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 ${A_\bullet}$ looks a lot different from ${A_\bullet}$, which a priori has much more structure. Nonetheless, we are going to see that it is possible to recover ${A_\bullet}$ 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 ${A_\bullet}$

$\displaystyle \bigoplus_{\phi: [n] \twoheadrightarrow [k]} NA_k \simeq A_n. \ \ \ \ \ (1)$

Here the map from a factor ${NA_k}$ corresponding to some ${\phi: [n] \twoheadrightarrow [k]}$ to ${A_n}$ is given by pulling back by ${\phi}$. 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) ${A_n = \bigoplus_{[n] \twoheadrightarrow [k]} NA_k}$. Given ${f: [m] \rightarrow [n]}$ and a factor ${NA_k}$ of ${A_n}$ (for some epimorphism ${\phi: [n] \twoheadrightarrow [k]}$), we want to know where ${f^*}$ takes ${NA_k}$ into ${A_m}$.

We can factor the composite ${[m] \rightarrow [n] \twoheadrightarrow [k]}$ as ${[m] \stackrel{\psi_1}{\twoheadrightarrow} [m'] \stackrel{\psi_2}{\hookrightarrow} [k]}$. It is easy to see that simplicial maps induced by injections in ${\Delta}$ preserve ${NA}$. There is a commutative diagram:

Now ${\psi_1}$ is an epimorphism. It follows that ${\psi_1^*: NA_{m'} \rightarrow A_m}$ is one of the maps in the canonical decomposition. It follows that we have a recipe for determining where the ${\phi}$-factor ${NA_k}$ of ${A_n}$ goes:

1. Consider the composite ${[m] \stackrel{f}{\rightarrow} [n] \stackrel{\phi}{\rightarrow} k}$, and factor this as a composite ${\psi_2 \circ \psi_1}$ with ${\psi_1: [m] \twoheadrightarrow [m']}$ an epimorphism and ${\psi_2: [m'] \hookrightarrow [k]}$ a monomorphism.
2. Then ${NA_k}$ (embedded in ${A_n}$ via ${\phi^*}$) gets sent to ${NA_{m'} \subset A_m}$ (embedded in ${A_m}$ via ${\psi_1^*}$).
3. The map ${NA_k \rightarrow NA_{m'}}$ is given by ${\psi_2^*}$.

The construction

Motivated by this, let us describe the inverse construction. Let ${C_*}$ be a chain complex (nonnegatively graded, as always). We define a simplicial abelian group ${\sigma C_\bullet}$ such that

$\displaystyle \sigma C_n = \bigoplus_{[n] \twoheadrightarrow [k]} C_k.$

The sum is taken over all surjections ${[n] \twoheadrightarrow [k]}$.

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 ${\psi_2^*}$ in the explanation of (3) above does not a priori make sense, let us note that if we restrict to the subcategory ${\Delta' \subset \Delta}$ consisting of injective maps, then the map ${[n] \mapsto C_n}$ becomes a contravariant functor in a natural way. Indeed, we let the map ${C_n \rightarrow C_m}$ induced by an injection ${[m] \hookrightarrow [n]}$ be zero unless ${m = n-1}$ and we are working with the map ${d^{n}}$, in which case we let the map ${C_n \rightarrow C_{n-1}}$ be the differential. Since ${C_*}$ 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 ${A_\bullet}$, then if the chain complex ${NA}$ is made into a contravariant functor ${\Delta' \rightarrow \mathbf{Ab}}$, we have gotten nothing new: we just recover the simplicial structure maps. Indeed, if ${\psi: [m] \hookrightarrow [n]}$ is an injection, then the map ${\psi^*: NA_n \rightarrow NA_m}$ is zero unless ${\psi = d^{n}}$ and ${m = n-1}$. Otherwise ${\psi}$ will contain a ${d^i}$ for some ${i < n}$, and the definition of ${NA}$ completes the proof. We thus see:

Lemma 1 Let ${C_*}$ be a chain complex. Then there is a functor from ${\Delta' \rightarrow \mathbf{Ab}}$ sending ${[n] \rightarrow C_n}$ and an injection ${[m] \hookrightarrow [n]}$ to zero unless ${m = n-1}$ and the injection is ${d^{n}}$, in which case it is the differential. If ${A_\bullet}$ is a simplicial abelian group, this construction agrees with the simplicial maps when restricted to ${NA_*}$.

Let us now, finally, show how to make ${\sigma C_\bullet}$ into a simplicial abelian group. Given some map ${[m] \rightarrow [n]}$ in ${\Delta}$, we map the individual terms as follows. Let ${\phi: [n] \twoheadrightarrow [k]}$ be an epimorphism in ${\Delta}$, giving a factor ${C_k \subset \sigma C_n}$. We then map

$\displaystyle C_k \rightarrow \sigma C_m = \bigoplus_{[m] \twoheadrightarrow [l] } C_l$

as follows. If ${[n] \twoheadrightarrow [k]}$ is the given surjection, then we have a map ${[m] \rightarrow [n] \rightarrow [k]}$, which we can factor as a composite ${[m] \twoheadrightarrow [m'] \stackrel{\psi}{\hookrightarrow} [k]}$, of a surjection and an injection. So we send ${C_k}$ (via ${\psi^*}$, which is defined by the functoriality) to ${C_{m'}}$, imbedded in ${\sigma C_m}$ as the factor corresponding to the surjection ${[m] \twoheadrightarrow [m']}$.

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 ${[m] \rightarrow [n] \rightarrow [p]}$, then the corresponding map ${\sigma C_p \rightarrow \sigma C_m}$ induced by the composite ${[p] \rightarrow [m]}$ is the composite of the maps ${\sigma C_p \rightarrow \sigma C_n \rightarrow \sigma C_m}$. So consider a factor of ${\sigma C_p}$ corresponding to a surjection ${[p] \twoheadrightarrow [p']}$. Now we can draw a commutative diagram in the simplex category ${\Delta}$:

A close look at this will establish the claim, since ${C_*}$ is a functor from the category of finite ordered sets and injective, order-preserving maps. As a result, we have (finally!) constructed our functor ${\sigma}$ from chain complexes to simplicial abelian groups. Note that there is a natural transformation

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

for any simplicial abelian group ${A_\bullet}$. On the ${n}$-simplices, this is the map

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

where the factor corresponding to ${\phi}$ is mapped to ${A_n}$ by pulling back by ${\phi}$. 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 ${\sigma}$ from chain complexes to simplicial abelian groups, and the natural transformation ${\sigma (N A_*)_\bullet \rightarrow A_\bullet}$ for any simplicial abelian group ${A_\bullet}$. We want to show that this is a quasi-inverse to ${N}$, 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 ${A_\bullet}$, 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.

This is going to take some work, and we are going to need first a simpler splitting that will, incidentally, show that ${NA_*}$ and ${(A/DA)_*}$ are isomorphic. We are going to prove the above result by induction, using:

Lemma 4 Let ${A_\bullet}$ be a simplicial abelian group. Then the map

$\displaystyle NA_n \oplus DA_n \rightarrow A_n$

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 ${k < n}$, we define ${N_k A_n = \bigcap_{0}^k \ker d_k}$ and ${D_k A_n}$ to be the group generated by the images of ${s_j(A_{n-1}), j \leq k}$. So these are partial versions of the ${NA_n, DA_n}$. The claim is that there is a natural splitting

$\displaystyle N_k A_n \oplus D_k A_n = A_n.$

When ${k=n-1}$, the result will be proved (note that ${D_{n-1}A_n}$ is the group generated by degenerate simplices because the degeneracies ${s_i: A_{n-1} \rightarrow A_{n}}$ only go up to ${n-1}$). When ${k=0}$, the splitting is

$\displaystyle \ker d_0 \oplus \mathrm{Im} s_0 = A_n.$

We can see this as follows. We have maps

Here ${s_0}$ is a split injection, with ${d_0}$ being a section. But in general, whenever ${i: A \rightarrow B}$ is a split injection with section ${q: B \rightarrow A}$, then ${B}$ splits as ${\ker q \oplus \mathrm{Im}}$. Now let us suppose we have established the splitting ${A_n = N_{k-1}A_n \oplus D_{k-1} A_n}$. We need to establish it for ${k}$. For this we will write some exact sequences.

1. We have a split exact sequence:

$\displaystyle 0 \rightarrow A_{n-1}/D_{k-1}A_{n-1} \stackrel{s_k}{\rightarrow} A_n/D_{k-1}A_n \rightarrow A_n/D_k A_n \rightarrow 0.\ \ \ \ \ (2)$

Indeed, exactness of this sequence will be clear once we show that is well-defined. But if ${j < k}$, then ${s_k s_j = s_j s_{k-1}}$, so ${s_k}$ sends ${D_{k-1}A_{n-1}}$ into ${D_{k-1} A_n}$. The splitting is given by ${d_k}$.

2. Similarly, we have a split exact sequence (where the simplicial identities show that ${s_k(N_{k-1} A_{n-1}) \subset N_{k-1} A_n}$)

$\displaystyle 0 \rightarrow N_{k-1} A_{n-1} \stackrel{s_k}{\rightarrow} N_{k-1} A_n \rightarrow N_{k} A_n \rightarrow 0. \ \ \ \ \ (3)$

This is perhaps less obvious. This is equivalent to the claim that the map

$\displaystyle N_k A_n \oplus N_{k-1}A_{n-1} \stackrel{(i, s_{k})}{\rightarrow} N_{k-1}A_n$

is an isomorphism. (Here ${i}$ denotes the inclusion.) We first claim that it is surjective. Indeed, if ${a \in N_{k-1}A_n}$, then ${a - s_k d_k a}$ lies in fact in ${N_k A_n}$. This is because ${d_k}$ is a section of the split injection ${s_k}$, and because ${d_j (a - s_k d_k a)}$ for ${j < k}$ by using the simplicial identities to move ${d_j}$ to the inside. Conversely, to see that it is injective, it suffices to note that if ${s_k b \in N_k A_n}$ for ${b \in N_{k-1} A_{n-1}}$, then ${b = 0}$; but ${b = d_k (s_k b) = 0}$ by definition of ${N_k A_n}$.

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 ${s_k d_k a}$ does not affect things modulo ${D_k A_n}$, 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

$\displaystyle NA_* \rightarrow (A/DA)_*$

is an isomorphism of chain complexes.

This is why we added the sign to the definition of the differential in constructing ${NA_*}$.

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.