Recall that we were in the middle of establishing 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 simplicial abelian group; on the other hand, we had defined a functor that amalgamated a chain complex into a simplicial abelian group. We were in the middle of proving that the two functors were quasi-inverse.

With the same notation as before, we were trying to prove:

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.

We have a natural map

which we need to prove is an isomorphism. This is a map of simplicial abelian groups.

Let us first show that is surjective. By induction on , we may assume that is surjective for smaller . Now splits as the sum of and . Clearly is in the image of (from the factor ). But by the inductive hypothesis, everything in is in the image of , and taking degeneracies now shows that anything in is in the image of . Thus is surjective.

Let us now show that is injective. Suppose a family is mapped to zero under this map; we must show that each is zero. By assumption, we have

Suppose some is nonzero. Note that is zero by the canonical splitting, since that is the only term that might not be in .

We shall now define an *ordering* on the surjections . Say that if for each . We can assume that is chosen minimal with respect to this (partial) ordering such that . Now choose a section which is maximal in that is not a section of any . If we think of as determining a partition of into subsets, then we have sending to the last element of the th subset of . Then is a section of , and of no other . If we apply to the equation , we find that

which implies by the inductive hypothesis (as ) that pulls back to zero.

But the component of the identity of this pull-back is just , from the choice of . This means that .

**Completion of the proof of Dold-Kan**

We thus have defined a functor from simplicial abelian groups to chain complexes. We have defined a functor in the opposite direction. We have, moreover, seen that the simplicial abelian group associated to for a simplicial abelian group is just itself, in view of the canonical decomposition of a simplicial abelian group. It suffices now, at least, to prove that the normalized chain complex associated to is just , for any chain complex . So we need to compute . In degree , this consists of elements of

that are killed by the . The claim is that this consists precisely of under the identity !

We can see this because we can show that by direct computation; if , then the map pulls down to via the functor induced; however, this functor induces zero on coface maps that are not the highest index. Conversely, we must show that .

To do this, we have to show that ; but we know that is a complement to the degeneracies. However, the occurring in the expression for are all (clearly) degeneracies. Thus our assertion is clear.

We have now fully proved the Dold-Kan correspondence. It is a fairly technical argument, and I think it was more technical in the old days, when people thought of simplicial sets in terms of generators and relations.

What we have yet to do is to see that the Moore complex of a simplicial abelian group is actually homotopy equivalent (naturally) to the normalized chain complex, and thus to obtain a relation between the homotopy groups of the abelianization of a simplicial set and its homology. With this in mind, we’ll be able to do fun things like derive non-additive functors, by replacing projective resolutions by *simplicial resolutions, *and this will be the key idea behind the construction of the cotangent complex.

## Leave a Reply