# Acyclic Models

#math #algtop

The theorem(s) of acyclic models are a rather powerful technique for proving that some functors defined on truncated chain complexes can be extended in higher dimensions, and that two such functors are homotopic, by proving it on a small class of “model” objects.

For some reason I only discovered this last year, and I always find myself forgetting the precise hypotheses and conclusion… Hopefully writing this blog post will fix them in my mind. My main reference will be:

• Samuel Eilenberg and Saunders MacLane. “Acyclic models”. In: Amer. J. Math. 75 (1953), pp. 189–199. ISSN: 0002-9327. JSTOR: 2372628. MR0052766.

Unless indicated otherwise, all the definitions, theorems… come from there.

## The theorem

### “Representable” functors

Let $\mathsf{C}$ be any category (for example topological spaces, simplicial set…). Let also $\mathcal{M} \subset \operatorname{ob} \mathsf{C}$ be a set of objects, called (acyclic) model objects (think about standard simplexes).

For a functor $T : \mathsf{C} \to \mathsf{Ab}$ to the category of abelian groups, let $\widetilde{T} : \mathsf{C} \to \mathsf{Ab}$ be the new functor defined as follows:

• for an object $A$, $\widetilde{T}(A)$ is the free abelian group generated by the set of pairs $(M, \varphi, m)$, where $M \in \mathcal{M}$, $\varphi : M \to A$ is a morphism, and $m \in T(M)$;
• for a morphism $\alpha : A \to B$, $\widetilde{T}(\alpha)$ maps $(M, \varphi, m)$ to $(M, \alpha \circ \varphi, m)$.

There is also a natural transformation $\Phi : \widetilde{T} \to T$ given by $\Phi_A(M, \varphi, m) = T(\varphi)(m)$. Morally, this is similar to the construction where you take an abelian group and you want to view it as a quotient of a free object: you take the free abelian group generated by all elements of your group, and you quotient by the relation $g \cdot g' \equiv gg'$. Here we want to view all the $T(A)$ as generated by the $T(M)$.

The functor $T$ is then said to be representable (the terminology is unfortunate…) if $\Phi$ has a right inverse, i.e. there is a natural transformation $\Psi : T \to \widetilde{T}$ such that $\Phi\Psi$ is the identity of $T$.

In other words, this means that for all $A \in \mathsf{C}$, $T(A)$ has a “basis” (consisting of models objects) which is described by $\Psi$: $\Psi_A : T(A) \to \widetilde{T}(A)$ sends $x \in T(A)$ to a sum $\sum_{\alpha} (M_\alpha, \varphi_\alpha, m_\alpha)$ such that

$$x = \sum_\alpha T(\varphi_\alpha)(m_\alpha).$$

When a functor is representable, we will be able to check equalities on $\mathcal{M}$, and equalities on $\mathsf{C}$ will follow.

### Chain complexes

Now suppose we are given a functor $K : \mathsf{C} \to \mathsf{Ch}$ from our category $\mathsf{C}$ to the category of nonnegatively graded chain complexes. We will denote $\partial_q : K_q(A) \to K_{q-1}(A)$ the differential.

If $K$ and $L$ are two such functors, then we can define maps $K \to L$ in the obvious way, and we can also define truncated maps in degrees $\le n$. Similarly given two maps, we can define homotopies between then, as well as truncated homotopies, which satisfy

$$\partial_{q+1} h_q + h_{q-1} \partial_q = g_q - f_q$$

in degrees $\le n$.

Theorem. Let $K,L : \mathsf{C} \to \mathsf{Ch}$ be two functors and $f : K \to L$ be a map truncated in degrees $< q$. If $K_q$ is representable and $H_{q-1}(L(M)) = 0$ for all $M \in \mathcal{M}$, then $f$ can be extended in dimension $q$.

Proof. We want to define $f_q : K_q \to L_q$ such that $\partial_q f_q f_{q-1} \partial_{q-1}$. For all models $M$ and all $m \in K_q(M)$, the element $f_{q-1}(\partial_q m)$ is a cycle in $L_{q-1}(M)$; but $L_{q-1}(M)$ is acyclic, thus we can choose some $\lambda(m) \in L_q(M)$ such that $\partial_q \lambda(m) = f_{q-1} \partial_q m$.

Now we let $\Lambda : \widetilde{K_q} \to L_q$ be defined by

$$\Lambda_A(M, \varphi, m) := L_q(\varphi)(\lambda(m)).$$

It’s easy to check that this is a natural transformation, and that $\partial_q \Lambda = f_{q-1} \partial_q \Phi$ (where $\Phi : \widetilde{K_q} \to K_q$ is the canonical morphism).

But since $K_q$ was representable, we can let $f_q = \Lambda \Psi$ where $\Psi$ is the right inverse to $\Phi$, and everything works out. ■

Do you see what we did (well, Eilenberg–MacLane did) here? It was easy to define an “ $f_q$-like” map on the free stuff generated by the models, because we could just choose an arbitrary lift for each element without bothering to check if they were compatible with each other. The “representability” of $K_q$ then provides a way to combine these all together. (This is also why we needed $\mathcal{M}$ to be a set – I don’t think it would have been possible to “choose” these lifts if it were a proper class…).

The proof of the following theorem is almost the same (exercise!).

Theorem. Let $K,L : \mathsf{C} \to \mathsf{Ch}$ be two functors, $f,g : K \to L$ two maps and $h : f \simeq g$ a homotopy truncated in degrees $< q$. If $K_q$ is representable and $H_q(L(M)) = 0$ for all $M \in \mathcal{M}$, then $h$ can be extended in dimension $q$.

These two theorems combine to give:

Theorem. Let $K,L : \mathsf{C} \to \mathsf{Ch}$ be two functors, and $f : K \to L$ be a map truncated in degrees $< q$. If $K_n$ is representable for all $n \ge q$ and $H_n(L(M)) = 0$ for all models and for all $n \ge q - 1$, then $f$ admits a unique (up to homotopy) extension in all degrees.

## Example: action of a monoid on a group

Let $W$ be a monoid and $\Lambda = \mathbb{Z}[W]$ be its algebra. We may view $W$ as a category $\mathsf{C}$ with a unique object and $\hom_{\mathsf{C}}(*,*) = W$. We set $\mathcal{M} = \operatorname{ob} \mathsf{C} = \{*\}$.

A functor $T : \mathsf{C} \to \mathsf{Ab}$ is the same thing as a group equipped with a left action of $W$, i.e. a left $\Lambda$-module ; the functor $\tilde{T}$ is the free abelian group on the set of pairs $W \times G$, with $W$ acting as $w \cdot (w',g) := (ww', g)$. In other words, it’s the free $\Lambda$-module on $G$. The functor $T$ is then representable iff $G$ is projective as a $\Lambda$-module.

## Normalization of singular chains

Let $S_* : \mathsf{Top} \to \mathsf{Ch}$ be the classical “singular chains” functor. The abelian group $S_n(X)$ is free on maps $\Delta^n \to X$, where $\Delta^n$ is the standard $n$-simplex:

$$\Delta^n = \{ (x_0, \dots, x_n) \in [0,1]^{n+1} \mid \sum x_i = 1 \},$$

and $\partial_n : S_n(X) \to S_{n-1}(X)$ is the usual differential $\partial_n = \sum_i (-1)^i d_i$. We consider the augmented version with $S_{-1}(X) = \mathbb{Z}$ and $\partial_0(x) = 1$ for all $x \in X = S_0(X)$.

Now $\Delta^\bullet$ is in fact a cosimplicial space, and so we get degeneracy maps $s_j : S_n(X) \to S_{n+1}(X)$. For a simplex $\sigma : \Delta^n \to X$, its degeneracy is given by:

$$s_j(\sigma)(x_0, \dots, x_{n+1}) = \sigma(x_0, \dots, x_j + x_{j+1}, \dots, x_{n+1}).$$

These, together with the $d_i$, satisfy the usual simplicial identities. One can then normalize the singular chains functor, either by letting

$$\bar{S}_{n+1}(X) = S_{n+1}(X) / \operatorname{im}(s_n : S_n(X) \to S_{n+1}(X))$$

(the complex “normalized at the top”) or by letting

$$S^N(X) = S_n(X) / \bigcup_{j=0}^{n-1} \operatorname{im}(s_j)$$

(the “normalized complex”).

In the following theorem, $S'$ is either $S^N$ or $\bar{S}$.

Theorem. Let $f : S \to S'$ be the quotient map. Then there is a map $g : S' \to S$ and homotopies $gf \simeq \operatorname{id}_S$ and $fg \simeq \operatorname{id}_{S'}$.

The proof directly uses acyclic models. The set of models can be taken to be the set of standard simplexes $\{ \Delta^n \mid n \ge 0 \}$ (Eilenberg–MacLane use the class of all contractible spaces, but it’s not a set and their proof only uses standard simplexes).

In degree $(-1)$, all three complexes are equal (which gives the base case for the induction). One needs to know that the homology of $\Delta^n$ is trivial (both the normalized and the unnormalized one), which can be proven directly; then one needs to prove that the three functors $S$, $S^N$ and $S'$ are representable, which is almost immediate from the definition.

The interesting thing (IMO) is that one can go through the proof and see that this yields a completely explicit homotopy equivalence between the unnormalized complex and the normalized ones.

## Eilenberg–Zilber theorem

This part is not from the article of Eilenberg–MacLane (but it’s nevertheless completely classical, see e.g. MacLane’s book Homology, chapter VIII.8).

Let $M$ and $N$ be two simplicial modules over some ring $R$. One can produce two chain complexes out this: either take the two Moore complexes $M_*$, $N_*$ and then take their tensor product, or take the diagonal of the product $(M \times N)_n = M_n \times N_n$ and then take the Moore complexes. We will denote the two complexes respectively as $M_* \otimes N_*$ and $(M \times N)_*$. Of course this yields two functors $(s\mathsf{Mod}_R)^2 \to \mathsf{Ch}$, to which we will apply the previous techniques.

The two complexes are both equal to $M_0 \otimes N_0$ is degree zero. This gives the base case for the induction (we can just take the map to be the identity). One can then choose as models the simplicial modules given by $\Delta^n \otimes R$, which can easily be proven to be acyclic. The adjunctions

$$\hom_{s\mathsf{Mod}_R}(\Delta^n \otimes R, M_\bullet) \cong \hom_{s\mathsf{Set}}(\Delta^n, M_\bullet) \cong M_n$$

can then be used to prove that both functors are representable. (I’m omitting a lot of computations here! Though most of it is straightforward.) The acyclic models technique then yields the equivalence:

$$(M \times N)_* \simeq M_* \otimes N_*.$$

Again, what’s really interesting is that both maps (and both homotopies!) can be described completely explicitly once you make the right choices (you need to go back to the proof of the first theorem to know what choices I’m talking about). The map $f : (M \times N)_* \to M_* \otimes N_*$ is known as the Alexander–Whitney map, and it is given by (for $a \in A_n$, $b \in B_n$):

$$f(a \times b) = \sum_{i = 0}^n \bar{d}^{n-i}a \otimes d_0^i b,$$

where $d_0 : B_k \to B_{k-1}$ is the $0$th face and $\bar{d} = d_k : A_k \to A_{k-1}$ is the “last” face.

Conversely, $g : M_* \otimes N_* \to (M \times N)_*$ is known as the Eilenberg–Zilber map, given by (for $a \in A_p$, $b \in B_q$):

$$g(a \otimes b) = \sum_{(\mu, \nu) \in \mathrm{Sh}_{p,q}} \pm s_{\nu} a \times s_{\mu} b.$$

The sum runs over all $(p,q)$-shuffles, with (for $p+q = n$):

$$\mathrm{Sh}_{p,q} = \{ (\mu, \nu) \in \{1,\dots,n\}^p \times \{1,\dots,n\}^q \mid \mu(1) < \dots < \mu(p), \nu(1) < \dots < \nu(q), \mu(i) \neq \nu(j) \}$$

and where $s_{\mu} = s_{\mu(p)} \circ \dots \circ s_{\mu(1)}$, and $s_{\nu} = s_{\nu(q)} \circ \dots \circ s_{\nu(1)}$. (The reader is encouraged to see explicitly what this all means in small cases, say $(p,q) = (1,2)$).

This provides a completely algebraic proof of the “topological” Eilenberg–Zilber theorem which states that, given two topological spaces $X$ and $Y$, one has $S_*(X \times Y) \simeq S_*(X) \otimes S_*(Y)$, which in turns is a key ingredient in the Künneth theorem (which now becomes a purely algebraic statement about the homology of the tensor product of two chain complexes).