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 ${C_*(X \times Y)}$ of a product ${X \times Y}$ to the singular complexes ${C_*(X), C_*(Y)}$). 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 1 Suppose ${f, g: X \rightarrow Y}$ are homotopic. Then the maps ${f_*, g_*: H_*(X) \rightarrow H_*(Y)}$ are equal.

One way to give an explicit proof is to argue geometrically, decomposing the space ${\Delta^n \times I}$ into a bunch of ${n}$-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 ${X \rightarrow X \times I}$ sending ${x \rightarrow (x,0), (x,1)}$ 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 ${X \rightarrow C_*(X)}$ which associates to each space ${X}$ 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 ${n}$, ${C_n(X)}$ is a free abelian group on the set

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

which is itself functorial in ${X}$. In particular, consider the distinguished element ${i_n: \Delta^n \rightarrow \Delta^n}$ which is the identity map, considered as an element of ${C_n(\Delta^n)}$. Then for each ${X}$, ${C_n(X)}$ is free on

$\displaystyle C_n(f)(i_n), \forall f: \Delta^n \rightarrow X.$

Here ${C_n(f)(i_n)}$ is ${f}$ applied to the element ${i_n \in C_n(\Delta^n)}$ by functoriality, i.e. ${f \circ i_n: \Delta^n \rightarrow X}$. 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

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

are naturally chain-homotopic. Since the maps ${C_*(X) \rightarrow C_*(X \times I)}$ are natural, we want a natural chain homotopy (i.e. functorial). If ${X \rightarrow Y}$, we want the diagram

to commute. One consequence of this is that if we have defined ${D}$ on ${C_*(\Delta^n) \rightarrow C_*(\Delta^n \times I)}$, the commutativity of the diagram will automatically define ${D}$ on any space ${X}$. So we just need to show that there is a chain homotopy ${C_*(\Delta^n) \rightarrow C_*(\Delta^n \times I)}$. In doing so, it is convenient to note that both are acyclic chain complexes because ${\Delta^n, \Delta^n\times I}$ are convex subsets of ${\mathbb{R}^n}$. You’ll note that I am slightly cheating: I’m using an (admittedly weak) corollary of the homotopy invariance to assert that the spaces ${\Delta^n, \Delta^n \times I}$ 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 2 A category with models is a category with a collection ${\mathcal{M}}$ 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 ${n}$-simplices ${\Delta^n}$ as models. Their homs into a space form a basis for ${C_*(X)}$. This notion will be formalized in the next definition.

Definition 3 Suppose ${\mathfrak{C}}$ is a category with models ${ \mathcal{M}}$ and ${F}$ is a covariant functor from ${\mathfrak{C}}$ to the category of chain complexes (in positive degree).

1. We say that ${F}$ is acyclic on ${\mathcal{M}}$ if for each ${M \in \mathcal{M}}$, ${F(M)}$ is an acyclic complex in positive dimensions. Note that we do not require ${F}$ to be acyclic at zero!
2. We say that ${F}$ is free on ${\mathcal{M}}$ if there are objects ${M_\alpha \in \mathcal{M}}$ (possibly with repetitions, possibly not including all of them) and elements ${m_\alpha \in F(M_\alpha)}$ such that if ${X \in \mathfrak{C}}$ is any object, then ${\{F(f)(m_\alpha)\}}$ for all ${\alpha}$ and ${f: M_\alpha \rightarrow X}$ forms a basis of ${F(X)}$.

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

Theorem 4 (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 ${\mathcal{M}}$.

1. Given a natural transformation ${H_0(F) \rightarrow H_0(G)}$, there is a natural transformation ${F \rightarrow G}$ inducing it.
2. Given two natural transformations ${F \rightarrow G}$ inducing the same ${H_0(F) \rightarrow H_0(G)}$, 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 ${P \rightarrow M}$ and a resolution ${Q \rightarrow N}$, a morphism ${M \rightarrow N}$ induces a unique-up-to-chain homotopy ${P \rightarrow Q}$; 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 ${\tau_0 : F_0 \rightarrow G_0}$ (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 ${F_0}$ is free with models ${\mathcal{M}}$, there are objects ${M^0_\alpha \in \mathcal{M}}$ and distinguished elements ${m^0_\alpha \in M^0_\alpha}$ such for each ${X}$, ${F_0(X)}$ is free on ${F_0(f)(m^0_\alpha)}$ for ${f \in \hom(M_\alpha, X)}$. Now if we define ${\tau_0}$ on each ${m^0_\alpha}$ (so that it will be an element of ${G_0(M_\alpha)}$, then naturality will define

$\displaystyle \tau_0( F_0(f)(m^0_\alpha)) = F_0(f) \tau_0(m^0_\alpha).$

By freeness, it will be possible to define ${\tau_0}$ on ${F_0(X)}$ for all ${X}$. Another (equivalent) way is to reason as follows. Consider the functor ${F': \mathfrak{C} \rightarrow \mathbf{Sets}}$ sending ${X \rightarrow \sqcup \hom(M^0_{\alpha}, X)}$. To hom out of this functor into ${G_0}$ is the same as homming out of the free abelian group ${\mathbb{Z}[\sqcup \hom(M^0_{\alpha}, X)]}$ into ${G_0}$. Same for doing it naturally. But studying the natural maps ${\mathbb{Z}[\sqcup \hom(M^0_{\alpha}, X)] \rightarrow G_0(X)}$ is the same as studying natural maps ${F_0 \rightarrow G_0}$. This is because, by assumption, ${F_0}$ and ${G_0}$ are naturally isomorphic—the isomorphism sending ${1_{M^0_\alpha}}$ to the distinguished element ${m^0_\alpha \in F_0(M^0_\alpha)}$. The Yoneda lemma says that homming ${F' \rightarrow G_0}$ is the same as picking elements of ${G_0(M^0_\alpha)}$. These correspond to the images of ${m^0_\alpha}$ from the induced map ${F_0 \rightarrow G_0}$. In either case, we see that it is sufficient to define the ${\tau(m^0_\alpha)}$, and abstract nonsense or Yoneda will take care of the rest. However, we have a natural transformation ${\overline{\tau}: H_0(F) \rightarrow H_0(G)}$. Let ${\tau(m^0_\alpha)}$ be something in ${G_0}$ that reduces to ${\overline{\tau}(\overline{m^0_\alpha})}$ for ${\overline{m^0_\alpha}}$ the reduction of ${m^0_\alpha}$. In this way, we get a natural ${\tau: F_0 \rightarrow G_0}$ such that for any ${m \in F_0(M)}$, we have that

$\displaystyle \overline{\tau}(\overline{m}) = \overline{\tau(m)}$

so that ${\tau}$ induces the natural transformation ${\overline{\tau}}$ on ${H_0}$. Now we need to define ${\tau}$ in higher dimensions. Suppose ${\tau}$ is defined as a natural transformation in dimensions less than ${n>0}$. We then need to define

$\displaystyle \tau: F_n \rightarrow G_n$

such that it commutes with the boundary maps of the complexes ${F_*, G_*}$. As before, we can reduce to defining it on suitable basis elements. Let ${\left\{M^n_\alpha\right\} \in M}$ be the models for ${F_n}$ with distinguished elements ${m^n_\alpha \in F_n(M^n_\alpha)}$. We just need to define each ${\tau(m^n_\alpha)}$. The key condition that is required is that

$\displaystyle d \tau(m^n_\alpha) = \tau( dm^n_\alpha).$

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

$\displaystyle d(\tau (d m^n_\alpha)) = \tau( d^2 m^n_\alpha) = 0$

since ${d^2 =0 }$. But ${G}$ is acyclic on the models ${\mathcal{M}}$. This means that we can find something in ${G_n(M^n_\alpha)}$ whose boundary is in fact ${ \tau( dm^n_\alpha)}$ as wanted. This we define to be ${\tau(m^n_\alpha)}$. In this way, inductively we get natural transformations ${F_n \rightarrow G_n}$ for all ${n \in \mathbb{Z}_{\geq 0}}$. Next, we need to prove the existence of a natural chain homotopy between any two ${\tau, \tau'}$. Again, we will use the same naturality trick. We will inductively construct a natural transformation

$\displaystyle D: F_n(X) \rightarrow G_{n+1}(X)$

such that

$\displaystyle Dd + dD = \tau - \tau'.$

First, we handle the case ${n=0}$. With the same notation as before, we want to define ${D(m^0_\alpha)}$ for each ${\alpha}$ such that

$\displaystyle d (D(m^0_\alpha)) = \tau(m_0^\alpha) - \tau'(m_0^\alpha).$

However, this last quantity maps to zero in ${H_0(G(X))}$ because ${\tau, \tau'}$ induce the same thing in zeroth homology. So we can choose ${D(m^0_\alpha) }$ in dimension zero, and this determines the natural map ${D: F_0 \rightarrow G_1}$ because ${F}$ is free. Now we assume that ${D}$ is defined in dimensions smaller than ${n}$. We will define ${D: F_n \rightarrow G_{n+1}}$ such that the key identity is satisfied. As before, it is sufficient to define

$\displaystyle D(m^n_\alpha) \in G_{n+1}(M^n_\alpha)$

where the ${M^n_\alpha, m^n_\alpha}$ are suitable models and distinguished elements. We want to solve in ${D(m^n_\alpha)}$:

$\displaystyle d D(m^n_\alpha) + D d( m^n_\alpha) = (\tau - \tau')(m^n_\alpha).$

Here only ${D(m^n_\alpha)}$ is undetermined; everything else is already known. We have to show that

$\displaystyle (\tau - \tau')(m^n_\alpha) - D d( m^n_\alpha)$

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

$\displaystyle d( (\tau - \tau')(m^n_\alpha) - D d( m^n_\alpha)) = (\tau - \tau') (dm^n_\alpha) - dDd(m^n_\alpha)$

since ${\tau, \tau'}$ are chain maps, and by the inductive hypothesis that ${D}$ satisfies the chain homotopy in smaller dimensions, this is

$\displaystyle (\tau - \tau') (dm^n_\alpha) -( (\tau - \tau')(dm^n_\alpha) - ddD(m^n_\alpha) )$

which is zero. As a result, we can define ${D}$ in dimension ${n}$, and the construction of the chain homotopy proceeds. $\Box$