x, t) → xt, and hence continuous as the restriction of the continuousmultiplication map Rn × R → Rn, (x, t) 7→ xt. It is straightforward to check that f is abijection, and hence it remains to show that its inverse is a bijection. We note that thedomain is not compact, but the closed cone (X × [0, 1])/(X ×{0}) on X = Sn−1 is compact(as quotient of the product of the spaces Sn−1 and [0, 1] which are compact by Heine-Borel).The map f extends in an obvious way to a continuous bijection from the closed cone to theclosed disk Dn. This extension then is a homeomorphism, since its domain is compact andits codomain Hausdorff. Restricting this homeomorphism to the open cone gives the desiredhomeomorphism.

Part (b). Let c ∈ CX be the cone point. To calculate the local homology groupsHq(CX,CX \ c) we will use the long exact homology sequence of the pair (CX,CX \ c):

// Hq(CX) // Hq(CX,CX \ c)∂ // Hq−1(CX \ c) // Hq−1(CX) //

We observe that the cone CX deformation retracts onto the cone point, and hence thereduced homology groups of CX vanish. This implies that the boundary map provides an

isomorphism between the q-th local homology group of CX at c and Hq−1(X). In particular,if CX is a manifold of dimension n, then

Hq−1(X) ∼= Hq(CX,CX \ c) ∼=

{Z q = n

0 q 6= n,

and hence the homology groups of X are isomorphic to those of Sn−1. �


1. For a polynomial p(z) = anzn + · · ·+ a0 of degree n with complex coefficients ai, let

p : CP1 −→ CP1 be given by [z0, z1] 7→ [anzn0 + an−1z

n−10 z1 + · · ·+ a0z

n1 , z

n1 ].

Using the fact that CP1 is homeomorphic to S2, we can interpret p as a map p : S2 → S2

which has a degree deg(p) ∈ Z.

(a) Show that deg(p) = n for the polynomial p(z) = zn.(b) Show that deg(q) = n for any polynomial q(z) of degree n. Hint: construct a homotopy

between q and p.


Proof. Part (a). We will calculate the degree of p : CP1 → CP1 as the sum of the localdegrees of p at the preimage points of a suitable point y ∈ CP1. For determining p−1(y) itis useful to note that the diagram

C� _i��

p // C� _i��

CP1 p // CP1

is commutative, where the vertical map i is the embedding given by mapping z ∈ C to[z, 1] ∈ CP1. If we chose y = 1 ∈ C ⊂ CP1, then p−1(y) = p−1(1) is the set {1, ζ, ζ2, . . . , ζn−1}of n-th roots of unity with ζ := e2πi/n.

To calculate the local degree deg(p, z) at a point z ∈ C, we will argue that the derivativeDpz : R2 → R2 is invertible and we use part (b) of problem # 3 in this assignment accordingto which deg(p, z) = sign det(Dpz). We observe that the map p(z) = zn is holomorphicand hence its derivative Dpz is the endomorphism of R2 = C given by multiplication by thecomplex number ∂p

∂z= nzn−1. This is a non-zero complex number for any z 6= 0. To calculate

det(Dpz), we write nzn−1 = a+ ib for a, b ∈ R. Using the basis {1, i} for C as a real vectorspace, we have

Dpz(1) = a+ ib Dpz(i) = i(a+ ib) = −b+ ia

and hence the matrix representing the linear map Dpz : R2 → R2 is(a −bb a

). It follows that

det(Dpz) = a2 + b2 > 0 for z 6= 0. In particular, deg(p, z) = 1 for any 0 6= z ∈ C. Hence

deg(p) =n−1∑k=0

deg(p, ζk) = n.

Part (b). Let q(z) be the polynomial q(z) = anzn + · · ·+ a0, an 6= 0. For t ∈ [0, 1] let qt(z)

be the polynomialqt(z) = anz

n + t(an−1zn−1 + · · ·+ a0).

We note that for any t this is again a polynomial of degree n. This guarantees that we canextend qt to a well-defined map qt : CP1 → CP1 by the formula above (note that it is essentialto assume that qt has degree n, since if the coefficient of zn is zero, then qt is not well-definedsince it would map [1, 0] ∈ CP1 to [0, 0] which is not a point of CP1!). This shows thatq = q1 is homotopic to q0 and hence these maps have the same degree. It remains to showthat there is a homotopy of polynomials of degree n between q0(z) = anz

n and p(z) = zn

to conclude that deg(q0) = deg(p) = n. Such a homotopy is given by pt(z) = etwzn, wherew ∈ C with ew = an ∈ C× and t ∈ [0, 1]. �

2. Prove the following statements for the local degree.

(a) Show that if f : Rn → Rn is a linear isometry (i.e., f belongs to the orthogonal group),then deg(f, 0) = deg(f|Sn−1). Hint: apply the statement deg(Σg) = deg(g) (which weproved in class) to g = f|Sn−1 , and apply the theorem expressing the degree of a map asa sum of local degrees to Σg.

(b) Show that if fn : Rn → Rn is the reflection map (x1, . . . , xn) 7→ (−x1, x2, . . . , xn),then deg(fn, 0) = −1. Hint: argue by induction over n, first using part (a) to showthe statement for n = 1. For the inductive step use part (a) again to argue thatdeg(fn+1, 0) = deg((fn+1)|Sn , en+1), where en+1 = (0, . . . , 0, 1) ∈ Sn. Then show thatdeg((fn+1)|Sn , en+1) = deg(fn, 0) by showing that (fn+1)|Sn near en+1 ∈ Sn corresponds


to fn near 0 ∈ Rn via a homeomorphism from a neighborhood of en+1 to a neighborhoodof 0.

Proof. Part (a). Following the hint,

deg(f|Sn−1) = deg(g) = deg(Σg) = deg(Σg, c−),

where Σg : ΣSn−1 → ΣSn−1 is the suspension of g, and c− = [x, 0] ∈ ΣSn−1 is the bottomcone point (I’m thinking of the suspension ΣSn−1 as the union of the two cones C− = {[x, t] ∈ΣSn−1 | t ∈ [0, 1/2]} with cone point c− and C− = {[x, t] ∈ ΣSn−1 | t ∈ [1/2, 1]} with conepoint c+ = [x, 1]). The last equality follows from the theorem expressing the degree of themap Σg as the sum of the local degree of Σg at the points of (Σg)−1(c−) = {c−}. To show

that deg(Σg, c−) = deg(f, 0) we will construct a homeomorphism h : C−≈−→ Dn which maps

c− to 0 such that the diagram

C−h

≈//

Σg|C−��

Dn

f|Dn

��C−

h

≈// Dn

commutes. The induced diagram of local homology groups

Hn(C−, C− \ c−)h∗∼=//

(Σg|C− )∗��

Hn(Dn, Dn \ 0)

(f|Dn )∗��

Hn(C−, C− \ c−)h∗∼=// Hn(Dn, Dn \ 0)

then implies deg(Σg, c−) = deg(f, 0). We define the map

h : C− −→ Dn by h([x, t]) = 2tx

This is evidently a continuous bijection. Since C− is compact (as quotient of the prod-uct of the compact spaces Sn−1 and [0, 1/2]) and Dn is Hausdorff, it follows that h is ahomeomorphism.

Part (b). We prove the statement deg(fn, 0) = −1 by induction over n. For n = 1 we have

deg(f1, 0) = deg((f1)|S0)

by part (a). To calculate the degree of g = (f1)|S0 we need to determine the map it induces

on H0(S0) = ker(H0(S0) → H0(pt)). An element of H0(X) for any space X is representedby a finite linear combination

∑i nixi of points xi ∈ X with coefficients ni ∈ Z. Such a sum

belongs to H0(X) if and only if∑

i ni = 0. In particular, X = S0 consists of the two points

x1 := 1 and x2 := −1, and the reduced homology group H0(S0) is the infinity cyclic groupgenerated by the homology class [x1 − x2] ∈ H0(S0). Since the map g permutes the twopoints x1, x2 we have

g∗([x1 − x2]) = [g(x1)− g(x2)] = [x2 − x1] = −[x1 − x2],

which proves that g has degree −1.we note that S0 = {±1} ⊂ R.


For the inductive step, let g : Sn → Sn be the restriction of fn+1 to Sn ⊂ Rn+1. Then wehave

deg(fn+1, 0) = deg(g) = deg(g, en+1) = deg(fn, 0).

Here the first equation is a consequence of part (a), the second equation follows again from theresult expressing the degree as a sum of local degrees at the points belonging to the preimageg−1(en+1) = {en+1}, en+1 = (0, . . . , 0, 1) ∈ Rn+1. The argument for the last equation is as inpart (a): we show that the map g restricted to a neighborhood of en+1 ∈ Sn is isomorphic tothe map fn restricted to a neighborhood of 0 ∈ Rn, namely the homeomorphism h from theupper hemisphere Dn

+ ⊂ Sn to the disk Dn given by (v1, . . . , vn, vn+1) 7→ (v1, . . . , vn) (with

inverse v = (v1, . . . , vn) 7→ (v1, . . . , vn,√

1− ||v||2)) maps en+1 to 0 and makes the diagram

Dn+

h

∼=//

g

��

Dn

fn��

Dn+

h

∼=// Dn

commutative. The induced diagram of local homology groups then implies

deg(g, en+1) = deg(fn, 0).

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3. Prove the following further statements for the local degree.

(a) Show that if f : Rn → Rn is a linear isomorphism, then deg(f, 0) = sign det(f). Hint:First show that the statement holds when f is reflection at the hypersurface v⊥ for anynon-zero v ∈ Rn. Then use the fact that any linear isometry f : Rn → Rn can be writtenas a composition of such reflections to argue that the statement holds for such f . Finallyshow that the statement holds for any isomorphism f by finding a path of isomorphismsft connecting f and a linear isometry.

(b) Let f : Rn → Rn be a continuous map which is differentiable at the point x0 ∈ Rn. LetDfx0 be the derivative at x0 (which is a linear map Dfx0 : Rn → Rn; the correspondingmatrix is the Jacobian of f at the point x0). Show that if Dfx0 is invertible, then

deg(f, x0) = sign det(Dfx0).

Hint: Use the assumption that f is differentiable at x0 to write f(x) in the form f(x) =f(x0) + Dfx0(x − x0) + e(x − x0) where the error term e(h) is o(h) for h → 0. Use thehomotopy ft(x) := f(x0) +Dfx0(x− x0) + te(x− x0) to argue (carefully!) that the localdegrees of f0 and f1 = f at x0 agree.

Proof. Part (a). As suggested by the hint, we first show deg(f, 0) = −1 if f is the reflectionRv at the hyperplane v⊥ orthogonal to a non-zero vector v ∈ Rn. By part (b) the statementholds for v = e1 = (1, 0, . . . , 0). We recall that the reflection

Rv : Rn → Rn is given explicitly by Rv(w) = w − 2〈w, v〉||v||2

v.

This shows that if v(t), t ∈ I is a path of non-zero vectors, then Rv(t) is a homotopy betweenRv(0) and Rv(1) and hence their local degrees at 0 agree. For any non-zero v ∈ Rn, n > 1,


with v⊥ 6= e⊥1 the linear path (1− t)e1 + tv does not go through zero (since v, e1 are linearlyindependent), and hence deg(Rv, 0) = deg(Re1 , 0) = −1.

For a general element of the orthogonal group O(n) consisting of linear isometries f : Rn →Rn, we use the fact that it can be written as a composition of reflections Rv1 ◦ · · · ◦ Rvk .Using the fact that the degree of a composition is the product of the degrees we have

deg(f, 0) = deg(Rv1 , 0) · · · deg(Rvk , 0) = (−1)k.

For the determinant we have the analogous equation

det(f) = det(Rv1) · · · det(Rvk) = (−1)k,

which implies that deg(f, 0) = det(f) for any f ∈ O(n).To prove the result for a general isomorphism f : Rn → Rn, it suffices to show that there

is a path f(t) in the group GLn(R) of isomorphism of Rn which connects f = f(0) with anelement f(1) ∈ O(n) ⊂ GLn(R). We note that the sign of det(f(t)) is independent of t,since it depends continuously on t, and det(f(t)) 6= 0 for all t ∈ [0, 1].

To construct the path, we identify linear maps Rn → Rn with n× n matrices in the usualway. We will write a matrix A in the form A = (a1, a2, . . . , an), where ai ∈ Rn are thecolumn vectors of the matrix. We recall that A belongs to GLn(R) if and only if the vectorsa1, . . . , an are linearly independent; A ∈ O(n) if and only if the vectors ai are unit vectorswhich are mutually perpendicular. Let pi : Rn → Rn be the orthogonal projection onto thesubspace spanned by a1, . . . , ai, and let p⊥i : Rn → Rn be the orthogonal projection onto theorthogonal complement of that subspace; in particular, we have v = p⊥i v+piv for all v ∈ Rn.Given a matrix A, inspired by the Gram-Schmidt procedure, let us define

At = (v1, p⊥1 v2 + tp1v2, . . . , p

⊥i−1vi + tpi−1vi + · · ·+ p⊥n−1vn + tpn−1vn)

We note that det(At) is independent of t, since the determinant depends linearly on eachcolumn vector and this determinant is zero if the vectors are linearly dependent. In particular,if A = A1 invertible, then so is At. The column vectors ofB = A0 are mutually perpendicular,but not necessarily of unit length. Now for B = (w1, . . . , wn) we define

Bt = ((1− t)w1 + tw1

||w1||, . . . , (1− t)wn + t

wn||wn||

),

which is a path connecting B = B0 with B1 ∈ O(n). Concatenating these paths, we obtainthe desired path from A to B0 ∈ O(n).

Part(b). To prove part (b), we note that the assumption that f is differentiable at x0 meansthat f(x) can be written in the form

f(x) = f(x0) +Dfx0(x− x0) + e(x− x0),

where the ‘error term’ e(h) is o(h) for h→ 0, which means

(4) limh→0

e(h)

|h|= 0.

We define

f(x)t := f(x0) +Dfx0(x− x0) + te(x− x0),

and want to argue that ft is a map of pairs

ft : (Bε(x0), Bε(x0) \ {x0}) −→ (Rn,Rn \ f(x0)),


for sufficiently small ε > 0, where Bε(x0) is the ball of radius ε around x0. In other words,we want to argue that f−1

t (f(x0))∩Bε(x0) = {x0}, or equivalently, that Dfx0(h) + te(h) 6= 0for all h with 0 < h < ε. The idea is to show that for 0 < h < ε the norm of Dfx0(h) is largecompared to the norm of te(h).

To make this precise, let m := minh∈Sn−1 ||Dfx0(h)||. We note that m > 0, since m =||Dfx0(h0)|| for some h0 ∈ Sn−1, and Dfx0(h0) 6= 0 due to our assumption that Dfx0 is

invertible. Now the statement (4) allows us to choose ε > 0 such that ||e(h)|||h| < m for

||h|| < ε. This implies that for 0 < ||h|| < ε we have

||te(h)|| ≤ ||e(t)|| < m||h|| ≤ ||Dfx0(h

||h||)|| ||h|| = ||Dfx0(h)||

and hence Dfx0(h) + te(h) 6= 0 as desired. We conclude that f = f1 is homotopic to g = f0

as maps from (Bε(x0), Bε(x0) \ {x0}) to (Rn,Rn \ f(x0)) and hence deg(f, x0) = deg(g, x0).

Finally, we want to compare deg(g, x0) and deg(D, 0), where D = Dfx0 . We note that byconstruction of g the following diagram is commutative

Bε(0)Dfx0 //

Tx0 ≈��

Rn

Tf(x0) ≈��

Bε(x0)g // Rn

where Tx0 , Tf(x0) are translation maps defined for v ∈ Rn by Tv : Rn → Rn, w 7→ w+ v. Thecorresponding diagram of local homology groups then implies deg(Dfx0) = deg(g, 0).

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4. Let (X, V,A) be a triple of topological spaces (i.e., A ⊂ V ⊂ X). Show that there is along exact sequence of homology groups

. . . −→ Hq(V,A) −→ Hq(X,A) −→ Hq(X, V )∂−→ Hq−1(V,A) −→ . . .

Hint: use the algebraic fact that a short exact sequence of chain complexes leads to a longexact sequence of homology groups.

Proof. Let i : (V,A)→ (X,A) and j : (X,A)→ (X, V ) be the maps of pairs induced by theinclusion map V → X resp. the identity on X. The induced maps on singular q-chains

Cq(V,A)iq // Cq(X,A)

jq // Cq(X, V )

Cq(V )/Cq(A) Cq(X)/Cq(A) Cq(X)/Cq(V )

form a short exact sequence since jq is an epimorphism whose kernel is equal to Cq(V )/Cq(A) ⊂Cq(X)/Cq(A).

This implies that

C∗(V,A)i∗ //C∗(X,A)

j∗ //C∗(X, V )

is a short exact sequence of chain complexes which implies the desired long exact sequenceof homology groups. �



1. Prove the following statement which is known as the 5-lemma. Suppose we have acommutative diagram of abelian groups and group homomorphisms

A1f1 //

h1��

A2f2 //

h2��

A3f3 //

h3��

A4f4 //

h4��

A5

h5��

B1g1 // B2

g2 // B3g3 // B4

g4 // B5

such that the rows are exact sequences. Show that if the vertical maps h1, h2, h4, and h5

are isomorphisms, then also the middle map h3 is an isomorphism.Remark: the assumptions that the maps h1, h2, h4, h5 are all isomorphisms are slightly

stronger than needed for the proof. What weaker assumptions will do?

Proof. We first prove that h3 is injective. Let a3 ∈ A3 such that h3a3 = 0 ∈ B3. Then by thecommutativity of the diagram, we have h4f3a3 = g3h3a3 = g3(0) = 0, so f3(a3) ∈ ker h4 = 0since h4 is injective. Hence, a3 ∈ ker f3 = im f2, so there is an a2 ∈ A2 such thatf2(a2) = a3. Again, by the commutativity of the diagram, we have g2h2(a2) = h3f2(a2) =h3(a3) = 0, so h2(a2) ∈ ker g2 = im g1. Thus, there is an element b1 ∈ B1 such thatg1(b1) = h2(a2). Moreover, since h1 is surjective, there is some a1 ∈ A1 with h1a1 = b1.It follows that h2f1a1 = g1h1a1 = g1b1 = h2a2. Thus, since h2 is injective, a2 = f1a1, soa3 = f2a2 = f2f1a1 = 0. Therefore, h3 is injectiv. To show surjectivity of h3, let b3 ∈ B3.Since h4 is surjective, there is some a4 ∈ A4 with h4a4 = g3b3 ∈ A4 and by commutativity,h5f4a4 = g4h4a4 = g4g3b3 = 0. Since h5 is injective, this implies f4a4 = 0. By exactness ofthe top row at A4, there is an a3 ∈ A3 such that f3(a3) = a4. Hence,

g3(h3(a3)− b3) = h4f3a3 − g3b3 = h4a4 − g3b3 = 0.

By the exactness of the lower row at B3, this implies that there exists b2 ∈ B2 such thatg2(b2) = h3(a3)− b3. Since h2 is surjective, there is some a2 ∈ A2 with h2a2 = b2 and hence

h3(a3 − f2a2) = h3a3 − h3f2a2 = h3a3 − g2h2a2 = b3,

which shows that b3 is in the image of h3. Since b3 was arbitrary, this shows that h3 issurjective.

We see that we’ve used the assumptions that h4, h2 are injective, and that h1 is surjectiveto show injectivity of h3. Our proof that h3 is surjective required the assumptions that h2,h4 are surjective, and that h5 is injective. So it is sufficient to assume that h2 and h4 areisomorphisms, that h1 is an epimorphism, and that h5 is a monomorphism. �

2. Show that the complex projective space CPn is a CW complex with one cell of dimension2i for 0 ≤ i ≤ n.

Proof. It suffices to show that CP n is obtained from CP n−1 by attaching a cell of dimension2n. Define

Φ: D2n −→ CP n

z = (z0, . . . , zn−1) 7→ [z0, z1, . . . , zn−1,√

1− ||z||2],


and let ϕ : S2n−1 → CP n−1 be the natural projection map given by (z0, . . . , zn−1) 7→ [z0, z1, . . . , zn−1].We note that these maps are compatible in the sense that the following diagram is commu-tative

S2n−1 ϕ //

j��

CP n−1

i��

D2n Φ // CP n,

where i, j are the obvious inclusion maps. It follows that the map

CP n−1 ∪ϕ D2n i∪Φ−→ CP n

is well-defined and continuous. We note that this map is surjective, since if [z0, . . . , zn] ∈ CP n

with zn 6= 0, then multiplying all components by the unit complex number z−1n ||zn||, we can

assume w.l.o.g. that zn is a positive real number. Injectivity is obvious and it follows thatf = i ∪ Φ is a continuous bijection. It follows that f is a homeomorphism since the domainof f is compact and its range is Hausdorff. �

3. Calculate Hq(Sm × Sn) for m ≥ n ≥ 1. Use the fact that if X, Y are CW complexes,

then the product X × Y has a CW structure whose cells correspond to products of cellsof X and Y . More precisely, if emα is an m-cell of X, and enβ is an n-cell of Y , then these

determine a m + n-cell of X × Y denoted emα × enβ. In particular, CCWq (X × Y ) is the free

abelian group generated by products cells emα × enβ with m + n = q. The cellular boundarymap is determined by the formula ∂(emα × enβ) = (∂emα )× enβ + (−1)memα × ∂(enβ) .

Proof. To calculate the homology of Sm × Sn, let us furnish the sphere Sm with the CWstructure consisting of one 0-cell e0 and one m-cell em. Then the product CW structure onSm×Sn has four cells e0× e0, em× e0, e0× en, em× en of dimension 0, m, n and m+n. Thecellular boundary map in the cellular chain complex of Sm and Sn is zero, and hence the‘product rule’ for the boundary map in CCW

∗ (Sm × Sn) implies that the cellular boundarymap for Sm×Sn is trivial. Hence Hq(S

m×Sn) ∼= CCWq (Sm×Sn) is a direct sum of as many

copies of Z as there are cells of dimension q. In particular, for m,n ≥ 1 and m > n we have

Hq(Sm × Sn) ∼=

{Z q = 0,m, n,m+ n

0 otherwise

If m = n we have:

Hq(Sn × Sn) ∼=

Z⊕ Z q = n

Z q = 0, 2n

0 otherwise

�

4. (a) Show that if 0 // Ck∂ // Ck−1

∂ // . . .∂ // C0

// 0 is a chain complex of

finitely generated abelian groups with homology groups Hq := Hq(C∗), then

k∑q=0

(−1)q rkCq =k∑q=0

(−1)q rkHq.


Hint: Use the fact that if 0 → A → B → C → 0 is a short exact sequence of finitelygenerated abelian groups, then rkB = rkA + rkC. Show that if Zq (resp. Bq) are theq-cycles (resp. q-boundaries) of C∗, then there are short exact sequences

0→ Zq → Cq → Bq−1 → 0 and 0→ Bq → Zq → Hq → 0.

(b) Show that if X is a finite CW complex, then

k∑q=0

(−1)q rkHq(X) =k∑q=0

(−1)q#{q-cells of X}.

The alternating sum on the left is known as the Euler characteristic of the topological spaceX, denoted χ(X). The statement above shows that if X is a surface, this definition agreeswith our previous definition as the alternating sum of the number of vertices, edges and facesof a pattern of polygons on X (since we can now interpret such a pattern as providing a CWstructure for the surface X).

(c) Let Σ be a surface of genus g, and let Σ be a surface which is a d-fold covering of Σ.

Show that the genus g of Σ is given by the formula g = dg − d+ 1.

(d) According to the formula above, the surface of genus 4 is a 3-fold covering of thesurface of genus 2; in particular, there is a free action of the cyclic group of order threeon the surface of genus 4. Draw a picture of a surface of genus 4 which has an obviousZ/3-symmetry.

Proof. Part (a). By definition of Bq−1 it is the image of the homomorphism ∂ : Cq → Cq−1;the kernel of the surjective map ∂ : Cq → Bq−1 is, again by definition, the group of q-cyclesZq. This gives the first exact sequence. The second sequence is exact since the homologygroup Hq by definition is the quotient group Zq/Bq. Now we calculate ranks

rkCq = rkZq + rkBq−1 = rkHq + rkBq + rkBq−1.

Hence∑q

(−1)q rkCq =∑q

(−1)q rkHq +∑q

(−1)q rkBq +∑q

(−1)q rkBq−1 =∑q

(−1)q rkHq,

since the last two sums cancel.

Part (b). Applying part (a) to the cellular chain complex CCW∗ (X), we have

k∑q=0

(−1)q rkHq(X) =k∑q=0

(−1)q rkHq(CCW∗ (X))

=k∑q=0

(−1)q rkCCWq (X) =

k∑q=0

(−1)q#{q-cells of X}.

Part (c). From class we know that the Euler characteristic χ(Σ) of a closed connectedsurface of genus g is given by

χ(Σ) = 2− 2g.


Moreover, by problem (2) of the homework assignment # 1, we know that χ(Σ) = dχ(Σ).It follows that

g = 1− χ(Σ))

2= 1− dχ(Σ)

2= 1− d(2− 2g)

2= 1− d+ dg.

Part (d). Here is the picture of a surface Σ of genus 4 with a rotational Z/3-symmetry.

Note that the Z/3-action on this surface is free, and that the quotient space Σ = Σ/Z/3 isa surface of genus 2. Thinking of the surface of genus 2 as the connected sum to two tori,

the circle separating the two tori is triply covered by the three circles in Σ which separatethe “central hole” from the three “outer holes”.

�


1. Calculate the homology groups of a product of two Moore spaces M(Z/k,m)×M(Z/`, n)for m ≥ n ≥ 1. Recall that the Moore space M(Z/k,m) is the space obtained from Sm byattaching an (m+ 1)-cell via an attaching map ϕ : Sm → Sm of degree k. Hint: Decomposethe cellular chain complex of M(Z/k,m)×M(Z/`, n) into a direct sum of chain complexes.

Proof. The Moore space M(Z/k,m) has three cells e0, em, em+1 whose dimensions are indi-cated by the superscripts. By construction, the Moore space M(Z/k,m) is obtained fromthe sphere Sm by attaching an m + 1-cell via an attaching map ϕ : Sm → Sm of degree k,and hence the boundary map ∂ in its cellular chain complex maps em+1 to kem. The product

M(Z/k,m)×M(Z/`, n)

has nine cells. To simplify matters, we note that the cell e0 does not ‘interact’ with the cellsem, em+1 in the cellular chain complex of M(Z/k,m) in the sense that CCW

∗ (M(Z/k,m))is a direct sum of two chain complexes; the first summand is generated by e0, the other is


generated by em and em+1. It follows that the chain complex of the product can be writtenas a direct sum of four chain complexes:

A∗ = 〈e0 × e0〉B∗ = 〈e0 × en, e0 × en+1〉C∗ = 〈em × e0, em+1 × e0〉D∗ = 〈em × en, em × en+1, em+1 × en, em+1 × en+1〉

The boundary maps of these chain complexes are readily determined by the product formulafor ∂. Also the homology groups of A∗, B∗ and C∗ are straightforward to determine and sowe only state the result:

Hi(A∗) =

{Z i = 0

0 i 6= 0

Hi(B∗) =

{Z/` i = n

0 i 6= n

Hi(C∗) =

{Z/k i = m

0 i 6= m

To calculate the homology groups of the chain complex D∗, we first determine the boundarymaps of D∗:

∂(em × en) = 0

∂(em+1 × en) = k(em × en)

∂(em × en+1) = (−1)m`(em × en)

∂(em+1 × en+1) = k(em × en+1) + (−1)m+1`(em+1 × en)

To determine the homology groups it is convenient to use a different basis for the chaingroup Dm+n+1. Let g = gcd(k, `), and write k = gk′, ` = g`′. Then k′ and `′ are relativelyprime and hence there are integers a, b such that ak′ + b`′ = 1. We note that the element∂(em+1 × en+1) is divisible by g. Let e1 ∈ Dm+n+1 be the element obtained by that division,i.e.,

e1 := k′(em × en+1) + (−1)m+1`′(em+1 × en)

and define

e2 := (−1)mb(em × en+1) + a(em+1 × en)

This is taylor-made so that the base change matrix(k′ (−1)mb

(−1)m+1`′ a

)has determinant one which implies that {e1, e2} is in fact a new basis for Dm+n+1. Theadvantage of this new basis is that the boundary map of the chain complex D∗ has the


following simple form

∂(em+1 × en+1) = ge1

∂(e1) = 0

∂(e2) = (−1)mb∂(em × en+1) + a∂(em+1 × en)

= (−1)mb(−1)m`(em × en) + ak(em × en)

= (b`+ ak)(em × en) = g(em × en)

It follows that

Hq(D∗) =

{Z/g q = m+ n,m+ n+ 1

0 otherwise

�

2. (a) Show that the relation “chain homotopic” is transitive. In other words, show that iff, g, h : C∗ → D∗ are chain maps, and f is chain homotopic to g, and g is chain homotopicto h, then f is chain homotopic to h.

(b) Let f, g : C∗ → D∗ and h, k : D∗ → E∗ be chain maps. Show that if f is chainhomotopic to g and h is chain homotopic to k, then h ◦ f is chain homotopic to k ◦ g.

Proof. Let S : Cq → Dq+1 be a chain homotopy from f to g and let T : Cq → Dq+1 be a chainhomotopy from g to h, that is,

∂S + S∂ = g − f and ∂T + T∂ = h− g.Then ∂(S + T ) + (S + T )∂ = ∂S + ∂T + S∂ + T∂ = g − f + h − g = h − f , which showsthat S + T : Cq → Dq+1 is a chain homotopy from f to h. This proves part (a).

To prove part (b), let S : Cq → Dq+1 be a chain homotopy from f to g, and let T : Dq →Eq+1 be a chain homotopy from h to k, that is

∂S + S∂ = g − f and ∂T + T∂ = k − h.Then

∂hS + hS∂ = h∂S + hS∂ = h(∂S + S∂) = h(g − f) = hg − hf∂Tg + Tg∂ = ∂Tg + T∂g = (∂T + T∂)g = (k − h)g = kg − hg.

Here both second equalities follow from our assumption that h (resp. g) are chain maps.Adding both terms, we obtain

∂(hS + Tg) + (hS + Tg)∂ = kg − hf,which proves that hS + Tg : Cq → Eq+1 is a chain homotopy from hf to kg. �

3. Let f, g : C∗ → D∗ be two chain maps which are chain homotopic. Show that the inducedmaps in homology f∗, g∗ : Hq(C∗)→ Hq(D∗) are equal.

Proof. Let T : Cq → Dq+1 be a chain homotopy from g to f , that is, ∂T + T∂ = f − g. Letz ∈ Zq(C∗) be a q-cycle representing the homology class [z] ∈ Hq(C∗). Then

f∗([z]) = [fz] = [∂Tz + T∂z + gz] = [∂Tz + gz] = [gz] = g∗([z]).

Here the second equation holds since ∂z = 0, and the third equation holds since adding theboundary ∂Tz to the cycle gz does not change the homology class it represents. �


4. In class we outlined the proof of the statement that the inclusion map ι : CU∗ (X)→ C∗(X)is a chain homotopy equivalence. Here CU∗ (X) is the subchain complex of C∗(X) generatedby all simplices which are contained in one of the open subsets of the open cover U . The ideais to use barycentric subdivision to construct a chain map ρ : C∗(X) → CU∗ (X) such thatρ◦ι = id and ι◦ρ is chain homotopic to id. The purpose of this problem is to provide some of

the steps in the proof of this statement. As in class let LC∗(Y ) ⊂ C∗(Y ) be the augmentedchain complex generated by affine linear simplices λ = [w0, . . . , wq] : ∆q → Y , wi ∈ Y for aconvex subset Y of some vector space. Recall that for any point b ∈ Y , we define the linear

map b : LCq(Y )→ LCq+1(Y ) which sends an affine linear simplex [w0, . . . , wq] to the “cone”[b, w0, . . . , wq]. Recall further that ∂b = id−b∂; in other words, b is a chain homotopy from0 to id.

(a) Show that the linear map S : LCq(Y ) → LCq(Y ) defined inductively by Sλ := λ forq = −1, 0 and Sλ := bλ(S∂λ) is a chain map. Here bλ = 1

q+1(w0, . . . , wq) ∈ Y is the

barycenter of the affine linear simplex λ = [w0, . . . , wq].

(b) Show that the linear map T : LCq(Y ) → LCq+1(Y ) defined inductively by Tλ := 0 forq = −1 and Tλ := bλ(λ− T∂λ) is a chain homotopy from S to id.

(c) Extend the subdivision operator S and the chain homotopy T to the singular chaincomplex of any topological space X by defining for any singular simplex σ : ∆q −→ X

Sσ = σ#(Sλq) and Tσ = σ#(Tλq),

where λq : ∆q −→ ∆q is the tautological affine q-simplex in ∆q given by the identitymap. Show that S : Cq(X) → Cq(X) is a chain map and T : Cq(X) → Cq+1(X) is achain homotopy from S to the identity.

Proof. Part (a). We will prove that ∂Sλ = S∂λ for all λ ∈ LCq(Y ) by induction over q.

So let us assume that this holds for all elements of LCk(Y ) for k < q and let λ ∈ LCq(Y ).Then

∂Sλ = ∂(bλS∂λ) = S∂λ− bλ∂S∂λ = S∂λ− bλ(S∂2λ) = S∂λ,

where the second equality follows from the equation ∂b = id−b∂, and the third equality

follows from our inductive assumption since ∂λ ∈ LCq−1(Y ).

Part (b). Again we use induction to prove this statement. So let us assume that we have

the equality ∂T + T∂ = id−S of endomorphisms of LCk(Y ) for k < q. Then for an affinelinear q-simplex λ = [w0, . . . , wq] we have

∂Tλ = ∂(bλ(λ− T∂λ)) = λ− T∂λ− bλ(∂λ− ∂T∂λ)

= λ− T∂λ− bλ(S∂λ− T∂2λ) = λ− T∂λ− bλS∂λ = λ− T∂λ− Sλ.

Here the second equality follows from the equation ∂b = id−b∂, the third equality from

our inductive assumption since ∂λ ∈ LCq−1(Y ), and the last equality from the inductivedefinition of Sλ.

Part (c). We observe first that the linear maps S, T constructed this way are natural inthe sense that for any map f : X → Y we have

Sf# = f#S and Tf# = f#T,


where f# : C∗(X) → C∗(Y ) is the chain map induced by f . To prove the first equation wecalculate for a q-simplex σ : ∆q → X

(5) Sf#σ = S(f ◦ σ) = (f ◦ σ)#Sλq = f#σ#Sλq = f#Sσ.

Here the first equality is the definition of f#(σ), the second and last is the definition of S ongeneral simplices and the third holds by the functor property of the assignment X 7→ C∗(X).The equation Tf# = f#T holds by the same argument, we just need to replace the letter Sby the letter T .

Next we prove that S is a chain map. Let σ : ∆q → X be a q-simplex. Then

∂Sσ = ∂σ#Sλq by definition of S on C∗(X)

= σ#∂Sλq since σ# is a chain map

= σ#S∂λq since S is a chain map on LC∗(∆q)

= Sσ#∂λq by equation (5)

= S∂σ#λq since σ# is a chain map

= S∂σ by definition of the induced map σ#

This shows that S is a chain map. Similarly,

∂Tσ = ∂σ#Tλq by definition of T on C∗(X)

= σ#∂Tλq since σ# is a chain map

= σ#(T∂λq − λq + Sλq) since ∂T + T∂ = id−S on LC∗(∆q)

= Tσ#∂λq − σ#λq + Sσ#λq by equation (5)

= T∂σ#λq − σ#λq + Sσ#λq since σ# is a chain map

= T∂σ − σ + Sσ by definition of the induced map σ#

�


1. This is a continuation of problem # 4 in the previous homework assignment. It finishesthe proof that the inclusion map ι : CU∗ (X)→ C∗(X) is a chain homotopy equivalence, andhence the proof of the Excision Theorem.

We recall that barycentric subdivision gives a chain map

S : C∗(X) −→ C∗(X)

on the singular chain complex C∗(X) of any topological space X, which is chain homotopicto the identity map via a chain homotopy T . In class we showed that Dm :=

∑0≤i<m TS

i isa chain homotopy from Sm := S ◦ · · · ◦ S︸︷︷︸

m

to id and that D : Cq(X) → Cq+1(X) defined by

Dσ := Dm(σ)σ for a q-simplex σ : ∆q → X is a chain homotopy from the chain map

ρ : Cq(X)→ Cq(X) defined by ρ(σ) := Sm(σ)σ +Dm(σ)(∂σ)−D(∂σ)

to the identity. Here m(σ) is the smallest integer m such that the iterated subdivision Sm(σ)belongs to CUq (X) ⊂ Cq(X).


Show that for any singular q-simplex σ its image ρ(σ) belongs to CU∗ (X). Hint: Write ∂σin the form

∑j(−1)jσj, where σj = σ ◦ [e0, . . . , ej, . . . , eq] is the j-th face of σ and argue that

Dm(σ)(σj)−D(σj) belongs to CU∗ (X).

Proof. We want to show that for any q-simplex σ : ∆q → X the chain

ρ(σ) = Sm(σ)σ +Dm(σ)(∂σ)−D(∂σ) ∈ Cq(X)

belongs to the U -small chains CUq (X). This is clear for the terms Sm(σ)σ by definition of

m(σ). To prove that Dm(σ)(∂σ)−D(∂σ) belongs to CU∗ (X), we write

∂σ =

q∑j=0

(−1)jσj

where σj = σ ◦ [e0, . . . , ej, . . . , wq] is the j-th face of σ, and will show that

Dm(σ)(σj)−D(σj) = Dm(σ)(σj)−Dm(σj)(σj)

belongs to CU∗ (X). We note that m(σj) ≤ m(σ), and hence

Dm(σ)(σj)−Dm(σj)(σj) =∑

0≤i<m(σ)

TSiσj −∑

0≤i<m(σj)

TSiσj

=

m(σ)−1∑i=m(σj)

TSiσj =

m(σ)−m(σj)−1∑k=0

TSkSm(σj)σj.

Since Sm(σj)σj belongs to CU∗ (X), and S, T : C∗(X)→ C∗(X) preserve the subchain complexCU∗ (X) ⊂ C∗(X), this shows that Dm(σ)(σj)−Dm(σj)(σj) belongs to CU∗ (X). �

2. (a) Show that Z/k ⊗ Z/` ∼= Z/ gcd(k, `).(b) Show that TorZi (Z, N) = 0 for i ≥ 1 and any Z-module N .

(c) Show that TorZi (Z/k,Z/`) ∼=

{Z/ gcd(k, `) i = 0, 1

0 i 6= 0, 1.

Proof. Proof of part (a). The homomorphism

Z/k ⊗ Z/` −→ Z/ gcd(k, `) given by m⊗ n 7→ mn

is a well defined surjective map. To show that this is in fact an isomorphism, it remains toshow that the element 1⊗ 1 ∈ Z/k ⊗ Z/` multiplied by g := gcd(k, `) is zero in Z/k ⊗ Z/`.To show this, write g in the form g = ak + b` for k, ` ∈ Z. Then

g(1⊗ 1) = ak(1⊗ 1) + b`(1⊗ 1) = a(k ⊗ 1) + b(1⊗ `) = 0.

Proof of part (b). Since Z is free as Z-module, a free resolution M∗ of Z is given by

Z M0 = Zεoo M1 = 0oo M2 = 0oo oo

Tensoring with N results in the chain complex

degree 0 1 2

M0 ⊗N = Z⊗N = N M1 ⊗N = 0oo M2 ⊗N = 0oo oo


In particular, TorZi (Z, N), the homology group of this chain complex in degree i, vanishesfor i ≥ 1.

Proof of part (c). A free resolution M∗ of Z/k is given by

Z/k M0 = Zεoo M1 = Zkoo M2 = 0oo oo

Tensoring M∗ with Z/` results in the chain complex

degree 0 1 2

Z⊗ Z/` Z⊗ Z/`k⊗idoo 0oo oo

Identifying the tensor product Z⊗Z/` with Z/` via the map given by Z⊗Z/` 3 m⊗ [n] 7→[mn] ∈ Z/`, the homomorphism k ⊗ id corresponds to the homomorphism k : Z/` → Z/`(given by multiplication by k). To determine TorZ1 (Z/k,Z/`) which is the kernel of this map,let g := gcd(k, `), and write k = gk′, ` = g`′ for integers k′, `′. Then multiplication by k′ isan automorphism of Z/`, and hence the kernel of multiplication by k equals the kernel ofmultiplication by g, which is generated by the element [`′] ∈ Z/` which generates a cyclicsubgroup of Z/` of order g. This shows TorZ1 (Z/k,Z/`) ∼= Z/g.

�

3. In this problem set, let us denote by Σk the connected sum of k copies of the real projectiveplane RP2, k ≥ 1. Compute the homology groups Hq(Σk;Z/2) in the following two differentways:(a) As the homology groups of CCW

∗ (Σk)⊗Z/2, where CCW∗ (Σk) is the cellular chain complex

associated to the standard CW structure on Σk obtained by regarding Σk as a quotient ofthe 2k-gon.(b) Via the Universal Coefficient Theorem, using the fact that

Hq(Σk) =

Z q = 0

Zk−1 ⊕ Z/2 q = 1

0 otherwise

.

Proof. Part (a). We recall that the connected sum of k copies of RP2 is homeomorphic tothe quotient obtained from a 2k-gon with edges labeled a1a1a2a2 . . . akak. As discussed in

class, we interpret this as a CW decomposition of Σk with 0-skeleton Σ(0)k given by the vertex

v. The 1-skeleton Σ(1)k consists of the quotient ∂P/ ∼ of the boundary ∂P of the polygon P

obtained by identifying those edges with the same label; this gives a bouquet∨ki=1 S

1i of k

circles labeled ai that have the point v in common. The open 2-cell e is the interior of thepolygon; its attaching map is given by the projection map

S1 = ∂P → ∂P/ ∼ = Σ(1)k .

We note that the composition of this map with the projection onto S1i has degree 2 (since this

loop runs twice through each of the circles S1). Hence according to our theorem describingthe boundary map ∂ of the cellular chain complex, we have

∂e = 2a1 + 2a2 + · · ·+ 2ak.


Summarizing, the cellular chain complex CCW∗ (Σk) looks as follows.

degree 0 1 2

CCW∗ (Σk) Zv Za1 ⊕ · · · ⊕ Zak

∂oo Ze∂oo

0 ai 2(a1 + · · ·+ ak)�oo e�oo

Tensoring with Z/2 we obtain the chain complex

CCW∗ (Σk)⊗ Z/2 ∼= Z/2 Z/2⊕ · · · ⊕ Z/2oo Z/2oo

with trivial differential. It follows that

Hq(Σk) =

Z/2 q = 0, 2

(Z/2)k q = 1

0 otherwise

Part (b). According to the Universal Coefficient Theorem,

H∗(Σk;Z/2) ∼= H∗(Σk)⊗ Z/2 ⊕ Tor1(H∗(Σk),Z/2).

The following table gives all relevant graded Z-modules by showing all their non-trivialhomogeneous pieces sorted by degree.

degree H∗(Σk) H∗(Σk)⊗ Z/2 Tor1(H∗(Σk),Z/2) H∗(Σk;Z/2)0 Z Z/2 0 Z/21 Zk−1 ⊕ Z/2 (Z/2)k 0 (Z/2)k

2 0 0 Z/2 Z/2

We note that the summand Z/2 in degree 1 ofH∗(Σ) leads to the summand Z/2 of Tor1(H∗(Σk),Z/2)of degree 2. �

4. (a) Let M , N be right R-modules, and let εM : M∗ →M , εN : N∗ → N be free resolutions.Show that if f : M → N is an R-linear map, then one can construct a morphism f∗ : M∗ → N∗of dg R-modules such that the diagram

(6) M

f��

M∗εMoo

f∗��

N N∗εNoo

is commutative. Hint: Unpacking the definitions, the morphism f∗ amounts to a collectionof R-linear maps fq : Mq → Nq such that the diagram

M

f��

M0

f0��

εMoo M1

f1��

oo M2

f2��

oo . . .oo

N N0εNoo N1

oo N2oo . . .oo


is commutative. Note that the top and bottom row are exact sequences of R-modules bydefinition of “free resolution”. Construct the maps fq inductively using the following propertyof a free module: if g : A→ B is an R-module map whose domain A is a free module, then gfactors through any R-linear surjection h : C → B; i.e., there is an R-linear map g : A→ Cmaking the following diagram commutative:

C

��h��

Ag //

g??

B

(b) Show that the R-linear chain map f∗ : M∗ → N∗ constructed in part (a) is unique up toR-linear chain homotopies, i.e., if f ′∗ : M∗ → N∗ is another solution to (a), show that thereis a chain homotopy T between them.

(c) Show that if M is a right R-module, and P is a left R-module, then the Z-moduleTorqR(M,P ) is independent of the choice of a free resolution of M in the sense that if M∗ →Mand M ′

∗ →M are free resolutions of M , then there is an isomorphism between Hq(M∗⊗RP )and Hq(M

′∗ ⊗R P ) which is natural in M and P .

Proof. Part (a). Since M0 is a free R-module, the module map f ◦ εM : M0 → N factorsthrough the surjective map εN : N0 → N ; i.e., there is an R-linear map f0 : M0 : N0 makingthe first square commutative. We will construct the fq’s by induction. Let us assumethat we already constructed R-linear maps f0, . . . , fq making all diagrams to the left of fqcommutative. We note that this implies in particular that fq maps ker(Mq → Mq−1) toker(Nq → Nq−1). Now we want to construct fq+1 such that the following diagram commutes:

ker(Mq →Mq−1)

fq��

Mq+1

fq+1

��

dMq+1oo

ker(Nq → Nq−1) Nq+1

dNq+1oo

This map exists sinceMq+1 is free and the map dNq+1 is surjective by exactness of the resolutionN∗ at Nq.Part (b). Assume that f ′∗ : M∗ → N∗ is another chain map lifting the map f . Our goal isto construct a chain homotopy T between them; i.e., we want R-linear maps Tq : Mq → Nq+1

with

(7) dNq+1Tq + Tq−1dMq = fq − f ′q

where the modules Mq, Nq are interpreted as the trivial modules for q < 0. We will constructthe Tq’s inductively. To construct T0, we note that

εN ◦ f0 = f ◦ εM = εN ◦ f ′0implies that the range of f0 − f ′0 is contained in ker εN , and hence there is a map T0 makingthe diagram

M0

f0−f ′0��

T0

""ker εN N1

dN1

oo


commutative since M0 is free and the horizontal map is surjective.Now let us assume that we have constructed T0, . . . , Tk−1 satisfying equation (7) for q < k.

To construct Tk, we consider this equation for q = k and put the term Tk−1dMk on the right

side of the above equation and try to solve for Tk. We note that the image of

g := fk − f ′k − Tk−1dMk : Mk → Nk

is contained in the kernel of dNk since

dNk (fk − f ′k − Tk−1dMk ) = fk−1d

Mk − f ′k−1d

Mk − dNk Tk−1d

Mk

=fk−1dMk − f ′k−1d

Mk − (Tk−2d

Mk−1d

Mk − fk−1d

Mk − f ′k−1d

Mk ) = 0

Here the first equation holds since f∗, f′∗ are chain maps, and the second equation follows

from the inductive assumption.Now we can construct Tk making the diagram

Mk

Tk

$$g��

ker dNk Nk+1dNk+1

oo

commutative, since Mk is free and dNk+1 is surjective onto the kernel of dNk by the exactnessof N∗.

Part (c). Let ε : M∗ → M and ε′ : M ′∗ → M be two free resolutions of M . Then applying

part (a) to the identity map id: M → M we obtain a chain map f∗ : M∗ → M ′∗ which is

a lift of f = id in the sense that the diagram (6) is commutative (for N = M , N∗ = M ′∗).

Reversing the roles of M∗ and M ′∗, we also obtain a chain map f ′∗ : M

′∗ →M∗ which is a lift

of the identity map of M . It follows that the composition

M∗f∗ // M ′

∗f ′∗ // M∗

is a chain map which is a lift of the identity map of M . We observe that the identity map ofM∗ is another choice of chain map M∗ →M∗ which is a lift of the identity on M , and henceby part (b), the composition f ′∗ ◦ f∗ is chain homotopic to the identity. Reversing the rolesof f∗ and f ′∗, the same argument shows that also the composition f∗ ◦ f ′∗ is chain homotopicto the identity map. In other words, f∗ : M∗ → M ′

∗ is a chain homotopy equivalence withchain homotopy inverse f ′∗.

It follows that f∗⊗ idP : M∗⊗R P →M ′∗⊗R P is a chain homotopy equivalence and hence

the induced map in homology

(8) Hq(M∗ ⊗R P )∼=−→ Hq(M

′∗ ⊗R P )

is the desired isomorphism between the tor-group defined using the resolution M∗ and thatdefined using M ′

∗.It remains to show that the isomorphism constructed is natural with respect to M and P .

Naturality in P is easier to show. Suppose that h : P → Q is an R-linear map. Then the


diagram of chain maps

M∗ ⊗R PidM∗ ⊗h

��

f∗⊗idP // M ′∗ ⊗R P

idM′∗⊗h

��M∗ ⊗R Q

f∗⊗idQ // M ′∗ ⊗R Q

is commutative. This implies that the induced of homology groups is commutative, whichproves that the isomorphism (8) is natural in P .

To prove naturality in M , suppose that h : M → N is an R-linear map. Let M∗, M′∗

be free resolutions of M , and let N∗, N′∗ be free resolutions of N . Let h∗ : M∗ → N∗ and

h′∗ : M′∗ → N ′∗ be chain maps which are lifts of h : M → N . The diagram of chain maps

M∗f∗ //

h∗��

M ′∗

h′∗��

N∗g∗ // N ′∗

is not necessarily commutative, but the two chain maps h′∗ ◦ f∗, g∗ ◦ h∗ are both lifts of theR-module map h : M → N , and hence are chain homotopic by part (b). In particular, aftertensoring with the identity on P we have a diagram of chain maps

M∗ ⊗R Pf∗⊗idP //

h∗⊗idP

��

M ′∗ ⊗R P

h′∗⊗idP

��N∗ ⊗R P

g∗⊗idP // N ′∗ ⊗R P

which commutes up to chain homotopy. It follows that the induced diagram of homologygroups commutes, which establishes the naturality of our isomorphism (8) in M . �


1. Let 0 → A → B → C → 0 be a short exact sequence of Z-modules. Show that for anytopological space X there is a corresponding long exact sequence of homology groups

// Hq(X;A) // Hq(X;B) // Hq(X;C)∂ // Hq−1(X;A) // Hq−1(X;B) // .

Hint: Recall that a short exact sequence of chain complexes induces a long exact sequenceof homology groups.

Proof. Tensoring the short exact sequence 0 → A → B → C → 0 with the chain complexC∗(X) leads to the sequence

(9) 0 // C∗(X)⊗ A // C∗(X)⊗B // C∗(X)⊗ C // 0

of chain complexes. Looking at the submodule of homogenous elements of degree q of thesechain complexes we have the sequence

(10) 0 // Cq(X)⊗ A // Cq(X)⊗B // Cq(X)⊗ C // 0


Since tensoring with any Z-module is right-exact, this sequence is exact except possibly atthe term Cq(X)⊗A. However, we recall that this half-exact short sequence can be continuedto a long exact sequence:

// Tor1R(Cq(X), C) // Cq(X)⊗ A // Cq(X)⊗B // Cq(X)⊗ C // 0

Since Cq(X) is a free Z-module, the term Tor1R(Cq(X), C) vanishes, and hence the sequence

(10) of Z-modules is exact. It follows that (9) is a short exact sequence of chain complexeswhich results in the desired long exact sequence of homology groups. �

2. Let Σk be the the connected sum of k copies of the real projective plane RP2. Calculatethe homology groups of Σk × Σ` in two ways:(a) As the homology groups of the cellular chain complex CCW

∗ (Σk × Σ`) ∼= CCW∗ (Σk) ⊗

CCW∗ (Σ`). Hint: decompose CCW

∗ (Σk) as a sum of simpler chain complexes.(b) Via the Kunneth Theorem.

Proof. Part (a). By inspection of the chain complex CCW∗ (Σk) written down explicitly in

part (a) of problem #3 in homework assignment # 9, we see that it decomposes as a directsum of sub chain complexes

CCW∗ (Σk) ∼= Z⊕ ΣZ⊕ · · · ⊕ ΣZ︸︷︷︸

k−1

⊕ΣMZ/2.

Here Z-modules are interpreted as a chain complex concentrated in degree 0 (like the sum-mand Z in the decomposition above, generated by the 0-cell v). For any chain complex C∗we denote by ΣC∗ the suspension of C∗, the dg Z-module obtained from C∗ by shifting alldegrees by declaring (ΣC∗)q := Cq−1; for example, ΣZ is a chain complex concentrated in de-gree +1. The k−1 summands ΣZ in the decomposition above are generated by a1, . . . , ak−1.Finally, MZ/` denotes the chain complex consisting of one copy of Z in degree 0 and 1 witha differential that is multiplication by `; MZ/` is trivial in all other degrees. The summandΣMZ/2 is generated by e in degree 2 and a1 + · · ·+ ak in degree 1.

Then we have the following isomorphisms of dg Z-modules:

CCW∗ (Σk)⊗ CCW

∗ (Σ`)

∼=

Z⊕ ΣZ⊕ · · · ⊕ ΣZ︸︷︷︸k−1

⊕ΣMZ/2

⊗Z⊕ ΣZ⊕ · · · ⊕ ΣZ︸︷︷︸

`−1

⊕ΣMZ/2

∼= Z⊕

⊕k+`−2

ΣZ ⊕⊕

2

ΣMZ/2 ⊕⊕

(k−1)(`−1)

Σ2Z ⊕⊕k+`−2

Σ2MZ/2 ⊕ ΣMZ/2⊗ ΣMZ/2

From homework problem (3) in assignment #8, we know that the tensor product ΣMZ/2⊗ΣMZ/2 can be decomposed as

ΣMZ/2⊗ ΣMZ/2 ∼= Σ2(MZ/2⊗MZ/2) ∼= Σ2(MZ/2⊕ ΣMZ/2) ∼= Σ2MZ/2 ⊕ Σ3MZ/2


From the resulting direct sum decomposition we can read off the homology groups of Σk×Σ`

to obtain

Hq(Σk × Σ`) ∼=

Z q = 0

Zk+`−2 ⊕ (Z/2)2 q = 1

Z(k−1)(`−1) ⊕ (Z/2)k+`−1 q = 2

Z/2 q = 3

Part (b). According to the Kunneth Theorem,

Hq(Σk × Σ`) ∼= (H∗(Σk)⊗H∗(Σ`))q ⊕ Tor1(H∗(Σk), H∗(Σ`))q−1,

where we use the shorthand notation

(H∗(Σk)⊗H∗(Σ`))q :=⊕

m+n=q

Hm(Σk)⊗Hn(Σ`)

Tor1(H∗(Σk), H∗(Σ`))q−1 :=⊕

m+n=q−1

Tor1(Hm(Σk), Hn(Σ`)).

The following table shows all these graded Z-modules.

q Hq(Σk) (H∗(Σk)⊗H∗(Σ`))q Tor1(H∗(Σk), H∗(Σ`))q−1 Hq(Σk × Σ`)0 Z Z 0 Z1 Zk−1 ⊕Z/2 Zk+`−2 ⊕ (Z/2)2 0 Zk+`−2 ⊕ (Z/2)2

2 0 Z(k−1)(`−1) ⊕ (Z/2)k+`−2 0 Z(k−1)(`−1) ⊕ (Z/2)k+`−1

3 0 0 Z/2 Z/2The third and fourth columns are obtained from the second by using linearity in each slotof −⊗− resp. Tor1(−,−) to write H∗(Σk)⊗H∗(Σ`) and Tor1(H∗(Σk), H∗(Σ`)) as a sum ofterms of the form M ⊗N (resp. Tor1(M,N) where M , N are Z or Z/2. �

3. For a topological space X with finitely generated homology groups its Poincare series isdefined to be the power series

P (X) :=∞∑q=0

rkHq(X)tq ∈ Z[[t]]

(a) Show that if X and Y are spaces with finitely generated homology groups, then so isX × Y with Poincare series given by

P (X × Y ) = P (X) · P (Y )

(b) Express the Poincare series P (CPn) and P (CPm × CPn) as rational functions of t (i.e.,as quotients of polynomials).(c) We recall that if X is of bounded finite type (that is, its homology groups are finitelygenerated and Hq(X) = 0 for q sufficiently large), then the Euler characteristic of X isdefined as the alternating sum

χ(X) =∞∑q=0

(−1)q rkHq(X)

of the ranks of the homology groups of X (note that this is a finite sum). How can the Eulercharacteristic χ(X) be expressed in terms of the Poincare series P (X)? Use part (a) to showthat χ(X × Y ) = χ(X) · χ(Y ) for bounded spaces X, Y of finite type.


Proof. Part (a). By the Kunneth Theorem,

Hq(X × Y ) ∼=⊕k+`=q

Hk(X)⊗H`(Y ) ⊕⊕

k+`=q−1

Tor1(Hk(X), H`(Y )).

In particular, Hq(X×Y ) is a finite direct sum of terms of tensor and tor-products of finitelygenerated Z-modules and hence it is finitely generated. Moreover, if Hq(X) vanishes forq > M and Hq(Y ) vanishes for q > N , then this isomorphism implies that Hq(X × Y ) = 0for q > M + N + 1 (we note that the Tor product of HM(X) and HN(Y ) can lead to anon-trivial homology group HM+N+1(X × Y ) as the example of problem #2 shows).

We recall that Tor1(M,N) is a torsion group for cyclic groups M , N and hence for finitelygenerated Z-modules M , N . It follows that the Tor-groups don’t contribute to the rank ofHq(X × Y ). Writing Pk(X) = rkHk(X) for the coefficient of the Poincare series P (X) weobtain

Pq(X × Y ) = rkHq(X × Y ) = rk(⊕k+`=q

Hk(X)⊗H`(Y ))

=∑k+`=q

rk(Hk(X)⊗H`(Y )) =∑k+`=q

rkHk(X) · rkH`(Y )

=∑k+`=q

Pk(X) · P`(Y ).

We recognize the last sum as the expression giving the coefficient of tq of the power seriesP (X) · P (Y ), which shows that P (X × Y ) = P (X) · P (Y ).Part (b). We recall that

Hq(CPn) =

{Z q = 0, 2, . . . , 2n

0 otherwise

and hence P (CPn) = 1 + t2 + t4 + · · ·+ t2n = 1−t2n+2

1−t2 . By part (a) it follows that

P (CPm × CPn) = P (CPm) · P (CPn) =(1− t2m+2)(1− t2n+2)

(1− t2)2.

Part (c). We note that the assumption that the homology group Hq(X) is zero for suffi-ciently large q means that the Poincare series P (X) is a polynomial. In particular, we canevaluate P (X) for any complex number t. For t = −1 we obtain

P (X)(−1) =∞∑q=0

rkHq(X)(−1)q = χ(X).

In particular,

χ(X × Y ) = P (X × Y )(−1) = (P (X) · P (Y ))(−1)

= (P (X)(−1)) · (P (Y )(−1)) = χ(X) · χ(Y ).

�

4. If X is a topological space of bounded finite type, we define a variant of the Eulercharacteristic of X by χ(X;Z/p) =

∑∞q=0(−1)q dimHq(X;Z/p) as the alternating sum of

the dimensions of the vector spaces Hq(X;Z/p). Show that χ(X;Z/p) = χ(X). Hint: usethe universal coefficent theorem.


Proof. By the Universal Coefficient Theorem,

Hq(X;Z/p) ∼= Hq(X)⊗ Z/p ⊕ Tor1(Hq−1(X),Z/p).The finitely generated Z-module Hq(X) can be decomposed as a direct sum

Hq(X) ∼= Aq ⊕Bq ⊕ Cq,where Aq is a sum of Z’s, Bq is a sum of finite cyclic groups whose order is a power of p,and Cq is a sum of finite cyclic groups of order prime to p. Let us write aq (resp. bq) for thenumber of summands in Aq (resp. Bq). We note that aq is the rank of Hq(X).

We note that

Z⊗ Z/p ∼= Z/p Z/pi ⊗ Z/p ∼= Z/p Tor1(Z/pi,Z/p) ∼= Z/p for i ≥ 1,

and hence the Universal Coefficient Theorem implies

dimHq(X;Z/p) = dim (Hq(X)⊗ Z/p) + dim Tor1(Hq−1(X),Z/p) = (aq + bq) + bq−1.

It follows that

χ(X;Z/p) =∞∑q=0

(−1)q(aq + bq + bq−1)

=∞∑q=0

(−1)qaq +∞∑q=0

(−1)qbq +∞∑q=0

(−1)qbq−1

=∞∑q=0

(−1)qaq =∞∑q=0

(−1)q rkHq(X) = χ(X).

�


1. Prove the algebraic version of the Kunneth Theorem. In other words, suppose R = Z ora field k (or more generally a PID if you want), C∗, D∗ are chain complexes of R-modules,and assume that all modules Cn are free. Show that there is a short exact sequence(11)

0 //⊕k+`=q

Hk(C∗)⊗R H`(D∗)× // Hq(C∗ ⊗R D∗) //

⊕k+`=q−1

TorR1 Hk(C∗), H`(D∗)) // 0

Discussing whether this sequence is split is not required. Hint: Follow the pattern we usedfor the proof the universal coefficient theorem by first showing that there is a short exactsequence of chain complexes

Z∗ ⊗R D∗ −→ C ∗ ⊗RD∗ −→ B∗−1 ⊗R D∗,then identifying the boundary map of the associated long exact sequence, and finally breakingit up into short exact sequences that can be identified with (2).

Proof. Let Bq ⊂ Zq be the q-boundaries (resp. the q-cycles) of the chain complex C∗. Thenwe have the short exact sequence

(12) 0 // Zq // Cq∂ // Bq−1

// 0.


This leads to a short exact sequence of chain complexes

0 // Z∗ // C∗∂ // B∗−1

// 0,

where

• Z∗ is the chain complex of R-modules whose degree q term is Zq and whose differentialis trivial.• B∗−1 is the chain complex of R-modules whose degree q term is Bq−1 and whose

differential is trivial.

Tensoring this sequence with the chain complex D∗ leads to the sequence of chain com-plexes

(13) 0 // Z∗ ⊗R D∗ // C∗ ⊗R D∗∂⊗id// B∗−1 ⊗R D∗ // 0.

We claim that this is an exact sequence. To show this, we note that (16) is a split exactsequence, since Bq−1 as a submodule of the free R-module Cq−1 and by our assumptions onR is indeed a free module. In particular, tensoring this exact sequence with any R-moduleresults again in a short exact sequence. Writing out the degree q part of the sequence (13)we obtain a sum of such sequences, which proves our claim.

Associated to this short exact sequence of chain complexes, we obtain the following longexact sequence in homology:

(14) . . . // Hq+1(B∗−1 ⊗R D∗)∂q+1 // Hq(Z∗ ⊗R D∗) //// Hq(C∗ ⊗R D∗) //

Hq(B∗−1 ⊗R D∗)∂q // Hq−1(Z∗ ⊗R D∗) // . . .

This long exact sequence implies that the following sequence is short exact:

(15) 0 // coker ∂q+1// Hq(C∗ ⊗R D∗) // ker ∂q // 0

In order to identify kernel and cokernel of ∂q+1, we first identify this map as follows.

Claim. There is a commutative diagram whose vertical maps are isomorphisms⊕k+`=q

Bk ⊗H`(D∗)

∼=

��

⊕ik⊗id

//⊕k+`=q

Zk ⊗H`(D∗)

∼=

��Hq+1(B∗−1 ⊗R D∗)

∂q+1 // Hq(Z∗ ⊗R D∗)

To prove the claim, we first define the vertical maps. The right hand side maps anelement zk ⊗ [d`] ∈ Zk ⊗H`(D∗), where d` is a cycle of degree ` in the chain complex D∗, to[zk ⊗ d`] ∈ Hq(Z∗ ⊗D∗). We note that the chain complex Z∗ ⊗R D∗ can be decomposed asthe direct sum of chain complexes of the form Zk ⊗D∗. Since Zk ⊂ Ck is a free module, theUniversal Coefficient Theorem implies that the map

Zk ⊗H`(D∗) −→ Hk+`(Zk ⊗D∗) zk ⊗ [d`] 7→ [zk ⊗ d`]is an isomorphism. The right vertical map is the sum of these isomorphisms, and hence anisomorphism. The left vertical map is defined in the same way: it sends bk ⊗ [c`] to [bk ⊗ c`],and it is an isomorphism by the same argument.


It remains to show that the boundary map sends [bk⊗c`] ∈ Hq+1(B∗−1⊗D∗) = Hq(B∗⊗D∗)to [bk ⊗ c`] ∈ Hq(Z∗ ⊗D∗). We recall that construction of the boundary map ∂ of the longexact homology sequence associated to a short exact sequence of chain complexes involves apull-push-pull procedure:

pull: For bk⊗d` ∈ Bk⊗D` we need to find a pre-image under the map ∂⊗id : C∗⊗D∗ →B∗ ⊗D∗. Since bk is a boundary, there is some ck+1 ∈ Ck+1 such that ∂(ck+1) = bk,and then (∂ ⊗ id)(ck+1 ⊗ d`) = bk ⊗ d`.

push: We apply the boundary map of the chain complex B∗ ⊗ D∗ to this pre-imageand obtain

∂(ck+1 ⊗ d`) = ∂ck+1 ⊗ d` + (−1)k+1ck+1 ⊗ ∂d` = bk ⊗ d`.pull: We observe that the element bk⊗d` is in the image of the inclusion map Z∗⊗D∗ ↪→C∗ ⊗D∗.

This shows that ∂([bk ⊗ c`]) = [bk ⊗ c`] and finishes the proof of the claim.

Thinking of ik : Bk → Zk as a free resolution of Hk(C∗), we recognize the cokernel of themap

ik ⊗ id : Bk ⊗H`(D∗) −→ Zk ⊗H`(D∗)

as Tor0(Hk(C∗), H`(D∗)) = Hk(C∗)⊗H`(D∗), and its kernel as Tor1(Hk(C∗), H`(D∗)). Sum-ming over all k we obtain the isomorphisms

coker ∂q+1∼=⊕k+`=q

Hk(C∗)⊗H`(D∗) ker ∂q+1∼=⊕k+`=q

Hk(C∗)⊗H`(D∗).

Hence substituting the terms coker ∂q+1 and ker ∂q in the short exact sequence (15) by thesesums give the Kunneth short exact sequence (16). �

2. (a) Calculate the abelian groups ExtqZ(Z,Z) and ExtqZ(Z,Z/t) for all q = 0, 1, 2 . . . .(b) Calculate the abelian groups ExtqZ(Z/s,Z) and ExtqZ(Z/s,Z/t) for all q = 0, 1, 2 . . . .

Proof. We recall that ExtqZ(M,N) for Z-modules M , N is defined as

ExtqZ(M,N) =

Hq

(δ // HomZ(Mq−1, N)

δ // HomZ(Mq, N)δ // HomZ(Mq+1, N)

δ //

),

where

M0 M1∂oo M2

∂oo ∂oo

is a free resolution of M , and δ = ∂∗ is the dual of ∂.

Part (a). Since Z is free as module over Z, a resolution M∗ of Z is simply given by M0 = Zand Mq = 0 for q > 0. It follows that

ExtqZ(Z, N) =

{HomZ(Z, N) = N q = 0

0 q > 0.

In particular,

ExtqZ(Z,Z) =

{Z q = 0

0 q > 0ExtqZ(Z,Z/t) =

{Z/t q = 0

0 q > 0.


Part (b). For M = Z/s, a free resolution is given by

M0 = Z M1 = Zsoo M2 = 0oo oo ,

where the map s : Z→ Z stands for multiplication by s.Applying the functor HomZ(−,Z) to the resolution M∗ we obtain the cochain complex

HomZ(Z,Z)s∗ // HomZ(Z,Z)

We note that HomZ(Z,Z) is isomorphic to Z, where the isomorphism is given by f 7→ f(1).This implies that the above cochain complex is isomorphic to

Z s // Z

and hence

ExtqZ(Z/s,Z) = Hq(HomZ(M∗,Z)) =

{Z/s q = 1

0 q 6= 1

For N = Z/t we obtain the cochain complex

HomZ(Z,Z/t) s∗ // HomZ(Z,Z/t)

We note that HomZ(Z,Z/t) is isomorphic to Z/t, where the isomorphism is given by f 7→f(1). This implies that the above cochain complex is isomorphic to

Z/t s // Z/t

We note that the cokernel of the map s is Z modulo the ideal generated by t and s. This isequal to the ideal generated by g = gcd(s, t) and hence the cokernel of the map s : Z/t→ Z/tis isomorphic to Z/g. Considering the order of the image, the domain and the kernel of themap s, we see that ker s has order g; as a subgroup of the cyclic group Z/t it is isomorphicto Z/g. We conclude:

ExtqZ(Z/s,Z/t) = Hq(HomZ(M∗,Z/t)) =

{Z/g q = 0, 1

0 q 6= 0, 1

where g = gcd(s, t). �

3. Let Σk be the connected sum of k copies of the real projective plane RP2. We recall thatthe homology groups of Σk are given by

Hq(Σk) =

Z q = 0

Zk−1 ⊕ Z/2 q = 1

0 q 6= 0, 1

(a) Calculate the cohomology groups Hq(Σk;Z).(b) Calculate the cohomology groups Hq(Σk;Z/2).

Proof. Part (a). By the Universal Coefficient Theorem for cohomology we have

(16) Hq(Σk;Z) ∼= Hom(Hq(Σk),Z)⊕ Ext1(Hq−1(Σk),Z).


The following table shows these groups:

q Hq(Σk) HomZ(Hq(Σk),Z) Ext1Z(Hq−1(Σk),Z) Hq(Σk;Z)

0 Z Z 0 Z1 Zk−1 ⊕Z/2 Zk−1 0 Zk−1

2 0 0 Z/2 Z/2q ≥ 3 0 0 0 0

Part (b). The Universal Coefficient Theorem calculating cohomology with Z/2-coefficientshas the same form as (16), we just need to replace each Z by Z/2. This results in thefollowing table.

q Hq(Σk) HomZ(Hq(Σk),Z/2) Ext1Z(Hq−1(Σk),Z/2) Hq(Σk;Z/2)

0 Z Z/2 0 Z/21 Zk−1 ⊕Z/2 (Z/2)k 0 (Z/2)k

2 0 0 Z/2 Z/2q ≥ 3 0 0 0 0

�

4. Recall that the homology groups of the lens space L2n−1(Z/k) = S2n−1/Z/k are given by

Hq(L2n−1(Z/k)) =

Z q = 0, 2n− 1

Z/k q = 1, 3, . . . 2n− 3

0 otherwise

Use the Universal Coefficient Theorem to calculate the cohomology groupsHq(L2n−1(Z/k);Z/p)for p a prime.

Proof. Let us write L := L2n−1(Z/k). The following table shows the contributions of theUniversal Coefficient Theorem to the cohomology groups H∗(L;Z/p). First we assume thatp divides k.

q Hq(L) Hom(Hq(L),Z/p) Ext1(Hq−1(L),Z/p) Hq(L;Z/p)0 Z Z/p 0 Z/p1 Z/k Z/p 0 Z/p2 0 0 Z/p Z/p3 Z/k Z/p 0 Z/p4 0 0 Z/p Z/p...

......

......

2n−2 0 0 Z/p Z/p2n−1 Z Z/p 0 Z/p

If p does not divide k, then all groups Ext1(Hq(L),Z/p) and Hom(Hq(L),Z/p) are zeroexcept for

Hom(H0(L),Z/p) = Hom(Z,Z/p) = Z/pHom(H2n−1(L),Z/p) = Hom(Z,Z/p) = Z/p.


It follows that if p does not divide k, then

Hq(L;Z/p) ∼=

{Z/p q = 0, 2n− 1

0 otherwise

�


1. (a) Show that the map

∪ : C∗(X)⊗ C∗(X) −→ C∗(X) given by φ⊗ ψ 7→ φ ∪ ψ

is a map of cochain homomorphism, that is, δ(φ ∪ ψ) = δφ ∪ ψ + (−1)|φ|φ ∪ δψ.(b). Show that the cup product is compatible with pull-back of cohomology classes in thesense that for a map f : X → Y and cohomology classes α ∈ Hk(Y ;R), β ∈ H l(Y ;R) wehave

f ∗(α ∪ β) = (f ∗α) ∪ (f ∗β).

Hint: Show first the analogous statement for cochains.

Proof. Part (a). For σ : ∆k+l+1 → X we have

(δϕ ∪ ψ)(σ) =k+1∑i=0

(−1)iϕ(σ ◦ [e0, . . . , ei, . . . , ek+1])ψ(σ ◦ [ek, . . . , ek+`+1])

(−1)k(ϕ ∪ δψ)(σ) =k+`+1∑i=k

(−1)iϕ(σ ◦ [e0, . . . , ek])ψ(σ ◦ [ek, . . . , ei, . . . , ek+`+1])

Adding these two expressions, the last term of the first sum cancels the first term of thesecond sum, and the remaining terms are exactly

δ(ϕ ∪ ψ) = (ϕ ∪ ψ)(∂σ)

since

∂σ =k+`+1∑i=0

(−1)iσ ◦ [e0, . . . , ei, . . . , ek+`+1].

Part (b). Let us write f# : C∗(Y ;R)→ C∗(X;R) for the cochain map induced by f ; i.e.,

(f#ϕ)(σ) = ϕ(f ◦ σ) for ϕ ∈ Cq(X;R), σ : ∆q → X

Then for ϕ ∈ Ck(Y ;R), ψ ∈ C`(Y ;R) and σ : ∆k+` → X we have

(f#(ϕ ∪ ψ))(σ) = (ϕ ∪ ψ)(f ◦ σ)

= ϕ(f ◦ σ ◦ [e0 . . . , ek])ψ(f ◦ σ ◦ [ek, . . . , ek+`])

= (f#ϕ)(σ ◦ [e0 . . . , ek])(f#ψ)(σ ◦ [ek, . . . , ek+`])

= (f#ϕ ∪ f#ψ)(σ)


If the cohomology classes α, β are represented by the cocycles ϕ resp. ψ, we obtain

f ∗(α ∪ β) = f ∗([ϕ ∪ ψ)]) = [f#(ϕ ∪ ψ)]

= [f#ϕ ∪ f#ψ] = [f#ϕ] ∪ [f#ψ]

= f ∗α ∪ f ∗β�

2. (a) Use the cellular chain complex of the Klein bottle K to determine its cohomologygroups Hq(K;Z/2).(b) Determine the cup products on cohomology with Z/2 coefficients. Hint: proceed similarlyto what we did in class for determining the cup products for the cohomology of the connectedsum of two tori.

Proof. Part (a). Our standard picture for the Kleinbottle

a //

b

��b

OO

a//

shows that K has a CW structure with one 0-cell (given by the common vertex v), two 1-cells(given by the edges a and b) and one 2-cell (given by the square f). Hence the cellular chaincomplex of K has the form

CCW0 (K) = Zv ∂←− CCW

1 (K) = Za⊕ Zb ∂←− CCW2 (K) = Zf

Looking at the degrees of the attaching maps we see that ∂f = 2b and ∂a = ∂b = 0. Hencethe cellular cochain complex C∗CW (K;Z/2) = Hom(CCW

∗ (K);Z/2) has the form

C0CW (K) = Z/2v∗ δ−→ C1

CW (K) = Z/2a∗ ⊕ Z/2b∗ δ−→ C2CW (K) = Z/2f ∗

Here v∗, a∗, b∗, f ∗ are the elements dual to v, a, b, f (i.e., a∗(a) = 1 ∈ Z/2, a∗(b) = 0, etc).We note that the coboudary maps δ are trivial. This is clear for δ : C0

CW → C1CW since it is

dual to ∂ : CCW1 → CCW

0 which is zero. Concerning δ : C1CW → C2

CW we have

δa∗(f) = a∗(∂f) = a∗(2b) = 0

δb∗(f) = b∗(∂f) = b∗(2b) = 2 = 0 ∈ Z/2

It follows that

Hq(K;Z/2) = Hq(C∗CW (K;Z/2) =

Z/2 q = 0, 2

Z/2⊕ Z/2 q = 1

0 otherwise

Part (b). To calculate the cup products, we choose a sub chain complex C∗(K)′ of the sin-gular chain comples C∗(K) to consist of the 0-simplices, 1-simplices and 2-simplices indicatedin the picture

a //c

��b

��b

OO

a//


Here we use the same convention as in class: the arrows allow us to identify the edges withaffine 1-simplices in the square; projection from the square to the Klein bottle then gives ussingular simplices a, b in K. The arrows allow us to identify triangles with affine 2-simplices(the 0-th vertex of a triangle is the vertex from which two edges are emanating, the 1-thvertex has one edge coming in and one edge going out and the 2-th vertex has two incomingedges). So we get two singular 2-simplices in K, say σ1 (corresponding to the top righttriangle) and σ2 (corresponding to the bottom left triangle). In particular:

1-front face of σ1 = a 1-back face of σ1= b

1-front face of σ2 = b 1-back face of σ2 = c

We note that σ1+σ2 is a cycle, and that given cochains ϕ, ψ ∈ C1(K;Z/2)′ := Hom(C1(K)′,Z/2)we have

(ϕ ∪ ψ)(σ1 + σ2) = ϕ(a)ψ(b) + ϕ(b)ψ(c)

Now we define cochains α, β ∈ C1(K;Z/2)′ by

α(a) = 1 α(b)= 0 α(c) = 1

β(a) = 0 β(b)= 1 β(c) = 1

We note that α and β are cocycles, since

(δα)(σ1) = α(∂σi) = α(a+ b+ c) = 0 ∈ Z/2and similarly for β. Moreover, α, β form a basis for H1(K;Z/2) since they are dual to thebasis {a, b} of H2(K;Z). We calculate:

(α ∪ α)(σ1 + σ2) = 1 · 0 + 0 · 1 = 0

(β ∪ β)(σ1 + σ2) = 0 · 1 + 1 · 1 = 1

(α ∪ β)(σ1 + σ2) = 1 · 1 + 0 · 1 = 1

(β ∪ α)(σ1 + σ2) = 0 · 0 + 1 · 1 = 1

This implies that β ∪ β = α ∪ β = β ∪ α is the non-zero element of H2(K;Z/2) ∼= Z/2.In particular, the cycle σ1 + σ2 represents a generator of H2(K) and hence the calculationabove shows that α ∪ α = 0. �

3. The Kummer surface is the submanifold of CP3 of (real) dimension 4 given by

K := {[z0, z1, z2, z3] ∈ CP3 | z40 + z4

1 + z42 + z4

3 = 0}It can be shown that

• K is simply connected (i.e., K is connected and its fundamental group π1(K) istrivial)• its Euler characteristic is given by χ(K) = 24.

Use these facts to calculate the homology groups Hq(K) and the cohomology groupsHq(K) for all q. Hint: use the Universal Coefficient Theorem and Poincare duality torelate homology and cohomology groups. Make sure to provide an argument for why theassumptions of the Poincare Duality Theorem are satisfied.


Proof. The fact that K is simply connected implies that H0(K) ∼= Z (since K is path con-nected), and that H1(K) = 0 (since H1(K) ∼= π1(K)ab = 0 by the Hurewicz isomorphism).

In order to apply Poincare duality we need to make sure that the manifold K is compactand oriented.

K is compact: The subspace

K := {(z0, z1, z2, z3) ∈ S7 ⊂ C4 | z40 + z4

1 + z42 + z4

3 = 0} ⊂ S7

is closed since K = f−0 for the continuous map f : S7 → C given by f(z0, z1, z2, z3) =z4

0 + z41 + z4

2 + z43 . Since S7 ⊂ C4 = R8 is compact by Heine-Borel, the closed

subspace K ⊂ S7 is compact. Since K is the image of K under the projection mapπ : S7 → CP3, it is compact.

K is orientable: We recall that an orientation for a manifold M is a section of thedouble covering M → M whose fiber Mx for a point x ∈ M is the set consisting ofthe two orientations of the tangent space TxM . If M is simply connected, like theKummer surface, this double covering is isomorphic to the trivial double covering.In particular, M is orientable.

Next we calculate the cohomology group Hq(K) in two ways in terms of the homologygroups of K:

• by Poincare duality, Hq(K) ∼= H4−q(K);• by the Universal Coefficient Theorem, Hq(K) ∼= Hom(Hq(K),Z)⊕Ext1

Z(Hq−1(K),Z).

The homology group H2(K) can be written in the form H2(K) = Zr ⊕ T , where T is atorsion group. It will be convenient to collect our knowledge about the various (co)homologygroups in the following table.

q Hq(K) Hq(K) ∼= H4−q(K) Hom(Hq(K),Z) Ext1Z(Hq−1(K),Z)

0 Z Z Z 01 0 0 0 02 Zr ⊕ T Zr Zr 03 04 Z

Here the colors of the entries indicate in which logical order we calculate these groups –black entries first, then blue entries, and finally red entries. The information we start with isH0(K) = Z, H1(K) = 0 and H2(K) = Zr⊕T (with r a to be determined integer, and T a tobe determined torsion group) in the second column of the table. These black entries in thesecond column immediately determine the black entries in the other columns of the table.By the UCT, the sum of the last two rows is isomorphic to the third row. In particular, weobtain the three blue entries in third row that way.

The Poincare duality isomorphism Hq(K) ∼= H4−q(K) for q = 0, 1 results in the twored entries in the second column. For q = 2, this isomorphism becomes an isomoprhismZr ∼= Zr ⊕ T , which implies that the torsion group T must be trivial. Finally, to determiner = rkH2(K), we use the fact that the Euler characteristic of K is 24, which implies theequation

24 = χ(K) =4∑q=0

(−1)q rkHq(K) = 1− 0 + r − 0 + 1 = 2 + r,


and hence r = 22. Summarizing our results, we have

Hq(K) =

Z q = 0, 4

Z22 q = 2

0 q 6= 0, 2, 4

�

4. The construction of the fundamental class of a closed oriented manifold is based on thefollowing

Proposition 2. Let M be a manifold of dimension n, and K ⊂M a compact subset. Thenthe homology Hi(M,M −K) is zero for i > n. An element α ∈ Hn(M,M −K) is trivial ifand only if it is in the kernel of the map

Hn(M,M −K) −→ Hn(M,M − x)

induced by the inclusion (M,M −K)→ (M,M − x) for every x ∈ K.

A crucial step in the proof of this proposition is to show that if this statement holds forcompact subsets K1, K2 and their intersection K1 ∩K2, then is also holds for K = K1 ∪K2.Prove this step. Hint: note that

(M,M −K) = (M,M −K1) ∩ (M,M −K2)

(M,M − (K1 ∩K2)) = (M,M −K1) ∪ (M,M −K2),

and use without proof that there is a version of the Meyer-Vietoris sequence for pairs ofspaces.

Proof. We will use the Meyer-Vietoris sequence for the suggested pairs of spaces which takesthe form

// Hi+1(M,M −K1 ∩K2)∂ // Hi(M,M −K)

// Hi(M,M −K1)⊕Hi(M,M −K2) //

This exact sequence implies that if the statement holds for K1, K2, and K1 ∩ K2, thenHi(M,M − K) vanishes for i > 0, since the adjacent terms in the long exact sequenceboth vanish. Now suppose that α ∈ Hn(M,M − K) which is in the kernel of the mapHn(M,M −K) −→ Hn(M,M − x) for every x ∈ K. We need to show α = 0.

Since the term to the left of Hn(M,M − K) in the exact sequence above vanishes, itfollows that the map Φ in the sequence is injective (for i = n), and hence it suffices to showthat Φ(α) = 0. Now by construction of the Meyer-Vietoris sequence, Φ(α) = (j1

∗(α), j2∗(α)),

wherejk : (M,M −K) −→ (M,M −Kk)

is the inclusion map of pairs. To show that jk∗ (α) = 0, suppose x ∈ Kk, and consider thefollowing commutative diagram of pairs

(M,M −K)jk //

((

(M,M −Kk)

��(M,M − x)


The corresponding diagram of homology groups of degree n shows that jk∗ (α) is in the kernelof the map Hn(M,M −Kk)→ Hn(M,M − x). Since this holds for every point x ∈ Kk, thisimplies that jk∗ (α) = 0. This holds for both, k = 1 and k = 2, and hence Φ(α) = 0. Theexact sequence then implies α = 0 as desired. �

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