Recall from last time that we were in the middle of proving the Dold-Kan correspondence, an important equivalence of categories between simplicial abelian groups and chain complexes. We defined three functors last time from simplicial abelian groups: the most obvious was the Moore complex, which just spliced all the components into one big chain complex with the differential the alternating sum of the face maps. But we noted that the functor one uses to construct this equivalence is ultimately either the normalized chain complex or the Moore complex modulo degeneracies.
Today, I’ll show that the two functors from simplicial abelian groups to chain complexes are in fact the same, through a decomposition that next time will let us construct the inverse functor. I’ll also construct the functor in the reverse direction. A minor word of warning: the argument in Goerss-Jardine (which seems to be the main source nowadays for this kind of material) has a small mistake! See their errata. This confused me for quite a while.
From chain complexes to simplicial groups
A priori, the normalized chain complex of a simplicial abelian group looks a lot different from
, which a priori has much more structure. Nonetheless, we are going to see that it is possible to recover
entirely from this chain complex. A key step in the proof of the Dold-Kan correspondence will be the establishment of the functorial decomposition for any simplicial abelian group
Here the map from a factor corresponding to some
to
is given by pulling back by
. We will establish this below.
Now, let us assume that (1) is true. Motivated by this, we shall define a functor from chain complexes to simplicial abelian groups. Let us now determine how the simplicial maps will play with the decomposition (which we are assuming) . Given
and a factor
of
(for some epimorphism
), we want to know where
takes
into
.
We can factor the composite as
. It is easy to see that simplicial maps induced by injections in
preserve
. There is a commutative diagram:
Now is an epimorphism. It follows that
is one of the maps in the canonical decomposition. It follows that we have a recipe for determining where the
-factor
of
goes:
- Consider the composite
, and factor this as a composite
with
an epimorphism and
a monomorphism.
- Then
(embedded in
via
) gets sent to
(embedded in
via
).
- The map
is given by
.
The construction
Motivated by this, let us describe the inverse construction. Let be a chain complex (nonnegatively graded, as always). We define a simplicial abelian group
such that
The sum is taken over all surjections .
We can make this into a simplicial abelian group using the above “recipe” describing how the canonical decomposition for a simplicial abelian group behaves, but there is a bit of subtlety. Since the in the explanation of (3) above does not a priori make sense, let us note that if we restrict to the subcategory
consisting of injective maps, then the map
becomes a contravariant functor in a natural way. Indeed, we let the map
induced by an injection
be zero unless
and we are working with the map
, in which case we let the map
be the differential. Since
is a chain complex, this is indeed a functor. So a chain complex gives an abelian presheaf on the “semi-simplicial” category.
Note that if we started with a simplicial abelian group , then if the chain complex
is made into a contravariant functor
, we have gotten nothing new: we just recover the simplicial structure maps. Indeed, if
is an injection, then the map
is zero unless
and
. Otherwise
will contain a
for some
, and the definition of
completes the proof. We thus see:
Lemma 1 Let
be a chain complex. Then there is a functor from
sending
and an injection
to zero unless
and the injection is
, in which case it is the differential. If
is a simplicial abelian group, this construction agrees with the simplicial maps when restricted to
.
Let us now, finally, show how to make into a simplicial abelian group. Given some map
in
, we map the individual terms as follows. Let
be an epimorphism in
, giving a factor
. We then map
as follows. If is the given surjection, then we have a map
, which we can factor as a composite
, of a surjection and an injection. So we send
(via
, which is defined by the functoriality) to
, imbedded in
as the factor corresponding to the surjection
.
Lemma 2 The above construction gives a functor from chain complexes to simplicial abelian groups.
In fact, the above construction will give a simplicial object from any semi-simplicial object. (A semi-simplicial object is a presheaf on the category of finite ordered sets and injective order-preserving maps.)
Proof: In other words, we need to show that if we have a composite , then the corresponding map
induced by the composite
is the composite of the maps
. So consider a factor of
corresponding to a surjection
. Now we can draw a commutative diagram in the simplex category
:
A close look at this will establish the claim, since is a functor from the category of finite ordered sets and injective, order-preserving maps. As a result, we have (finally!) constructed our functor
from chain complexes to simplicial abelian groups. Note that there is a natural transformation
for any simplicial abelian group . On the
-simplices, this is the map
where the factor corresponding to is mapped to
by pulling back by
. This is the map discussed above. It is immediate from the definition that this is a simplicial map. The crux of the proof of the Dold-Kan correspondence is that this is an isomorphism.
The canonical splitting
We have just defined the functor from chain complexes to simplicial abelian groups, and the natural transformation
for any simplicial abelian group
. We want to show that this is a quasi-inverse to
, that is, the above natural transformation is an isomorphism. Thus we need to show:
Proposition 3 (One half of Dold-Kan) For a simplicial abelian group
, we have for each
, an isomorphism of abelian groups
Here the map is given by sending a summand
to
via the pull-back by the term
. Alternatively, the morphism of simplicial abelian groups
is an isomorphism.
This is going to take some work, and we are going to need first a simpler splitting that will, incidentally, show that and
are isomorphic. We are going to prove the above result by induction, using:
Lemma 4 Let
be a simplicial abelian group. Then the map
is an isomorphism.
So we have a canonical splitting of each term of a simplicial abelian group. This splitting is into the degenerate simplices (or rather, their linear combinations) and the ones almost all of whose faces are zero.
Proof: We shall prove this by induction. Namely, for each , we define
and
to be the group generated by the images of
. So these are partial versions of the
. The claim is that there is a natural splitting
When , the result will be proved (note that
is the group generated by degenerate simplices because the degeneracies
only go up to
). When
, the splitting is
We can see this as follows. We have maps
Here is a split injection, with
being a section. But in general, whenever
is a split injection with section
, then
splits as
. Now let us suppose we have established the splitting
. We need to establish it for
. For this we will write some exact sequences.
- We have a split exact sequence:
Indeed, exactness of this sequence will be clear once we show that is well-defined. But if
, then
, so
sends
into
. The splitting is given by
.
- Similarly, we have a split exact sequence (where the simplicial identities show that
)
This is perhaps less obvious. This is equivalent to the claim that the map
is an isomorphism. (Here
denotes the inclusion.) We first claim that it is surjective. Indeed, if
, then
lies in fact in
. This is because
is a section of the split injection
, and because
for
by using the simplicial identities to move
to the inside. Conversely, to see that it is injective, it suffices to note that if
for
, then
; but
by definition of
.
Now we are going to fit the exact sequences (2) and (3) into a diagram:
It is clear that this diagram commutes. The first square consists of the natural inclusions and projections, so it is obvious. For the second square, the extra term does not affect things modulo
, so it commutes as well. Since both rows are exact and the first two columns are isomorphisms by the inductive hypothesis, so is the third.
Corollary 5 The map
is an isomorphism of chain complexes.
This is why we added the sign to the definition of the differential in constructing .
So we are halfway there to proving the Dold-Kan correspondence; we are almost ready to see that these functors are quasi-inverse. But this post is already long, so the completion of the proof will be deferred until next time.
May 6, 2011 at 3:45 pm
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