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 5Suppose are homotopic. Then the maps are equal.

*Proof:* Suppose is a homotopy with . Then the maps factor as

so if we show that the inclusions sending to induce equal maps on homology, we will be done.

Write for simplicity. For each space , the maps are *natural*. More precisely, if are the two functors , then are natural transformations between them.

So we have two maps induced by , and these are natural transformations between the functors . They induce the same map on zeroth homology because and are connected by a path in and so the zero-singular simplices differ by the boundary of a 1-simplex. Moreover, the functor is free on the objects with the canonical -simplices .

Moreover, is acyclic in positive dimensions on these simplices because is contractible for any . So the two maps are naturally chain homotopic by the acyclic model theorem, and thus induce the same maps on homology. This completes the proof.

**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 . Let’s recall what this means.

Definition 6Anaugmented chain complex(of abelian groups) is a chain complex together with an epimorphism such that is zero. The map is called theaugmentation. A morphism of augmented chain complexes is a morphism of chain complexes that preserves the augmentation on .

We note that the chain complexes are augmented. Indeed, an element in is just a formal sum of points, and the augmentation sends this to the total number of points in the formal sum; is sent to . Note that is functorial as a functor from to the category of augmented chain complexes. We say that an augmented chain complex is **exact** if the complex is exact. This is what happens, for instance, for the augmented chain complex when is contractible.

Theorem 7 (Acyclic model theorem)Let be functors from into the category of chain complexes. Suppose is free on the models and is acyclic on (i.e. is exact for as an augmented complex).

There is a natural augmentation-preserving transformation inducing the identity map .Given two natural augmentation-preserving transformations , there is a natural chain homotopy between them.

So, with the augmentation hypothesis added, we have now eliminated all mention of and ! This is helpful, but first we need to prove this: *Proof:* We start by defining a morphism between . We will define a map commuting with the augmentations and such that sends boundaries in into zeros in , so that induces a natural map .

By freeness, we can choose models and such that the form a basis for for any as ranges over . It is sufficient to define to define . For this, we choose the unique with augmentation one. There is uniqueness because is acyclic. If denotes the augmentation, send .

Then by freeness, this becomes a natural, augmentation-preserving map ; since the were unique, it is clear that it is the only such . We just need to check that vanishes on . But anything coming out of has augmentation zero, so sends it to something in with augmentation zero. If is any one of the models, then we’re done.

More precisely, let ; then (where is the augmentation from ) so that by acyclicity if . Otherwise, however, let . Then is a sum of various elements for by freeness. Then by naturality the boundary of each of these is annilhilated by . So is annihilated by . It follows that we have defined a natural, augmentation-preserving ; 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 (necessarily augmentation-preserving), and that this is unique up to chain homotopy. And this completes the proof. So this allows us to simplify the proof of homotopy invariance, which revolved around showing that the two maps

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 8Suppose is a category with models and are functors from to the category of augmented chain complexes. Suppose are both free and acyclic on . Then are naturally chain equivalent (as augmented chain complexes).

*Proof:* Indeed, the previous result gives augmentation-preserving natural maps and . Their composites and are, by the second part, each chain equivalent to and . And that is it.

**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 . With coefficients in a field, the implication is that

though in general there are torsion terms that appear.

Theorem 9 (Eilenberg-Zilber)Let be topological spaces. Then there is a natural (in and ) chain equivalence

where denotes the product complex.

*Proof:* Indeed, consider the category of pairs of topological spaces and pairs of continuous maps. Then

are both functors (call them ) from to the category of augmented chain complexes. Consider the set of pairs . I claim that both functors are free and acyclic on .

Acyclicity is easiest to check: and are all acyclic (as augmented complexes!) because these are contractible spaces. Acyclicity of the last implies that is acyclic on . 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 is also acyclic on . We next need to check freeness; first we do this for . Consider the elements which is the singular -chain

that is the diagonal map.

I claim that the form a “basis” for . The reason is that is free on

which in turn is equal to

The map from sends to the pull-back via of . So is free on the , hence on the bigger set . It is slightly easier to see that is free on . Indeed, is free on . This is isomorphic to

So is also free on . Now the acyclic model business immediately implies that are naturally chain-equivalent, proving the Eilenberg-Zilber theorem. As another example:

Theorem 10Let be topological spaces. Then the following diagram of complexes commutes up to chain homotopy:

where the top map is induced by the homeomorphism , while the bottom sends .

*Proof:* Note that if are chain complexes, then the natural isomorphism

sends to ; 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 to the category of augmented chain complexes. By freeness and acyclicity, the diagram must have to commute up to chain homotopy!

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