I hope I’ll get a chance to continue with blogging about descent soon; for now, I’m swamped with other things and mildly distracted by algebraic topology.

There are various theorems in algebraic topology whose proofs can require significant computation. For instance, the homotopy invariance of singular homology, the Eilenberg-Zilber theorem (which relates the singular chain complex of a product to the singular complexes ). On the other hand, there is also a strictly categorical framework in which these theorems may be proved. This is the method of acyclic models, to which the present post is dedicated. Let us start with the first example.

Theorem 1Suppose are homotopic. Then the maps are equal.

One way to give an explicit proof is to argue geometrically, decomposing the space into a bunch of -simplices. I always found this confusing. So I will explain how category theory does this magically and gives a natural chain homotopy. To start, note that it is enough to show that the two inclusions sending induce the same maps on homology. This is a standard argument.

The first idea in proving this is to step back from homology. Before homology, one has a functor which associates to each space its singular chain complex; this is a functor from the category of topological spaces to the category of chain complexes. It is a very special functor, though. For each , is a free abelian group on the set

which is itself functorial in . In particular, consider the distinguished element which is the identity map, considered as an element of . Then for each , is free on

Here is applied to the element by functoriality, i.e. . In particular, we have managed to express something very simple behind categorical language. Now, however, it will prove handy. Anyway, our goal is to show that the two maps

are naturally chain-homotopic. Since the maps are natural, we want a *natural* chain homotopy (i.e. functorial). If , we want the diagram

to commute. One consequence of this is that if we have defined on , the commutativity of the diagram will automatically define on *any* space . So we just need to show that there is a chain homotopy . In doing so, it is convenient to note that both are acyclic chain complexes because are convex subsets of . You’ll note that I am slightly cheating: I’m using an (admittedly weak) corollary of the homotopy invariance to assert that the spaces have trivial homology. But this can be easily proved by an explicit convexity argument, though I will not go into it here. So far, I wanted to motivate why this abstract nonsense actually helps. The point of naturality is that it lets us reduce to much simpler cases. This is the idea behind the method of acyclic models, which I will now go into formally.

**1. Acyclic models **

Definition 2Acategory with modelsis a category with a collection of objects of the category.

The point of the models is that they will have some special property relative to some functor. The example to keep in mind in what follows is the category of topological spaces with the standard -simplices as models. Their homs into a space form a basis for . This notion will be formalized in the next definition.

Definition 3Suppose is a category with models and is a covariant functor from to the category of chain complexes (in positive degree).

We say that isacyclic onif for each , is an acyclic complex in positive dimensions. Note that we do not require to be acyclic at zero!We say that isfree onif there are objects (possibly with repetitions, possibly not including all of them) and elements such that if is any object, then for all and forms a basis of .

The acyclicity condition is straightforward. For instance, we know it is true for on when the models are the -simplices. There the caveat that the complex has to be acyclic in positive dimensions was necessary. We also know that acyclicity is true for . The freeness condition is a generalization of the naturality discussion above. It means that often you just have to define things on the to define things everywhere. Also, I think of it as a sort of representability condition. In particular, has a functorial basis which is the set sum of a bunch of representable functors (represented by the ). We now have the main theorem:

Theorem 4 (Acyclic model theorem)Let be functors from into the category of chain complexes. Suppose is free on the models and is acyclic on .

Given a natural transformation , there is a natural transformation inducing it.Given two natural transformations inducing the same , there is a natural chain homotopy between them.

The closed analogy that I can think of the theorem on projective resolutions. If you have a projective resolution and a resolution , a morphism induces a unique-up-to-chain homotopy ; this is a basic result needed to show that derived functors are, well, functors. *Proof:* We start with the first part. First, we will define a morphism (I am using subscripts to mean the corresponding part of a complex) of functors . I will write this out in detail; for higher indices, it will be more brief. Since is free with models , there are objects and distinguished elements such for each , is free on for . Now if we define on each (so that it will be an element of , then naturality will define

By *freeness*, it will be possible to define on for all . Another (equivalent) way is to reason as follows. Consider the functor sending . To hom out of this functor into is the same as homming out of the free abelian group into . Same for doing it naturally. But studying the natural maps is the same as studying natural maps . This is because, by assumption, and are naturally isomorphic—the isomorphism sending to the distinguished element . The Yoneda lemma says that homming is the same as picking elements of . These correspond to the images of from the induced map . In either case, we see that it is sufficient to define the , and abstract nonsense or Yoneda will take care of the rest. However, we have a natural transformation . Let be something in that reduces to for the reduction of . In this way, we get a natural such that for any , we have that

so that induces the natural transformation on . Now we need to define in higher dimensions. Suppose is defined as a natural transformation in dimensions less than . We then need to define

such that it commutes with the boundary maps of the complexes . As before, we can reduce to defining it on suitable basis elements. Let be the models for with distinguished elements . We just need to define each . The key condition that is required is that

This will imply, by naturality, that when is made into a functor, that it commutes with the boundary maps on *any* object. But by induction, in smaller degrees is already defined and is a chain map, so

since . But is acyclic on the models . This means that we can find something in whose boundary is in fact as wanted. This we define to be . In this way, inductively we get natural transformations for all . Next, we need to prove the existence of a natural chain homotopy between any two . Again, we will use the same naturality trick. We will inductively construct a natural transformation

such that

First, we handle the case . With the same notation as before, we want to define for each such that

However, this last quantity maps to zero in because induce the same thing in zeroth homology. So we can choose in dimension zero, and this determines the natural map because is free. Now we assume that is defined in dimensions smaller than . We will define such that the key identity is satisfied. As before, it is sufficient to define

where the are suitable models and distinguished elements. We want to solve in :

Here only is undetermined; everything else is already known. We have to show that

is a coboundary, for then it will be possible to solve the above equation for each and thus define in dimension . By acyclicity, it is enough to show that this last quantity has dimension zero. But

since are chain maps, and by the inductive hypothesis that satisfies the chain homotopy in smaller dimensions, this is

which is zero. As a result, we can define in dimension , and the construction of the chain homotopy proceeds.

September 12, 2010 at 12:31 pm

[...] singular homology by Akhil Mathew 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 are homotopic. Then [...]