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 1Let 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 2The 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 4Let 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 5The 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

[…] a crucial equivalence of categories between simplicial abelian groups and chain complexes. Last time, we had defined the two functors: on the one hand, we had the normalized chain complex of a […]