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 $\sigma$ 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 ${A_\bullet}$, we have for each ${n}$, an isomorphism of abelian groups

$\displaystyle \bigoplus_{\phi: [n] \twoheadrightarrow [k]} NA_k \simeq A_n.$

Here the map is given by sending a summand ${NA_k}$ to ${A_n}$ via the pull-back by the term ${\phi: [n] \twoheadrightarrow [k]}$. Alternatively, the morphism of simplicial abelian groups

$\displaystyle \sigma (N A_*)_\bullet \rightarrow A_\bullet$

is an isomorphism.

We have a natural map

$\displaystyle \Phi_n: \bigoplus_{\phi: [n] \twoheadrightarrow [k]} NA_k \rightarrow A_n,$

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

Let us first show that ${\Phi_n: (\sigma NA_*)_n \rightarrow A_n}$ is surjective. By induction on ${n}$, we may assume that ${\Phi_m: (\sigma NA_*)_m \rightarrow A_m}$ is surjective for smaller ${m. Now ${A_n}$ splits as the sum of ${NA_n}$ and ${DA_n}$. Clearly ${NA_n}$ is in the image of ${\Phi_n}$ (from the factor ${NA_n}$). But by the inductive hypothesis, everything in ${A_{n-1}}$ is in the image of ${\Phi_{n-1}}$, and taking degeneracies now shows that anything in ${DA_n}$ is in the image of ${\Phi_n}$. Thus ${\Phi_n}$ is surjective.

Let us now show that ${\Phi_n}$ is injective. Suppose a family ${(a_\phi) \in \bigoplus_{\phi: [n] \twoheadrightarrow [k]} NA_k}$ is mapped to zero under this map; we must show that each ${a_\phi }$ is zero. By assumption, we have

$\displaystyle \sum_{\phi: [n] \twoheadrightarrow [k]} \phi^* a_\phi = 0 \in A_n.$

Suppose some ${a_\phi}$ is nonzero. Note that ${a_{1: [n] \rightarrow [n]}}$ is zero by the canonical splitting, since that is the only term that might not be in ${DA_n}$.

We shall now define an ordering on the surjections ${[n] \twoheadrightarrow [k]}$. Say that ${\phi_1 \leq \phi_2}$ if ${\phi_1(a) \leq \phi_2(a)}$ for each ${a \in [n]}$. We can assume that ${\phi}$ is chosen minimal with respect to this (partial) ordering such that ${a_\phi \neq 0}$. Now choose a section ${\psi: [k] \hookrightarrow [n]}$ which is maximal in that ${\psi}$ is not a section of any ${\phi' > \phi}$. If we think of ${\phi}$ as determining a partition of ${[n]}$ into ${k}$ subsets, then we have ${\psi}$ sending ${i \in [k]}$ to the last element of the ${i}$th subset of ${[n]}$. Then ${\psi}$ is a section of ${\phi}$, and of no other ${\phi' < \phi}$. If we apply ${\psi^*}$ to the equation ${(a_\phi) = 0}$, we find that

$\displaystyle \Phi_k((\psi^* a_\phi)) = 0$

which implies by the inductive hypothesis (as ${k < n}$) that ${\psi^* }$ pulls back ${(a_\phi) \in (\sigma NA_*)_n}$ to zero.

But the component of the identity ${[k] \rightarrow [k]}$ of this pull-back is just ${a_\phi}$, from the choice of ${\psi}$. This means that ${a_\phi = 0}$.

Completion of the proof of Dold-Kan

We thus have defined a functor ${N}$ from simplicial abelian groups to chain complexes. We have defined a functor ${\sigma}$ in the opposite direction. We have, moreover, seen that the simplicial abelian group associated to ${NA}$ for ${A_\bullet}$ a simplicial abelian group is just ${A_\bullet}$ 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 ${\sigma C_\bullet}$ is just ${C_*}$, for any chain complex ${C_*}$. So we need to compute ${N(\sigma C_\bullet)}$. In degree ${n}$, this consists of elements of

$\displaystyle \bigoplus_{[n] \twoheadrightarrow [k]} C_k$

that are killed by the ${d_i, i < n}$. The claim is that this consists precisely of ${C_n}$ under the identity ${[n] \twoheadrightarrow [n]}$!

We can see this because we can show that ${C_n \subset N(\sigma C)_n}$ by direct computation; if ${i < n}$, then the map ${d^i: [n-1] \rightarrow [n] \twoheadrightarrow [n]}$ pulls ${C_n}$ down to ${C_{n-1}}$ via the functor ${\Delta' \rightarrow \mathbf{Ab}}$ induced; however, this functor induces zero on coface maps that are not the highest index. Conversely, we must show that ${N(\sigma C_\bullet)_n \subset C_n}$.

To do this, we have to show that ${C_n \subset N(\sigma C_\bullet)_n}$; but we know that ${N(\sigma C)_n}$ is a complement to the degeneracies. However, the ${C_k, k < n}$ occurring in the expression for ${(\sigma C)_n}$ 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.