After the effort invested in proving the general theorem on acyclic models, it is time to apply it to topology. First, let us prove:

Theorem 5 Suppose ${f, g: X \rightarrow Y}$ are homotopic. Then the maps ${f_*, g_*: H_*(X) \rightarrow H_*(Y)}$ are equal.

Proof: Suppose ${H: X \times [0,1]}$ is a homotopy with ${H(\cdot, 0) = f, H(\cdot, 1) = g}$. Then the maps ${f,g}$ factor as

$\displaystyle X \rightrightarrows^{i_0}_{i_1} X \times [0,1] \rightarrow^H Y$

so if we show that the inclusions ${i_0,i_1}$ sending ${x \in X}$ to ${(x, 0), (x,1)}$ induce equal maps on homology, we will be done.

Write ${I = [0,1]}$ for simplicity. For each space ${X}$, the maps ${i_0, i_1: X \rightarrow X \times I}$ are natural. More precisely, if ${X \rightarrow X, X \rightarrow X \times I}$ are the two functors ${\mathbf{Top} \rightarrow \mathbf{Top}}$, then ${i_0, i_1}$ are natural transformations between them.

So we have two maps ${C_*(X) \rightarrow C_*(X \times I)}$ induced by ${i_0,i_1}$, and these are natural transformations between the functors ${X \rightarrow C_*(X), X \rightarrow C_*(X \times I)}$. They induce the same map on zeroth homology because ${(x, 0)}$ and ${(x,1)}$ are connected by a path in ${X \times I}$ and so the zero-singular simplices ${(x,0), (x,1)}$ differ by the boundary of a 1-simplex. Moreover, the functor ${X \rightarrow C_*(X)}$ is free on the objects ${\Delta^n}$ with the canonical ${n}$-simplices ${1_{\Delta^n}:\Delta^n \rightarrow \Delta^n}$.

Moreover, ${X \rightarrow C_*(X \times I)}$ is acyclic in positive dimensions on these simplices because ${\Delta^n \times I}$ is contractible for any ${I}$. So the two maps ${C_*(X) \rightarrow C_*(X \times I)}$ are naturally chain homotopic by the acyclic model theorem, and thus induce the same maps on homology. This completes the proof. $\Box$

3. The acyclic model theorem for augmented chain complexes

That was convenient. Yet there was something annoying. We had to show that the maps on zeroth homology were the same, which required some (not much) topological reasoning. We want to eliminate that. We can do so, because there is a canonical augmentation on the chain complexes ${C_*(X)}$. Let’s recall what this means.

Definition 6 An augmented chain complex (of abelian groups) is a chain complex ${C_*}$ together with an epimorphism ${C_0 \rightarrow \mathbb{Z}}$ such that ${C_1 \rightarrow C_0 \rightarrow \mathbb{Z}}$ is zero. The map ${C_0 \rightarrow \mathbb{Z}}$ is called the augmentation. A morphism of augmented chain complexes ${C_* \rightarrow D_*}$ is a morphism of chain complexes that preserves the augmentation on ${C_0, D_0}$.

We note that the chain complexes ${C_*(X)}$ are augmented. Indeed, an element in ${C_0(X)}$ is just a formal sum of points, and the augmentation sends this to the total number of points in the formal sum; ${\sum n_i p_i}$ is sent to ${\sum n_i}$. Note that ${C_*}$ is functorial as a functor from ${\mathbf{Top}}$ to the category of augmented chain complexes. We say that an augmented chain complex is exact if the complex ${C_n \rightarrow C_{n-1} \rightarrow \dots \rightarrow C_1 \rightarrow C_0 \rightarrow \mathbb{Z} \rightarrow 0}$ is exact. This is what happens, for instance, for the augmented chain complex ${C_*(X)}$ when ${X}$ is contractible.

Theorem 7 (Acyclic model theorem) Let ${F, G}$ be functors from ${\mathfrak{C}}$ into the category of chain complexes. Suppose ${F}$ is free on the models ${\mathcal{M}}$ and ${G}$ is acyclic on ${\mathcal{M}}$ (i.e. ${G(M)}$ is exact for ${M \in \mathcal{M}}$ as an augmented complex).

1. There is a natural augmentation-preserving transformation ${F \rightarrow G}$ inducing the identity map ${\mathbb{Z} \rightarrow \mathbb{Z}}$.
2. Given two natural augmentation-preserving transformations ${F \rightarrow G}$, there is a natural chain homotopy between them.

So, with the augmentation hypothesis added, we have now eliminated all mention of ${H_0(F)}$ and ${H_0(G)}$! This is helpful, but first we need to prove this: Proof: We start by defining a morphism between ${H_0(F), H_0(G)}$. We will define a map ${\tau: F_0 \rightarrow H_0(G)}$ commuting with the augmentations and such that ${\tau}$ sends boundaries in ${F_0}$ into zeros in ${G_0}$, so that ${\tau}$ induces a natural map ${H_0(F) \rightarrow H_0(G)}$.

By freeness, we can choose models ${M^0_\alpha}$ and ${m^0_\alpha \in F_0(M^0_\alpha)}$ such that the ${F_0(f)(m_0)}$ form a basis for ${F_0(X)}$ for any ${X}$ as ${f}$ ranges over ${\hom(M^0_\alpha, X)}$. It is sufficient to define ${\tau(m^0_\alpha)}$ to define ${\tau: F_0 \rightarrow H_0(G)}$. For this, we choose the unique ${n^0_\alpha \in H_0(G(M^0_\alpha))}$ with augmentation one. There is uniqueness because ${G(M^0_\alpha)}$ is acyclic. If ${\epsilon: F_0 \rightarrow \mathbb{Z}}$ denotes the augmentation, send ${m^0_\alpha \rightarrow \epsilon(m^0_\alpha) n^0_\alpha \in H_0(G(M^0_\alpha))}$.

Then by freeness, this becomes a natural, augmentation-preserving map ${\tau: F_0 \rightarrow H_0(G)}$; since the ${n^0_\alpha}$ were unique, it is clear that it is the only such ${F_0 \rightarrow H_0(G)}$. We just need to check that ${\tau}$ vanishes on ${d F_1}$. But anything coming out of ${d F_1(M)}$ has augmentation zero, so ${\tau}$ sends it to something in ${H_0(G(M))}$ with augmentation zero. If ${M}$ is any one of the models, then we’re done.

More precisely, let ${x \in dF_1(M)}$; then ${\epsilon_G\tau(x) = 0}$ (where ${\epsilon_G}$ is the augmentation from ${G}$) so that ${\tau(x) = 0}$ by acyclicity if ${M \in \mathcal{M}}$. Otherwise, however, let ${x = dy}$. Then ${y}$ is a sum of various elements ${y_\beta \in F_1(f)(M'_\beta)}$ for ${M'_\beta \in \mathcal{M}}$ by freeness. Then by naturality the boundary of each of these ${y_\beta}$ is annilhilated by ${\tau}$. So ${dy}$ is annihilated by ${\tau}$. It follows that we have defined a natural, augmentation-preserving ${H_0(F) \rightarrow H_0(G)}$; it is also clear that this the unique one. The standard acyclic model theorem states that there is a natural map of functors of complexes ${F \rightarrow G}$ (necessarily augmentation-preserving), and that this is unique up to chain homotopy. And this completes the proof. $\Box$ So this allows us to simplify the proof of homotopy invariance, which revolved around showing that the two maps

$\displaystyle C_*(X) \rightarrow C_*(X \times I);$

however, we know that they are both augmentation-preserving maps and that the first functor is free, while the second is acyclic. So this immediately implies the existence of a natural chain homotopy. We don’t have to argue about zeroth homology, because the existence of the augmentation takes care of that for us. Incidentally, the following corollary is very easy but highly useful:

Corollary 8 Suppose ${\mathfrak{C}}$ is a category with models ${\mathcal{M}}$ and ${F,G}$ are functors from ${\mathfrak{C}}$ to the category of augmented chain complexes. Suppose ${F,G}$ are both free and acyclic on ${\mathcal{M}}$. Then ${F,G}$ are naturally chain equivalent (as augmented chain complexes).

Proof: Indeed, the previous result gives augmentation-preserving natural maps ${F \rightarrow G}$ and ${G \rightarrow F}$. Their composites ${F \rightarrow G \rightarrow F}$ and ${G \rightarrow F \rightarrow G}$ are, by the second part, each chain equivalent to ${F \rightarrow F}$ and ${G \rightarrow G}$. And that is it. $\Box$

4. The Eilenberg-Zilber theorem

The Eilenberg-Zilber theorem (together with the Kunneth formula) tells you how to compute the homology of a product space ${X \times Y}$. With coefficients in a field, the implication is that

$\displaystyle H_n(X \times Y, F) = \bigoplus_{i+j = n} H_i(X, F) \otimes H_j(Y, F) ,$

though in general there are torsion terms that appear.

Theorem 9 (Eilenberg-Zilber) Let ${X,Y}$ be topological spaces. Then there is a natural (in ${X}$ and ${Y}$) chain equivalence$\displaystyle C_*(X \times Y) \simeq C_*(X) \otimes C_*(Y)$

where ${C_*(X) \otimes C_*(Y)}$ denotes the product complex.

Proof: Indeed, consider the category ${\mathfrak{C}}$ of pairs of topological spaces ${(X,Y)}$ and pairs of continuous maps. Then

$\displaystyle (X,Y) \rightarrow C_*(X \otimes Y) \quad\mathrm{and} \quad (X,Y) \rightarrow C_*(X) \otimes C_*(Y)$

are both functors (call them ${F,G}$) from ${\mathfrak{C}}$ to the category of augmented chain complexes. Consider the set of pairs ${\mathcal{M}=(\Delta^m, \Delta^n)_{m,n \in \mathbb{Z}_{\geq 0}}}$. I claim that both functors are free and acyclic on ${\mathcal{M}}$.

Acyclicity is easiest to check: ${C_*(\Delta^n), C_*(\Delta^m),}$ and ${C_*(\Delta^m \times \Delta^n)}$ are all acyclic (as augmented complexes!) because these are contractible spaces. Acyclicity of the last implies that ${F}$ is acyclic on ${\mathcal{M}}$. Moreover, the Kunneth theorem implies that the tensor product of two augmented free and acyclic complexes which are acyclic is itself acyclic (when considered as an augmented complex). So ${G}$ is also acyclic on ${\mathcal{M}}$. We next need to check freeness; first we do this for ${F}$. Consider the elements ${e_n \in F_n((\Delta^n, \Delta^n))}$ which is the singular ${n}$-chain

$\displaystyle \Delta^n \rightarrow \Delta^n \times \Delta^n$

that is the diagonal map.

I claim that the ${e_n}$ form a “basis” for ${F}$. The reason is that ${F_n((X,Y))}$ is free on

$\displaystyle \hom_{\mathbf{Top}}(\Delta^n, X \times Y) = \hom_{\mathbf{Top}}(\Delta^n, X) \times \hom_{\mathbf{Top}}(\Delta^n, Y)$

which in turn is equal to

$\displaystyle \hom_{\mathfrak{C}}((\Delta^n , \Delta^n), (X,Y)).$

The map from ${\hom_{\mathfrak{C}}((\Delta^n , \Delta^n) \rightarrow F_n(X \times Y)}$ sends ${\Delta^n \rightarrow X, \Delta^n \rightarrow Y}$ to the pull-back via ${e_n}$ of ${\Delta^n \times \Delta^n \rightarrow X \times Y}$. So ${F}$ is free on the ${(\Delta^n, \Delta^n)}$, hence on the bigger set ${\mathcal{M}}$. It is slightly easier to see that ${G}$ is free on ${\mathcal{M}}$. Indeed, ${G_n}$ is free on ${\sqcup_{p+q=n}\hom_{\mathbf{Top}}(\Delta^p, X) \times \hom_{\mathbf{Top}}(\Delta^q, Y)}$. This is isomorphic to

$\displaystyle \sqcup_{p+q=n} \hom_{\mathfrak{C}}((\Delta^p, \Delta^q), (X,Y)).$

So ${G}$ is also free on ${\mathcal{M}}$. Now the acyclic model business immediately implies that ${F,G}$ are naturally chain-equivalent, proving the Eilenberg-Zilber theorem. $\Box$ As another example:

Theorem 10 Let ${X,Y}$ be topological spaces. Then the following diagram of complexes commutes up to chain homotopy:

where the top map is induced by the homeomorphism ${X \times Y \rightarrow Y \times X}$, while the bottom sends ${s \otimes t \rightarrow (-1)^{\deg t \deg s}t \otimes s}$.

Proof: Note that if ${E_*, F_*}$ are chain complexes, then the natural isomorphism

$\displaystyle E_* \otimes F_* \simeq F_* \otimes E_*$

sends ${e \otimes f }$ to ${(-1)^{\deg e \deg f} f \otimes e}$; this is necessary for commutation with the boundary maps to happen. It is now clear that the each of the maps in the diagram above is a natural transformation of functors from ${\mathbf{Top} \times \mathbf{Top}}$ to the category of augmented chain complexes. By freeness and acyclicity, the diagram must have to commute up to chain homotopy! $\Box$