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 1 Suppose
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 2 A category with models is 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 3 Suppose
is a category with models
and
is a covariant functor from
to the category of chain complexes (in positive degree).
- We say that
is acyclic on
if for each
,
is an acyclic complex in positive dimensions. Note that we do not require
to be acyclic at zero!
- We say that
is free on
if 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 […]
March 16, 2015 at 2:21 pm
I have a question regarding the excerpt “…the commutativity of the diagram will automatically define D on any space…”
The proofs I’ve seen of this fact seem to be centered around following the identity singular simplex around, and using naturality to determine the value of the chain homotopy on other singular simplices. This “following the identity around” stuff reminds me of the Yoneda lemma. What is (if there is) the connection?