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.
Let be any category (for example topological spaces, simplicial set…). Let also be a set of objects, called (acyclic) model objects (think about standard simplexes).
For a functor to the category of abelian groups, let be the new functor defined as follows:
- for an object , is the free abelian group generated by the set of pairs , where , is a morphism, and ;
- for a morphism , maps to .
There is also a natural transformation given by . 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 . Here we want to view all the as generated by the .
The functor is then said to be representable (the terminology is unfortunate…) if has a right inverse, i.e. there is a natural transformation such that is the identity of .
In other words, this means that for all , has a “basis” (consisting of models objects) which is described by : sends to a sum such that
When a functor is representable, we will be able to check equalities on , and equalities on will follow.
Now suppose we are given a functor from our category to the category of nonnegatively graded chain complexes. We will denote the differential.
If and are two such functors, then we can define maps in the obvious way, and we can also define truncated maps in degrees . Similarly given two maps, we can define homotopies between then, as well as truncated homotopies, which satisfy
in degrees .
Theorem. Let be two functors and be a map truncated in degrees . If is representable and for all , then can be extended in dimension .
Proof. We want to define such that . For all models and all , the element is a cycle in ; but is acyclic, thus we can choose some such that .
Now we let be defined by
It’s easy to check that this is a natural transformation, and that (where is the canonical morphism).
But since was representable, we can let where is the right inverse to , and everything works out. ■
Do you see what we did (well, Eilenberg–MacLane did) here? It was easy to define an “-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 then provides a way to combine these all together. (This is also why we needed 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 be two functors, two maps and a homotopy truncated in degrees . If is representable and for all , then can be extended in dimension .
These two theorems combine to give:
Theorem. Let be two functors, and be a map truncated in degrees . If is representable for all and for all models and for all , then admits a unique (up to homotopy) extension in all degrees.
Example: action of a monoid on a group
Let be a monoid and be its algebra. We may view as a category with a unique object and . We set .
A functor is the same thing as a group equipped with a left action of , i.e. a left -module ; the functor is the free abelian group on the set of pairs , with acting as . In other words, it’s the free -module on . The functor is then representable iff is projective as a -module.
Normalization of singular chains
Let be the classical “singular chains” functor. The abelian group is free on maps , where is the standard -simplex:
and is the usual differential . We consider the augmented version with and for all .
Now is in fact a cosimplicial space, and so we get degeneracy maps . For a simplex , its degeneracy is given by:
These, together with the , satisfy the usual simplicial identities. One can then normalize the singular chains functor, either by letting
(the complex “normalized at the top”) or by letting
(the “normalized complex”).
In the following theorem, is either or .
Theorem. Let be the quotient map. Then there is a map and homotopies and .
The proof directly uses acyclic models. The set of models can be taken to be the set of standard simplexes (Eilenberg–MacLane use the class of all contractible spaces, but it’s not a set and their proof only uses standard simplexes).
In degree , all three complexes are equal (which gives the base case for the induction). One needs to know that the homology of is trivial (both the normalized and the unnormalized one), which can be proven directly; then one needs to prove that the three functors , and 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.
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 and be two simplicial modules over some ring . One can produce two chain complexes out this: either take the two Moore complexes , and then take their tensor product, or take the diagonal of the product and then take the Moore complexes. We will denote the two complexes respectively as and . Of course this yields two functors , to which we will apply the previous techniques.
The two complexes are both equal to 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 , which can easily be proven to be acyclic. The adjunctions
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:
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 is known as the Alexander–Whitney map, and it is given by (for , ):
where is the th face and is the “last” face.
Conversely, is known as the Eilenberg–Zilber map, given by (for , ):
The sum runs over all -shuffles, with (for ):
and where , and . (The reader is encouraged to see explicitly what this all means in small cases, say ).
This provides a completely algebraic proof of the “topological” Eilenberg–Zilber theorem which states that, given two topological spaces and , one has , 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).