121B: ALGEBRAIC TOPOLOGY

1. Introduction and examples

In the first week I am doing the definition of simplicial cohomology and some examples.

1.1. ∆-complex. Here is a formal definition of a ∆-complex.

Definition 1.1. A ∆-complex X is a sequence of setsX0, X1, · · · (Xn is the set of n-simpliciesof X) and mappings ∂i : Xn → Xn−1 for i = 0, · · · , n (∂i(σ) is the ith face of σ given bydeleting the ith vertex) so that:

∂i∂j = ∂j−1∂i

if i < j. (If you delete the ith vertex then the jth vertex becomes the j − 1st vertex.)

Problem: Find a formula for the ith vertex of σ ∈ Xn.

Answer:

A ∆-complex X is an (ordered) simplicial complex if

(1) Every n-simplex σ ∈ Xn has n+ 1 distinct vertices.(2) σ ∈ Xn is uniquely determined by its set of vertices.

Under these conditions we can write

σ = [v0, v1, · · · , vn]The ith boundary map is given by

∂i[v0, v1, · · · , vn] = [v0, v1, · · · , vi, · · · , vn]Hatcher uses this notation even if the vertices are all equal. This is meant to be an intuitivedescription of σ and its faces.

1.2. ∆-cohomology. If G is an additive group and X is a ∆ complex then an n-cochainon X with coefficients in G is any mapping φ : Xn → G. These form a group ∆n(X;G) bypointwise addition: (φ+ ψ)(x) = φ(x) + ψ(x). Let δ : ∆n(X;G)→ ∆n+1(X;G) be given byδφ(x) =

∑(−1)iφ(∂ix). In Hatcher’s notation:

(δφ)[v0, v1, · · · , vn] =n∑i=0

(−1)iφ[v0, · · · , vi, · · · , vn]

Elements in the kernel of δ : ∆n(X;G) → ∆n+1(X;G) are called n-cocycles on X withcoefficients in G and elements in the image are the n+ 1 coboundaries.Zn(X;G) = {φ ∈ ∆n(X;G) | δφ = 0}Bn(X;G) = {φ ∈ ∆n(X;G) |φ = δψ for some ψ ∈ ∆n−1(X;G)}We have the inclusion of additive groups: Bn(X;G) ⊆ Zn(X;G) ⊆ ∆n(X;G). The nth

cohomology group of X with coefficients in G is defined to be the quotient:

Hn(X;G) :=Zn(X;G)

Bn(X;G)1

Elements of Hn(X;G) are written

[φ] = φ+Bn(X;G)

1.3. Cohomology of a sphere. The problem is to compute the cohomology of the sphere.A ∆-complex model for the n− 1 sphere is ∂∆n. I started by pointing out that the reducedhomology of the n-simplex is trivial since ∆n is contractible.

Theorem 1.2. H∗(∆n) = 0.

We actually need the algebraic version of this. Namely, the augmented cellular chaincomplex is contractible.

Lemma 1.3. An exact sequence of free abelian groups is chain contractible.

You are supposed to do this as homework. You are given an exact sequence:

0← C−1∂←− C0

∂←− C1∂←− C2 ← · · ·

where all groups Cn are free abelian. You need to find a chain contraction, i.e., a sequenceof homomorphisms hn : Cn → Cn+1 so that

h∂ + ∂h = id

You may use the fact that every subgroup of a free abelian group is free.

Corollary 1.4. The augmented chain complex of ∆n is chain contractible.

I pointed out that the first and last terms of the cellular chain complex of ∆n are both Z:

0← Z← C0(∆n)← C1(∆

n)← · · · ← Cn(∆n)︸ ︷︷ ︸

=Z

← 0

To finish we need the following lemma:

Lemma 1.5. If C∗ is a contractible chain complex then the cochain complex C∗(C∗;G) isalso contractible for any (additive) group G.

The cochain complex C∗(C∗;G) has groups

Ck(C∗;G) := Hom(Ck, G)

connected by a sequence of coboundary maps

Ck(C∗;G) = Hom(Ck, G)δ−→ Hom(Ck+1, G) = Ck+1(C∗;G)

given by the dual of (precomposition with) the boundary operator: δ(φ) = ∂∗(φ) = φ ◦ ∂

Ck+1 δ(φ) = φ∂ ∈ Ck+1(C∗;G)

Ck G φ ∈ Ck(C∗;G)

@@

@@R

δ(φ)

?

∂

-φ

6δ

Proof. The chain contraction for C∗(C∗;G) is given by the dual h∗ of the chain contractionh for C∗:

h∗δ + δh∗ = (∂h+ h∂)∗ = id∗ = id

This uses the fact that Hom(−, G) is a contravariant functor. �2

Theorem 1.6. The cohomology of ∂∆n is given by Hk(∂∆n;G) = G for k = 0, n − 1 andHk(∂∆;G) = 0 otherwise.

Proof. We have the following exact sequence:

0→ Hom(Z, G)→ C0(C∗;G)δ0−→ C1(C∗;G)→ · · · δn−2−−→ Cn−1(C∗;G)︸ ︷︷ ︸

∗

→ Cn(C∗;G)→ 0

The middle terms (*) are the same as the cochain complex ∆∗(∂∆n;G). Since the entiresequence is exact, the subsequence (*) is exact except at the beginning and end where

H0(∂∆n;G) = ker δ0 = Hom(Z, G) ∼= G

andHn−1(∂∆n;G) = coker δn−1 = Hom(Cn(∆

n), G) = Hom(Z, G) ∼= G

The exactness at other points gives Hk(∂∆n;G) = 0 for k 6= 0, n− 1. �

2. Universal coefficient theorem

Hatcher proves the UCT twice: once conceptually and again computationally (the tradi-tional method). I’ll just do the first one. Here is the outline.

Lemma 2.1. Two free chain complexes C∗, D∗ are homotopy equivalent iff they have thesame (i.e., isomorphic) homologies.

In HW1 you are doing the easiest case when the homology is trivial.

Lemma 2.2. Homology commutes with direct sum: Hk(⊕

αCα∗ ) ∼=

⊕αHk(C

α∗ )

In the case when each homology group Hk(C∗) if finitely generated, it is the direct sumof Z’s and Z/n’s. So, we only need to do two examples:

C∗ = · · · ← 0← Z︸︷︷︸k

← 0← · · ·

with homology Hk(C∗) = Z and cohomology Hk(C∗;G) = G and

D∗ = · · · ← 0← Z︸︷︷︸k

n←− Z︸︷︷︸k+1

← 0← · · ·

with homology Hk(C∗) = Zn and cohomology given by the following cochain complex:

C∗(D∗;G) = · · · → 0→ G︸︷︷︸k

n−→ G︸︷︷︸k+1

→ 0→ · · ·

Which has cohomology:Hk(D∗;G) = {g ∈ G |ng = 0}

Hk+1(D∗;G) = G/nG

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The first group is naturally isomorphic to Hom(Zn, G). The second group is isomorphicto Zn ⊗G but the isomorphism is not natural since Zn ⊗G is covariant in both variables.

2.1. Reduction to the free resolution case. Lemma 2.2 is obvious is you know what thesymbols mean. The direct sum of chain complexes is given by

(C∗ ⊕D∗)n = Cn ⊕Dn

(C∗ ⊕D∗)n−1 = Cn−1 ⊕Dn−1

?∂⊕n =

?

∂Cn ⊕∂D

n

Or, more generally, (⊕

Cα)n =⊕

Cαn with boundary map defined coordinatewise:

∂⊕(· · · , xα, xβ, · · · ) = (· · · , ∂xα, ∂xβ, · · · )

The kernel and cokernels of ∂⊕ are also given coordinatewise:

ker ∂⊕ = ker ∂C ⊕ ker ∂D = ⊕Zαn

coker ∂⊕ = coker ∂C ⊕ coker ∂D = ⊕Bαn

with homology

Hn(⊕

Cα∗ ) =

⊕Zαn/

⊕Bαn∼=

⊕Hn(C

α∗ )

Lemma 2.1 has an obvious part and a nonobvious part. The obvious part, or at least thepart that you are already supposed to know, is the following.

(1) (⇒) Homotopy equivalent chain complexes have isomorphic homologies.(2) Next, C ' D implies that the cochain complexes are homotopy equivalent:

C∗(C;G) = Hom(C∗, G) ' C∗(D;G) = Hom(D∗, G)

(3) This implies that H∗(C∗;G) ∼= H∗(D∗;G).

Therefore, once we prove Lemma 2.1 we will know the following.

Theorem 2.3. The homology of a free chain complex determines its cohomology with anycoefficients.

We broke the (nontrivial part of) Lemma 2.1 into two parts and proved the first part.Step 1. Every free chain complex C∗ is isomorphic to a direct sum of free chain complexes

Cn∗ of the following form:

(2.1) Cn∗ : · · · ← 0← Dn

∂n+1←−−− Dn+1 ← Dn+2 ← 0 · · ·

where D = Cn∗ is exact except possibly at Dn. If the n-homology is H then we have a long

exact sequence:

(2.2) 0← Hε←− Dn

∂n+1←−−− Dn+1 ← Dn+2 ← 0 · · ·

The sequence (2.1) is called a free resolution of H.Step 2. Any two free resolutions of H are chain homotopy equivalent.To do step 1, we broke up any free chain complex

C∗ : · · · ← Cn−1∂n←− Cn

∂n+1←−−− Cn+1 ← · · ·4

at the point Cn. This uses the following short exact sequence:

0← Bn−1∂n←− Cn ←↩ Zn ← 0

The exactness of this sequence follows from the definitions of the symbols. Now use the factthat

Proposition 2.4. Every subgroup of a free abelian group is free abelian.

Thus Bn−1 is free which means that the above short exact sequence splits. I.e., there is ahomomorphism s : Bn−1 → Cn so that ∂ns = idB. This makes Cn = sBn−1 ⊕ Zn. But, theimage of ∂n+1 : Cn+1 → Cn lies in Zn and ∂n : Cn → Cn−1 is zero on Zn. Therefore the chaincomplex C∗ is isomorphic to the direct sum of the chain complexes:

· · · ← Cn−1∂n←− sBn−1 ← 0← · · ·

· · · ← 0← Zn∂n+1←−−− Cn+1 ← · · ·

The first sequence is exact at degree n and above. The second sequence is zero in degrees < n.So, if we do this at every place where the homology of C∗ is nonzero, we get a decompositionof C∗ as desired.

The traditional method is to do the above construction at every Cn, replacing each Cnwith sBn−1 ⊕ Zn to get a direct sum of the free chain complexes:

· · · ← 0← Zn∂n+1←−−− sBn ← 0← · · ·

2.2. Free resolutions and Ext. It remains to explain the following important theorem.

Theorem 2.5. Any abelian group has a free resolution and any two free resolutions arechain homotopy equivalent.

Serge Lang, in his book Algebra, said “Take any book on homological algebra, and proveall the theorems without looking at the proofs in that book.” Homological algebra hasadvanced a lot in 40 years. However, it is probably good advice to prove the basic theoremsyourself. You just need to know the exact statements of the theorems.

Proof. The existence of a free resolution is easy. Suppose that G is any abelian group. Thenthere exists a free abelian group F which maps onto G. (You could, e.g., take F to be thefree group generated by all the elements of G.) The kernel of the epimorphism F → G is asubgroup R of F which is also free since all subgroups of free abelian groups are free. Then

(2.3) 0← G← F ← R← 0

is a free resolution of G. This is called a free presentation of G. (A free resolution withCn = 0 for n ≥ 2.)

The following lemma shows the uniqueness of free resolutions up to homotopy: Given anytwo free resolutions C∗, D∗ of the same group G, the identity mapping G → G is coveredby chain maps f∗ : C∗ → D∗ and g∗ : D∗ → C∗. The composition g∗f∗ : C∗ → C∗ must behomotopic to the identity chain map C∗ → C∗ by the lemma since both cover the identitymap G→ G. �

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Lemma 2.6. Suppose that G,H are abelian groups with free resolutions:

0← Gε←− C0

∂←− C1 ← · · ·

0← Hε←− D0

∂′←− D1 ← · · ·Suppose that f : G→ H is any homomorphism. Then there exist a chain map f∗ : C∗ → D∗so that εf0 = fε : C0 → H. (We say that f∗ covers f .) Furthermore, any two such chainmaps are chain homotopy equivalent.

Proof. Since C0 is free, the homomorphism fε : C0 → H lifts to a homomorphism f0 : C0 →D0. The homomorphism f0∂ : C1 → D0 has image in ker ε = im ∂′. So it lifts to a mapf1 : C1 → D1, and so on.

If there are two chain maps f∗, g∗ covering the same homomorphism f : G→ H then thedifference f∗− g∗ covers 0 : G→ H. So, f0− g0 has image in ker ε = im ∂′ and therefore liftsto a homorphism h0 : C0 → D1, etc. The mappings hn give a chain homotopy h : f∗ ' g∗.I.e.,

∂′n+1hn + hn−1∂n = fn − gn�

Now we can define the functor Ext = Ext1Z.

Definition 2.7. If G,H are abelian groups then ExtnZ(H,G) is defined to be the nth coho-mology of the cochain complex1 Hom(C∗, G) where C∗ is any free resolution of H.

By Theorem 2.5 we can choose any C∗. In particular, we can take a free presentation:

0← H ← C0f←− C1 ← 0

Then Hom(C∗, G) is the chain complex:

0→ Hom(C0, G)f#

−→ Hom(C1, G)→ 0

I am using the symbol f# to indicate the dual of f given by pre-composition with f becausef ∗ is overused. Since this cochain complex has only two terms, it only has cohomology indegrees 0, 1. In degree 0, the kernel of f# is obviously isomorphic to Hom(H,G) since anymapping C0 → G which is zero on C1 factors uniquely through the cokernel of f : C1 → C0.

Proposition 2.8. ExtnZ(H,G) = 0 for n ≥ 2 and Ext0Z(H,G) = Hom(H,G).

This means that the only interesting term is Ext1Z which we call simply Ext

Ext(H,G) := Ext1Z(H,G) = coker f#

With the following theorem we can compute Ext(H,G) for any finitely generated abeliangroup H.

Theorem 2.9. For any abelian group G we have:

(1) Ext(H1 ⊕H2, G) ∼= Ext(H1, G)⊕ Ext(H2, G)(2) Ext(Z, G) = 0(3) Ext(Zn, G) ∼= G/nG

1A cochain complex is a sequence of abelian groups Cn and homomorphisms δ : Cn → Cn+1 so thatδ2 = 0. The cohomology of C∗ is the kernel of δ modulo the image of δ.

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2.3. UCT. Now that we have the definitions, we can state and prove the universal coefficienttheorem for cohomology.

Theorem 2.10. If C∗ is a free chain complex and G is an abelian group then we have anatural exact sequence:

0→ Ext(Hn−1(C∗), G)→ Hn(C∗;G)→ Hom(Hn(C∗), G)→ 0

Furthermore the sequence splits, but not naturally.

Proof. We know that C∗ is a direct sum of chain complexes C∗ =⊕

k C(k)∗ where C

(k)∗ is a

free resolution of Hk(C∗) shifted up in degree by k. So,

Hn(C∗;G) ∼=∏k

Hn(C(k)∗ ;G) =

∏k

Extn−kZ (Hk(C∗), G)

Since Extn−kZ = 0 except when k = n, n− 1 we have a finite product (= direct sum):

Hn(C∗;G) ∼= Ext0Z(Hn(C∗), G)⊕ Ext1

Z(Hn−1(C∗), G)

= Hom(Hn(C∗), G)⊕ Ext(Hn−1(C∗), G)

A more detailed description of this isomorphism shows which part is natural. I am skippingthis part. The important part is the naturality of the evaluation map

ev : Hn(C∗;G)→ Hom(Hn(C∗), G)

This map has the following natural description. Let f ∈ Hom(Cn, G) be a cocycle represent-ing a cohomology class [f ] ∈ Hn(C∗;G) and let z ∈ Cn be a cycle representing the homologyclass [z] ∈ Hn(C∗). Then ev[f ] is the homomorphism Hn(C∗)→ G which sends [z] to f(z).If f ′ = f + δg and z′ = z + ∂y are any other choices for f, z then

f ′(z′) = (f + g∂)(z + ∂y) = f(z) + f∂(y) + g∂z + g∂2y = f(z)

since f∂ = δf = 0, ∂z = 0 and ∂2 = 0. So, ev is independent of all choices, i.e., it iswell-defined. “Naturality” of ev is obvious if you know what it means. �

2.4. Problems. Problem 1. Find a free presentations of Zn and show that

Ext(Zn, G) ∼= G/nG

Conclude that G/6G ∼= G/2G⊕G/3G.Problem 2. Since Z6

∼= Z2 ⊕ Z3 we get two free presentations of Z6. Find an explicitchain homotopy equivalence between these free resolutions.

Problem 3. What does it mean that the evaluation map

ev : Hn(C∗;G)→ Hom(Hn(C∗), G)

is natural?Homework 2 This is the first problem in HW2. Find a free presentation for the additive

groupZ[1

p] = {a

b∈ Q | a, b ∈ Z and b is a power of p}

(p is prime) and use it to compute Ext(Z[1p],Z).

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3. Cohomology of spaces

Using the universal coefficient theorem and our knowledge of homology, we can computethe cohomology of spaces. The general properties of cohomology follow from the corre-sponding property for homology by duality. The evaluation map is also a consequence ofduality:

computation of Hn ⇒UCT computation of Hn

properties of Hn ⇒Hom properties of Hn

3.1. Singular cohomology. This is derived from singular homology. If X is a topologicalspace then recall that C∗(X) is the singular chain complex of X. Cn(X) is the free abeliangroup generated by the set of all continuous mappings σ : ∆n → X. These are called singularn-simplices in X and ∂ : Cn(X)→ Cn−1(X) is given by alternating sum of faces.

Definition 3.1. The singular cochain complex of a spaces X with coefficients in G is thesequence of groups Cn(X;G) := Hom(Cn(X);G) with coboundary map δ : Cn(X;G) →Cn+1(X;G) given by δ = ∂#.

δ(φ)σ = φ(∂σ) =n∑i=0

φ(σ | [v0, · · · , vi, · · · , vn])

3.1.1. UCT and computations. Since C∗(X) is a free chain complex, the UCT for free chaincomplexes (Theorem 2.10) implies the UCT for spaces.

Theorem 3.2 (UCT for spaces). For any space X and any additive group G we have anatural exact sequence:

0→ Ext(Hn−1(X), G)→ Hn(X;G)h−→ Hom(Hn(X), G)→ 0

where h is the evaluation map. Furthermore this sequence splits, but not naturally.

The evaluation map h is given by h[φ]([z]) = φ(z). In the special case when G = F is afield (e.g., F = Zp,Q,R,C) there is a simpler theorem:

Theorem 3.3 (UCT over a field). The evaluation map gives a natural isomorphism:

h : Hn(X;F )≈−→ Hom(Hn(X;F ), F )

This does not follow from the UCT for free modules (= vector spaces) over the field Finstead of free abelian groups and the fact that ExtnF (V,W ) = 0 for all n > 0.

Example 3.4. Take the projective plane X = P 2 = RP 2. We know that H0(P2) =

Z, H1(P 2) = Z2 and Hn(P 2) = 0 for n ≥ 2. So, the UCT tells us that

(0) H0(P 2;G) ∼= Hom(H0(P2), G) = G.

(1) H1(P 2;G) ∼= Hom(H1(P2), G) ⊕ Ext(H0(P

2);G) ∼= Hom(Z2, G). This is the groupof elements of G of order 2 (or 1).

(2) H2(P 2;G) ∼= Ext(H1(P2), G) = Ext(Z2, G) = G/2G.

In the special case when G = Z2 this gives H0 = H1 = H2 = Z2 and Hn = 0 for n > 2. Thisagrees with the second UCT since Hn(P

2; Z2) = Z2 for n = 0, 1, 2.

Problem Describe H0(X;G) for any X,G.8

3.1.2. Properties of cohomology. If X is a ∆-complex or a CW-complex, the cellular chaincomplex ∆∗(X) has the same homology as the singular chain complex C∗(X). Therefore,they are homotopy equivalent:

∆∗(X) ' C∗(X).

So, the cochain complexes are homotopy equivalent:

∆∗(X;G) = Hom(∆∗(X), G) ' Hom(C∗(X);G) = C∗(X;G)

This proves the following theorem.

Theorem 3.5. If X is a ∆-complex or a CW-complex, its cellular cohomology H∗∆(X;G) is

isomorphic to its singular cohomology H∗(X;G).

Example 3.6. The cellular chain complex of P 2 using the CW decomposition P 2 = ∗∪e1∪e2is given by:

0← Z 0←− Z 2←− Z← 0.

The cellular cochain complex is ∆∗(P 2;G) = Hom(∆∗(P2), G):

0→ G0−→ G

2−→ G→ 0.

If f : X → Y is a continuous mapping then we have a chain map f# : C∗(X) → C∗(Y )given by left composition with f : f#(σ) = f ◦ σ : ∆→ Y . This induces a cochain map:

f# : C∗(Y ;G)→ C∗(X;G)

given by right composition with f : f#(φ) = φ ◦ f . So,

f#(φ)(σ) = φ(f ◦ σ) = φ(f#(σ))

The cochain map f# induces a map (homomorphism) in cohomology:

f ∗ : H∗(Y ;G)→ H∗(X;G)

The key point is that it goes the wrong way (Y to X). So, cohomology is a contravariantfunctor.

Theorem 3.7. If f, g : X → Y are homotopic then they induce the same map in cohomology:

f ∗ = g∗ : H∗(Y ;G)→ H∗(X;G)

Proof. We know that f#, g# : C∗(X) → C∗(Y ) are chain homotopic. Let h be a chainhomotopy. Thus

∂h+ h∂ = g# − f#

Applying the functor Hom(−, G) to everything we get:

h#δ + δh# = g# − f#

Therefore, h# gives a homotopy f# ' g#. So, f#, g# induce the same map in cohomology.In case these symbols don’t mean anything to you, this follows from the previous equationby duality:

(h#δ + δh#)(φ) = φ ◦ ∂ ◦ h+ φ ◦ h ◦ ∂ = φ(∂h+ h∂)

= φ(g# − f#) = φ ◦ g + φ ◦ f = (g# − f#)(φ)

(h#δ)(φ) = φ ◦ ∂ ◦ h : Cn(X)h−→ Cn+1(Y )

∂−→ Cn(Y )φ−→ G

�

HW2(b) Do problem 11 on p. 205.9

3.2. Relative cohomology and long exact sequences. The definition of the relativecohomology is dual to relative homology and the long exact cohomology sequence of a pair isgiven by the same algebraic lemma, namely that a short exact sequence of cochain complexesgives a long exact sequence in cohomology. (Some authors use the formal trick: Cn = C−nto convert a cochain complex into a chain complex.)

If A is a subspace of X then we have the short exact sequence of free chain complexes:

0→ C∗(A)j#−→ C∗(X)→ C∗(X,A)→ 0

It is important that C∗(X,A) is a free chain complex: Cn(X,A) is the free abelian groupgenerated by singular n-simplices σ : ∆n → X whose images do not lie in A.

When we hom this into a group G we get a short exact sequence of cochain complexes:

(3.1) 0→ C∗(X,A;G)→ C∗(X;G)j#−→ C∗(A;G)→ 0

The relative cohomology groups H∗(X,A;G) are the cohomology groups of the cochain com-plex C∗(X,A;G).

3.2.1. Interpretation of relative cohomology. Cohomology is nicer than homology in manyways. For example Cn(X,A) = Cn(X)/Cn(A) is a quotient group and elements of Cn(X,A)are cosets of Cn(A). But Cn(X,A;G) is naturally a subset of Cn(X;G). (It is the kernel ofthe restriction map j#.) Therefore, we have the following interpretation of relative cochains,cocycles and coboundaries.

(1) A relative cochain φ ∈ Cn(X,A;G) is a function which assigns to each singularsimplex σ : ∆n → X an element φ(σ) ∈ G so that φ(σ) = 0 if σ(∆n) ⊂ A.

(2) φ is a relative cocycle if δφ = 0. I.e.,

Zn(X,A;G) = Cn(X,A;G) ∩ Zn(X;G)

(3) φ is a relative coboundary if φ = δψ for some ψ ∈ Cn−1(X,A;G).

3.2.2. Computation of relative cohomology. Since C∗(X,A) is a free chain complex, its co-homology can be computed using the UCT.

Theorem 3.8. There is a natural short exact sequence which splits unnaturally:

0→ Ext(Hn−1(X,A), G)→ Hn(X,A;G)h−→ Hom(Hn(X,A), G)→ 0

where h is the evaluation map:

h([φ])(z + Cn(A) +Bn(X)) = φ(z)

φ(z) ∈ G is well defined: φ = 0 on Cn(A) by definition of a relative cochain and φ = 0 onBn(X) since δφ = 0.

For example, if X is obtained from a space A by attaching r number of n-cells en1 , · · · , enkalong maps ηi : Sn−1 → A then Hn(X,A) = Zr and Hi(X,A) = 0 for i 6= n. So,

Hn(X,A;G) ∼= Hom(Zr, G) = Gr

3.2.3. Long exact sequences. The short exact sequence of cochain complexes (3.1) gives along exact sequence in cohomology, just as in the homology case:

· · · → Hn(X,A;G)→ Hn(X;G)∗−→ Hn(A;G)

δ−→ Hn+1(X,A;G)→ · · ·The mapping δ is given by diagram chasing, just like before.

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3.2.4. Problems.

(1) What is the definition of the connecting homomorphism δ in the long exact sequence?(2) Find an example of the previous example where δ : Hn−1(A;G) → Hn(X,A;G) is

nonzero.(3) Show that the composition

Ext(Hn−2(A), G)→ Hn−1(A;G)δ−→ Hn(X,A;G)

is zero when X = A ∪ en1 ∪ · · · ∪ enr .We did (1) and (3) in class. Do (2) for homework.

11

4. Cup product

If the coefficient group is (the underlying additive group of) a ring R. Thus we have amultiplication map µ : R ⊗ R → R which is associative with unit. Then the cohomologygroups Hn(X;R) form a graded ring

⊕Hn(X;R). It is very important that this ring

structure is defined at the cochain level. I.e.,⊕

Cn(X;R) has an associative ring structure.The multiplication is call cup product.

4.1. Cup product at the cochain level. Here is my favorite formula for the cup product:

φ ∪ ψ = µ ◦ (φ⊗ ψ) ◦∆

Here φ ∈ Ck(X;R), ψ ∈ C`(X;R) and φ ∪ ψ ∈ Ck+`(X;R) = Hom(Ck+`(X), R) is given bythe composition

Ck+`(X)∆k,`−−→ Ck(X)⊗ C`(X)

φ⊗ψ−−→ R⊗R µ−→ R

4.1.1. The diagonal map ∆. The diagonal map is given by “front face” and “back face.”

∆k,`(σ) = fkσ ⊗ b`σ

where fkσ is the front k-face of σ given by

fkσ = σ|[v0, v1, · · · , vk]

and b`σ is the back `-face of σ given by

b`σ = σ|[vk, · · · , vk+`]

Thus, the cup product is given by:

(φ ∪ ψ)(σ) = µ(φ⊗ ψ)∆(σ) = µ(φ⊗ ψ)(fkσ ⊗ b`σ) = φ(fkσ)ψ(b`σ)

= φ(σ|[v0, · · · , vk])ψ(σ|[vk, · · · , vk+`])

Lemma 4.1. We have the following identities of operators on Ci+j+k(X).

(1) fifi+j = fi: the front i-face of the front i+ j-face is the front i-face.(2) bkbj+k = bk: the back face of the back face is the back face.(3) bjfi+j = fjbj+k: this is the middle face mjσ = σ|[vi, · · · , vi+j].

In terms of the diagonal maps defined on Ci+j+k(X) these three formulas give one formula:

(∆ij ⊗ id) ◦∆i+j,k = (id⊗∆jk) ◦∆i,j+k

When we apply these operators to σ we get:

fi(fi+jσ)⊗ bj(fi+jσ)⊗ bkσ = fiσ ⊗ fj(bj+kσ)⊗ bk(bj+kσ)

When we sum over all i, j, k we get:

(∆⊗ id)∆ = (id⊗∆)∆

i.e., ∆ is coassociative.12

4.1.2. Properties of cup product. The coassociativity of ∆ (and the associativity of µ) implythat cup product is associative at the cochain level.

Theorem 4.2. If α ∈ Ci(X;R), β ∈ Cj(X;R), γ ∈ Ck(X;R) then

(α ∪ β) ∪ γ = α ∪ (β ∪ γ)

Proof. Associativity of

(α ∪ β) ∪ γ = µ((α ∪ β)⊗ γ)∆ = µ(µ⊗ id)(α⊗ β ⊗ γ)(∆⊗ id)∆α ∪ (β ∪ γ) = µ(α⊗ (β ∪ γ))∆ = µ(id⊗ µ)(α⊗ β ⊗ γ)(id⊗∆)∆

But µ(µ⊗ id) = µ(id⊗µ) by associativity of µ and (∆⊗ id)∆ = (id⊗∆)∆ by coassociativityof ∆ �

Problem Show that the augmentation map ε : C0(X) → R (given by composing theusual augmentation map ε : C0(X)→ Z with the unique ring homomorphism Z→ R) is theunit for cup product, i.e.

ε ∪ φ = φ = φ ∪ εProblem Show that cup product is natural, i.e., given any continuous mapping f : X →

Y and φ ∈ Ck(Y ;R), ψ ∈ C`(Y ;R) we have:

f#(φ ∪ ψ) = f#(φ) ∪ f#(ψ)

In other words, the following diagram commutes.

Ck(X;R)⊗ C`(X;R) Ck(Y ;R)⊗ C`(Y ;R)

Ck|`(X;R) Ck+`(Y ;R)

?∪X

�f#⊗f#

?

∪Y

�f#

By looking at this diagram and the definition of ∪ we realized that we need to show:

φ(ffkσ)ψ(fb`σ) = φ(fk(fσ))ψ(b`(fσ))

I.e., we need to know that ffkσ = fk(fσ) and fb`σ = b`(fσ). To show the first one welooked at another diagram:

∆k ∆k+`

X Y

-fk

@@

@@@R

fkσ

?

σ

@@

@@@R

fσ

-f

We realized that the front k-face of a singular k + ` simplex σ : ∆k+` → X is actuallya composition: fkσ = σ ◦ fk where fk : ∆k → ∆k+` is the inclusion of the front k-face.Therefore:

f(fkσ) = f ◦ (σ ◦ fk) = (f ◦ σ) ◦ fk = fk(fσ).

13

4.1.3. Derivation. The main property of cup product is that δ is a graded derivation:

Theorem 4.3.δ(φ ∪ ψ) = δφ ∪ ψ + (−1)kφ ∪ δψ

if φ ∈ Ck(X;R).

Proof. Apply both sides to any σ : ∆k+`+1 → X.

(4.1) δ(φ ∪ ψ)σ = (φ ∪ ψ)∂σ =k+`+1∑i=0

(−1)i(φ ∪ ψ)∂iσ =k+`+1∑i=0

(−1)iφ(fk∂iσ)ψ(b`∂iσ)

This expression has k + `+ 2 terms.

(4.2) (δφ ∪ ψ)σ = δφ(fk+1σ)ψ(b`σ) =k+1∑i=0

(−1)iφ(∂ifk+1σ)ψ(b`σ)

But ∂ifk+1σ = fk∂iσ and b`σ = σ|[vk+1, · · · , vk+`+1] = b`∂iσ for i ≤ k. So, the terms withi ≤ k in (4.2) are equal to the first k + 1 terms in (4.1).

(4.3) (−1)k(φ ∪ δψ)σ = (−1)kφ(fkσ)δψ(b`+1σ) =`+1∑j=0

(−1)k+jφ(fkσ)ψ(∂jb`+1σ)

Here ∂jb`+1σ = b`∂j+kσ and, for j ≥ 1, fkσ = fk∂j+kσ. So, the terms with j ≥ 1 in (4.3) areequal to the last `+ 1 terms in (4.1).

Finally, the last term of (4.2) cancels the first terms of (4.3). They are

φ(fkσ)ψ(b`σ)

with opposite signs. �

4.2. Cup product in cohomology. Theorem 4.3 implies that cup product induces a mul-tiplication in cohomology.

(1) If φ, ψ are both cocycles then φ ∪ ψ is a cocycle.(2) If φ is a coboundary φ = δα and ψ is a cocycle then

δ(α ∪ ψ) = δα ∪ ψ ± α ∪ δψ = φ ∪ ψSo, φ ∪ ψ is a coboundary.

(3) Similarly, if φ is a cocycle and ψ is a coboundary, then φ ∪ ψ is a coboundary.

Corollary 4.4. Cup product induces a multiplication in cohomology:

∪ : Hk(X;R)⊗H`(X;R)→ Hk+`(X;R)

by [φ] ∪ [ψ] = [φ ∪ ψ]. Furthermore, this multiplication is associative with unit. So,⊕Hn(X;R)

is a graded ring for any associative ring R.

The word graded ring means that all elements are sums of homogeneous elements and theproduct of a homogeneous element of degree k with a homogeneous element of degree ` isa homogeneous element of degree k + `. E.g., polynomial rings are graded but power seriesrings are not.

14

4.2.1. Note on tensor product. The coefficient ring R will be suppressed from the notation.Also, we will from now on, assume that R is commutative.

I actually only defined cup product on the Cartesian product

Ck(X)× C`(X)∪−→ Ck+`(X)

However, the cup product is biadditive, i.e., (φ1+φ2)∪ψ = φ1∪ψ+φ2∪ψ and φ∪(ψ1+ψ2) =φ ∪ ψ1 + φ ∪ ψ2. So, it induces a map on the tensor product

Ck(X)⊗ C`(X)∪−→ Ck+`(X)

It is also R-bilinear in the sense that it also satisfies (rφ)∪ ψ = r(φ∪ ψ) = φ∪ (rψ). So, weget:

Ck(X)⊗R C`(X)∪−→ Ck+`(X)

4.2.2. Cup product on relative cohomology. If A ⊆ B ⊆ X then the cup product gives amap:

Ck(X,A)⊗R C`(X,B)∪−→ Ck+`(X,B)

We analyzed why this is true.

(1) If φ ∈ Ck(X,A) then φ(σ) = 0 whenever σ(∆k) ⊆ A.(2) If ψ ∈ C`(X,B) then ψ(σ) = 0 whenever σ(∆`) ⊆ B.(3) So, (φ ∪ ψ)σ = φ(fkσ)ψ(b`σ) = 0 if σ(∆`) ⊆ B.

It would also work to put Ck+`(X,A) instead of Ck+`(X,B). I said that it was usually“better” to use B instead of A. In cohomology we get:

Hk(X,A)⊗R H`(X,B)∪−→ Hk+`(X,B)

4.2.3. Cross product.

Definition 4.5. For a Cartesian product X × Y the cross product

× : Hk(X;R)⊗R H`(Y ;R)→ Hk+`(X × Y ;R)

is defined byα× β = p∗X(α) ∪ p∗Y (β)

where p∗X , p∗Y are the maps in cohomology induced by the projection maps pX : X × Y → X

and pY : X × Y → Y .

The cross product is defined at the cochain level by

φ× ψ = p#X(φ) ∪ p#

Y (ψ)

15

5. Kunneth formula

The Kunneth formula gives the homology and cohomology of a product of spaces. In myapproach to this I used universal objects which are called “acyclic models” and the “Yonedaelement.” I used the algebraic part of the Kunneth formula to prove the geometric part.

5.1. Logic. To dispel the appearance of a circular argument, I explained the logic of theproof. The proof is divided into two parts which are independent. The first part is ageometric theorem called the Eilenberg-Zilber Theorem.

Theorem 5.1 (Thm A: Eilenberg-Zilber). The singular chain complex of a product spaceX×Y is naturally homotopy equivalent to the tensor product of the singular chain complexesof the factors:

C∗(X × Y ) ' C∗(X)⊗ C∗(Y )

The second part (Theorem B) computes the homology and cohomology of a tensor productof free chain complexes. This is done in two parts B1, B2. We only need the first part toprove part A (Thm B1 ⇒ Thm A).

Theorem 5.2 (Thm B1). Suppose that C∗, D∗ are free chain complexes so that the homologyof D∗ is free abelian. Then the homology of C∗ ⊗ C∗ is isomorphic to the tensor product ofthe homologies of C∗, D∗:

Hn(C∗ ⊗D∗) ∼=⊕k+`=n

Hk(C∗)⊗H`(D∗)

We will discuss the general formula (Theorem B2) later.

5.2. Tensor product of chain complexes. We went over the definition of the tensorproduct of chain complexes and proved Theorem B1.

Definition 5.3. If C∗, D∗ are chain complexes with differentials ∂C , ∂D then the chain com-plex C∗ ⊗D∗ is defined as follows. The groups in the chain complex are:

(C∗ ⊗D∗)n :=⊕k+`=n

The boundary map ∂⊗ : Ck ⊗D` → (C∗ ⊗D∗)k+`−1 is given by

∂⊗(x⊗ y) = ∂Cx⊗ y + (−1)kx⊗ ∂Dy

The sign (−1)k insures that (∂⊗)2 = 0.

Lemma 5.4. If C∗ ' C ′∗ and D∗ ' D′

∗ then C∗ ⊗D∗ ' C ′∗ ⊗D′

∗.

Proof of Theorem B1. If H∗(D∗) is free abelian then D∗ ' H∗(D∗) are homotopic free chaincomplexes (the boundary map being zero in H∗(D∗)). Therefore, by the lemma above,C∗ ⊗D∗ ' C∗ ⊗H∗(D∗) with homology

Hn(C∗ ⊗D∗) ∼= Hn(C∗ ⊗H∗(D∗)) ∼=⊕k+`=n

Hk(C∗;H`(D∗)) ∼=⊕k+`=n

Hk(C∗)⊗H`(D∗)

�16

5.3. Skew-commutativity of cup product. The first theorem we proved using TheoremB1 and the universal-Yoneda element argument was the skew commutativity of cup product.

Theorem 5.5. If R is commutative and α ∈ Hk(X;R), β ∈ H`(X;R) then

α ∪ β = (−1)k`β ∪ α

Proof. I made the definition:

α ∪′ β = (−1)k`β ∪ αWe want to show that ∪′ = ∪. To show this I went to the cocycles. Suppose that φ, ψ arecocycles representing α, β and σ : ∆k+` → X. Then

(φ ∪ ψ)(σ) = φ(fkσ)ψ(b`σ)

(φ ∪′ ψ)(σ) = (−1)k`(ψ ∪ φ)(σ) = (−1)k`φ(bkσ)ψ(f`σ)

Next, I eliminate all three variables (φ, ψ, σ) by using my favorite formula:

φ ∪ ψ = µ(φ⊗ ψ)∆k.`

φ ∪′ ψ = µ(φ⊗ ψ)∆′k.`

where ∆k,`σ = fkσ ⊗ b`σ and

∆′k,`σ = (−1)k`bkσ ⊗ f`σ

So, it suffices to show that ∆ ' ∆′. This follows from the following two lemmas.

Lemma 5.6. ∆ and ∆′ are natural augmented chain maps

C∗(X)→ C∗(X)⊗ C∗(X)

The fact that they are natural and augmented is obvious. (They are augmented sinceε = (ε⊗ ε)∆0,0 = (ε⊗ ε)∆′

0,0.) A calculation shows they are chain maps.

Lemma 5.7. Any two natural augmented chain maps C∗(X)→ C∗(X)⊗C∗(X) are naturallychain homotopic.

Suppose that f, g : C∗(X) → C∗(X) ⊗ C∗(X) are two natural augmented chain maps.Then, the homotopy that we want is a sequence of maps

hXn : Cn(X)→ (C∗(X)⊗ C∗(X))n+1

satisfying two condition

(1) ∂⊗hn + hn−1∂ = gn − fn(2) hXn is natural, i.e., for any map f : X → Y we get a commuting square.

Since Cn(X) is freely generated by σ : ∆n → X, it suffices to find an equation for hXn (σ). Thisis the point where I used the universal argument. The mapping hXn exists in the universalcase when X = ∆n.

h∆n

: C∗(∆n)→ C∗(∆

n)⊗ C∗(∆n)

This is because ∆n is contractible and thus C∗(∆n) and C∗(∆

n) ⊗ C∗(∆n) are acyclic (byB1 (Theorem 5.2). Thus, they are both free resolutions of Z and any two free resolutionsof the same abelian group are chain homotopy equivalent by augmented chain maps whichare unique up to homotopy. Furthermore, the maps h∆n

m in the homotopy are constructedinductively on m ≥ 0.

17

The logic is that, by induction on n, the maps hXn−1 exist for all X and are natural. Then,h∆n

n exists. We apply this map to the Yoneda element

id∆n : ∆n → ∆n ∈ Cn(∆n)

Naturality of hXn means that we want the following diagram to commute:

id∆n ∈ Cn(∆n) (C∗(∆n)⊗ C∗(∆n))n+1

σ ∈ Cn(X) (C∗(X)⊗ C∗(X))n+1

-h∆n

n

?

σ#

?

(σ⊗σ)#

-hX

n

This forces the following formula which we took as the definition of fXn (σ):

fXn (σ) := (σ ⊗ σ)∗(f∆n

n (id∆n)

In other words, fXn (σ) is determined for all X and all σ by the value of hn in the universalcase on the Yoneda element. �

18

5.4. Homotopy invariance. The second example of acyclic models and the Yoneda elementwas the homotopy invariance of homology and cohomology.

Theorem 5.8. Homotopic mappings f ' g : X → Y induce chain homotopic maps on thesingular chain complexes:

f# ' g# : C∗ → D∗

By an easy argument which I explained before, this theorem implies that homology andcohomology are homotopy invariant:

Corollary 5.9. Homotopic maps f ' g : X → Y induce the same map in homology andcohomology with arbitrary coefficients. (f∗ = g∗ and f ∗ = g∗.)

Thus it suffices to prove Theorem 5.8.

5.4.1. Eliminate Y . We want to eliminate X and Y . I.e., instead of proving the theorem forall X and all Y it suffices to prove it for a particular case where it is obvious. The first stepis to eliminate Y as an independent variable.

Problem. It suffices to consider the special case when Y = X × I, f = j0 and g = j1where jt : X → X × I is the mapping jt(x) = (x, t). Why?

Answer. Suppose that h : X × I → Y is a homotopy from f to g. Then f = h ◦ j0 andg = h ◦ j1. Therefore, f# = h# ◦ (j0)#

f# : C∗(X)(j0)#−−−→ C∗(X × I)

h#−→ C∗(Y )

and g# = h# ◦ (j1)#. Now suppose that J is a chain homotopy from (j0)# to (j1)#. Then∂J + J∂ = (j1)# − (j0)# and h#J is a homotopy from f# to g#. The reason is that h# is achain map:

∂(h#J) + (h#J)∂ = h#∂J + h#J∂ = h#(∂J + J∂) = h#((j1)# − (j0)#) = g# − f#

5.4.2. Universal example. We want to construct a chain homotopy between the two chainmaps (j0)#, (j1)# : C∗(X)→ C∗(X×I) for all spaces X. This will consist of homomorphisms

hXn : Cn(X)→ Cn+1(X × I)for all spaces X and all n ≥ 0 satisfying the equation:

(5.1) ∂hXn + hXn−1∂ = (j1)# − (j0)#

First of all, since Cn(X) is freely generated by the singular n simplices σ : ∆n → X, thehomomorphism hXn is determined by the elements hXn (σ) ∈ Cn+1(X×I) which can be chosenarbitrarily. However, we also need hXn to satisfy (5.1) and we want it to be natural. Inparticular we want the following diagram to commute for all σ : ∆n → X.

id∆n ∈ Cn(∆n) Cn+1(∆n × I)

σ ∈ Cn(X) Cn+1(X × I)

-h∆n

n

?

σ#

?

(σ×idI)#

-hX

n

Here ∆n is the universal object and id∆n ∈ Cn(∆n) is the Yoneda element.19

We assume by induction on n that hXn−1 has been constructed for all X so that it is naturaland satisfies (5.1). Since ∆n and ∆n × I are contractible, the chain complexes C∗(∆

n) andC∗(∆

n × I) acyclic. Therefore, h∆n

n exists satisfying (5.1). Choose one. Then hXn can bedefined for all X by

hXn (σ) := (σ × idI)#h∆n

n (id∆n)

It is easy to see that this is natural and satisfies (5.1). This completes the proof of Theorem5.8.

5.5. Eilenberg-Zilber. The last example of acyclic models is the Eilenberg-Zilber Theorem:

C∗(X)⊗ C∗(Y ) ' C∗(X × Y )

We need to construct natural chain maps

f : C∗(X)⊗ C∗(Y )→ C∗(X × Y )

and g going the other way so that the compositions f ◦ g, g ◦f are homotopic to the identity.First we construct f .

5.5.1. Construction of f . We want to construct homomorphisms

fX,Yk,` : Ck(X)⊗ C`(Y )→ Ck+`(X × Y )

for all X, Y and all k, ` ≥ 0 satisfying two properties:

(1) f is a chain map: ∂f = f∂.(2) f is natural.

The group Ck(X)⊗C`(Y ) is freely generated by the elements σ ⊗ τ where σ : ∆k → X andτ : ∆` → Y . The universal example is given by X = ∆k and Y = ∆`. The Yoneda elementis

id∆k ⊗ id∆` ∈ Ck(∆k)⊗ C`(∆`)

Naturality of f requires that the following diagram commute.

id∆k ⊗ id∆` ∈ Ck(∆k)⊗ C`(∆`) Ck+`(∆k ×∆`)

σ ⊗ τ ∈ Ck(X)⊗ C`(Y ) Ck+`(X × Y )?

-f∆k,∆`

k.`

?

σ#⊗τ#

?

(σ×τ)#

-fX,Y

k,`

Since C∗(∆k) ⊗ C∗(∆

`) and C∗(∆k × ∆`) are acyclic, the chain map f∆k,∆`

exists and isunique up to homotopy. Furthermore the construction of this chain map is by inductionon degree. Therefore, we may assume by induction on κ + ` that fX,Yk,` exists in all smaller

degree sums for all X, Y and we may choose f∆k,∆`

k,` so that ∂f = f∂. Then fX,Yk,` can bedefined for all X, Y by

fX,Yk,` (σ ⊗ τ) := (σ × τ)#f∆k,∆`

k,` (id∆k ⊗ id∆`)

20

5.5.2. Construction of g. Now we want to construct a natural chain map g : C∗(X × Y )→C∗(X)⊗ C∗(Y ). This is a sequence of homomorphisms

gn : Cn(X × Y )→ (C∗(X)⊗ C∗(Y ))n

satisfying ∂⊗gn = gn−1∂. Here the universal example is X = Y = ∆n and the Yonedaelement is the diagonal map ∆ : ∆n → ∆n×∆n. The reason that this is the Yoneda elementis the following.

A generator of Cn(X × Y ) is a singular n simplex σ : ∆n → X × Y . But this is given bythe pair of maps σX = pX ◦ σ and σY = pY ◦ σ since

σ(t) = (σX(t), σY (t))

The pair of maps σX , σY sends the diagonal ∆ to σ = (σX , σY ).

∆ ∈ Cn(∆n ×∆n) (C∗(∆n)⊗ C∗(∆n))n

σ = (σX , σY ) ∈ Cn(X × Y ) (C∗(X)⊗ C∗(Y ))n)?

-g∆

n,∆n

n

?

(σX×σY )#

?

(σX)#⊗(σY )#

-gX,Y

n

As before, we assume by induction on n that a natural map gX,Yn−1 has been constructed forall X, Y . Then the chain complexes at the top are acyclic, the map g∆n,∆n

n exists and wemay define gX,Yn by

gX,Yn (σX , σY ) := ((σX)# ⊗ (σY )#) g∆n,∆n

n (∆)

5.5.3. The homotopies. What I didn’t have time to do in class was the homotopies. I’llshow that fg : C∗(X × Y )→ C∗(X × Y ) is homotopic to the identity map. The other way(gf ' id) is similar. The precise statement is the following.

Lemma 5.10. Any two natural chain maps C∗(X × Y )→ C∗(X × Y ) are homotopic.

Proof. We want to construct a natural chain homotopy

hX,Yn : Cn(X × Y )→ Cn+1(X × Y )

hn−1 exists for allX, Y by induction and hn exists for the universal exampleX = ∆n, Y = ∆n.The value of h∆n,∆n

n on the Yoneda element ∆ determines hX,Yn :

hX,Yn (σX , σY ) := (σX × σY )#h∆n,∆n

n (∆).

�

5.6. The Yoneda element in general. The Yoneda element is, by definition, always equalto the identity map of the universal object (or it correspond to the identity map). (Whatabout ∆ ∈ Cn(∆n×∆n)?) The theorem that we saw several examples of is called the YonedaLemma. It is considered to be trivial. There are two variations of the Yoneda Lemma, onein the category of sets and one in the category of abelian groups.

21

5.6.1. Set theoretic Yoneda Lemma.

Definition 5.11. Suppose the C is any category and U is any fixed object of C. (U will bethe universal object.) For any other object X of C let F (X) = MorC(U,X) be the set of allmorphisms from U to X in the category C. Then idU ∈ F (U) is the Yoneda element for F .

A functor F ′ : C → Sets is called representable if there is a natural isomorphism F ′(X) ∼=MorC(U,X). The element of F ′(U) corresponding to the identity map on U is called theYoneda element of F ′.

There is a similar definition for contravariant functors: F (X) = MorC(X,U) is the rep-resentable contravariant functor with universal object U and Yoneda element idU ∈ F (U).

Theorem 5.12 (Yoneda Lemma). Suppose that G : C → Sets is any functor and F is asabove. Then, a natural map φX : F (X)→ G(X) is given uniquely by

φX(σ : U → X) := σ∗(x)

where x = φU(idU) is an arbitrary fixed element of G(U).

This theorem is usually stated as follows: There is a bijection between the set of naturaltransformations φ : F → G and the set of elements of G(U).

5.6.2. Two examples. Here are two examples in the homotopy category. First, the categories.Let H be the category whose objects are all nice spaces (those homotopy equivalent to ∆-complexes) and whose morphisms are homotopy classes of maps [f ] : X → Y . Thus:

MorH(X,Y ) = [X, Y ]

where [X, Y ] is the standard notation for the set of homotopy classes of maps X → Y .Let H0 be the category of pointed nice topological spaces (X, x0) and pointed homotopy

classes of maps [f ] : (X, x0) → (Y, y0). This means that f(x0) = y0 and f ' g if thehomotopy h is also pointed (h(x0, t) = y0 for all t ∈ I). So,

MorH0(X, Y ) = [X, Y ]0

The base point is understood and indicated by the subscript 0.

Example 5.13. Take C = H0 with universal object U = (Sn, ∗). Then πn(X, x0) :=[(Sn, ∗), (X, x0)]0. The Yoneda element is [idSn ] ∈ πn(Sn). The Yoneda Lemma says, e.g.,that the set of all natural maps πn(X)→ πm(X) are in 1-1 corespondence with the elementof G(U) = πm(Sn).

Example 5.14. Take C = H with some universal object U . Then F (X) = [X,U ] is arepresentable contravariant functor with Yoneda element [idU ] ∈ F (U). If G(X) = [X,V ] isanother representable functor then the Yoneda Lemma says that the set of natural transfor-mations F (X)→ G(X) is in 1-1 correspondence with the elements of G(U) = [U, V ].

An important special case of this example is singular cohomology:

Hn(X;G) ∼= [X,K(G, n)]

The Yoneda Lemma says that the set of all natural maps Hn(X;G)→ Hm(X;H) is in 1-1correspondence with the elements of Hm(K(G, n)).

Here is an example. Cup square is a natural map Hn(X;R)→ H2n(X;R). By the YonedaLemma, it corresponds to some element of H2n(K(R, n);R). What is it?

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5.6.3. Algebraic Yoneda Lemma. The version of the Yoneda Lemma that we have been usingis the following.

Lemma 5.15. Suppose that F : C → Sets is a representable functor F (X) = MorC(U,X).For each X in C let C(X) is the free abelian group generated by F (X). Then C is a functorC → Ab and, for any other functor G : C → Ab, there is a 1-1 correspondence between theset of natural transformations φ : C → G and the elements of G(U). We say that C is freeon the universal object U .

Proof. Any natural map φX : C(X)→ G(X) is given uniquely by

φX(f : U → X) = f∗(x)

where x = φU(idU) is an arbitrary element of G(U). �

The main example is the singular chain complex C∗(X). For each n ≥ 0, Cn(X) is freeon the universal object ∆n. The Eilenberg-Zilber Theorem is an example of the followinggeneral theorem.

Theorem 5.16 (Acyclic Model Theorem). Suppose that C∗(X), D∗(X) are augmented chaincomplexes defined naturally for all objects X of some category C. Suppose that Cn(X), Dn(X)are free with universal objects Un, Vn. If C∗(Un), C∗(Vn), D∗(Un), D∗(Vn) are acyclic for alln then C∗(X) ' D∗(X).

This also holds in the case when Cn(X), Dn(X) are direct sums of free functors whoseuniversal objects are acyclic in both C∗ and D∗. Universal objects are also called models.

The Acyclic Model Theorem follows from the following lemma. This is also the lemmathat we actually used several times.

Lemma 5.17. Suppose that C∗(X), D∗(X) are augmented chain complexes defined naturallyfor all objects X of some category C. Suppose that Cn(X) are free on models Un. If D∗(Un)are acyclic for all n then there is a natural augmented chain map C∗(X)→ D∗(X) and anytwo such chain maps are naturally chain homotopic.

Again, this also holds in the case when Cn(X) is a direct sum of free functors on modelsUn,i as long as D∗(Un,i) are all acyclic.

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