TITLE: A more natural proof of Dold-Kan?
QUESTION [36 upvotes]: The Dold-Kan correspondence gives an equivalence of categories between $SAb$, the category of simplicial abelian groups, and $Ch_{\geq 0}$, the category of non-negatively graded chain complexes of abelian groups. 
Recall that the equivalence in the direction $C: SAb \rightarrow Ch_{\geq 0}$ is most naturally given $C(A)_{n}=A_n/(s_{0}A_{n-1}+\cdots s_{n-1}A_{n-1})$ (simplices modulo degenerate simplices) and taking the differential to be $d=\sum (-1)^{i}d_{i}$. The equivalence in the other direction $\Gamma: Ch_{\geq 0} \rightarrow SAb$ is most naturally given by $\Gamma(C)_{n}=Ch_{\geq 0}(C(\mathbf{Z}\Delta[n]),C)$ (the group of chain maps) where $\Gamma(C)$ gets a simplicial structure by pulling back the comsimplicial structure on the collection of complexes $C(\mathbf{Z}\Delta[n])$. 
The reason that I say this is the most natural is because fits the standard pattern of constructing an adjunction between a presheaf category and another category. Namely, there is a functor $\Delta \rightarrow Ch_{\geq 0}$ sending
$[n]$ to the complex of non-degenerate simplices in $\Delta[n]$ with differential given as the alternating sum of face maps. Then $C: SAb \rightarrow Ch_{\geq 0}$ is the left Kan extension of $\Delta \rightarrow Ch_{\geq 0}$ and $\Gamma$ is the obvious right adjoint to that (it can't be anything else). (Once upon a time I learned this kind of construction from Maclane-Moerdijk, but it can also be found on the nLab :http://ncatlab.org/nlab/show/nerve+and+realization.)
One would then like to show that $C$ and $\Gamma$ are inverse equivalences by showing that the (obvious) unit $Id \rightarrow \Gamma C$ and counit $C\Gamma \rightarrow Id$ are isomorphisms.
By constrast, in every textbook treatment that I have seen, one constructs instead of a quotient complex $C$ an isomorphic subcomplex $N$, complementary to degenerate simplices, but you can do this in at least two different ways. Moreover, one constructs an isomorphism
$\oplus_{[n] \rightarrow [k]} NA_{k} \rightarrow A_{n}$, where the sum is over surjections $[n] \rightarrow [k]$ in $\Delta$, and defines $\Gamma$ similarly, as $\Gamma(C)=\oplus_{[n] \rightarrow [k]} C_{k}$. Ultimately, it seems that one then shows $\Gamma N(A)\simeq A$ for $A \in SAb$ and $N\Gamma(C) \simeq C$ for $C \in Ch_{\geq 0}$, but the isomorphism $\Gamma N(A)\simeq A$ is somehow in the wrong direction (before you know it's an isomorphism), since a priori you would only expect $N,\Gamma$ to form an adjunction and the isomorphism should be given by the unit $A \simeq \Gamma N(A)$.
So my question is whether there is a 'more natural' treatment somewhere in the literature, phrased in terms of the quotient complex $C$, rather than the subcomplex $N$, and in terms of $\Gamma$ given by the obvious right adjoint, not as a sum over surjections, which at first seems to be pulled out of a hat. Of course one can unwind the usual proof and see that everything matches up. For example, one should be able to identify $Ch_{\geq 0}(C(\mathbf{Z}\Delta[n],C)$ with $\oplus_{[n]\rightarrow [k]} C_{k}$, but one can see at least two ways to do this.  
If there is no such treatment, can one explain how to construct one or why one shouldn't bother?

REPLY [25 votes]: The proof of the Dold-Kan theorem basically amounts to the following.  Let $\mathbb{Z}\Delta$ denote the pre-additive category generated by the simplicial indexing category, so that $s\mathrm{Ab}=\mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})$, the category of additive functors.
Let $\mathcal{C}$ be the "Karoubi envelope" of $\mathbb{Z}\Delta$, the category obtained by freely adjoining splittings of idempotents.  You can identify the Karoubi envelope with a full subcategory $s\mathrm{Ab}$, namely the closure of the image of the Yoneda embedding under splitting idempotents.
Then it is trivial that $\mathrm{Fun}^{\mathrm{add}}(\mathcal{C}^\mathrm{op},\mathrm{Ab}) \approx \mathrm{Fun}^{\mathrm{add}}(\mathbb{Z}\Delta^\mathrm{op}, \mathrm{Ab})=s\mathrm{Ab}$.  The Dold-Kan theorem amounts to observing that every object in $\mathcal{C}$ is a direct sum of objects $G(n)$, and that the full subcategory of $G(n)$s is the "indexing category" for chain complexes.  You also have that $G(n)\approx \mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$.
The usual formulas for normalized chains arise by explicitly identifying $G(n)$ with either a (split) subobject or a (split) quotient object of $\mathbb{Z}\Delta[n]$.  In fact, there is only one way (up to sign) to identify $G(n)$ as a split subobject of $\mathbb{Z}\Delta[n]$, and this translates into the "quotient by degeneracies" construction of normalized chains.  There are several ways to identify $G(n)$ as a split quotient: the usual presentation of normalized chains as subobjects is induced by the projection $\Delta[n]\to\Lambda^0[n]$.  
Of course, you could also be perverse: $G(n)$ is a summand of $\mathbb{Z}\Delta[k]$ for any $k\geq n$, so you could define normalized chains using your favorite idenification of $G(n)$ as a summand of $\mathbb{Z}\Delta[n+42]$, if you want!
Added.  Let me try to answer Karol's question, by giving a proof which doesn't use much computation.  
Set $F(n):=\mathbb{Z}\Delta[n]$ and  $G(n):=\mathbb{Z}\Delta[n]/\mathbb{Z}\Lambda^0[n]$. The key observation is to show that the projection $F(n)\to G(n)$ admits a section.  I don't have a clever way to do that ... you just have to write down the formula.  (Which is: send the "generator" $\langle0,1,\dots,n\rangle$ of $G(n)$ to $\sum \pm\langle 0,a_1,\dots,a_n\rangle$, where the sum is over sequences with $a_k\in \{k-1,k\}$, and the sign is the parity of $\lvert\{k\,|\,a_k\neq k\}\rvert$.)  The nice thing is that the section turns out to be unique (this is hardly obvious to begin with).
Now filter $\Delta[n]$ by subcomplexes $\Delta^k[n]=$ the union of all $k$-simplices which contain the vertex $0$.  It's easy to see that
$$
\mathbb{Z}\Delta^k[n] /\mathbb{Z}\Delta^{k-1}[n] \approx G(k)^{\oplus \binom{n}{k}}.
$$
But we just showed that the $G(k)$ are projective, so the filtration lifts to a direct sum decomposition $F(n)\approx \bigoplus_k G(k)^{\oplus\binom{n}{k}}$.  
Now you want to compute $\hom(G(m), G(n))$.  Here is a trick.  It's clear that $\hom(F(m),F(n))=\mathbb{Z}^{\binom{n+m+1}{m+1}}$.  Since the $G(k)$ are summands of the $F(n)$, we have $\hom(G(i),G(j))\approx \mathbb{Z}^{a_{i,j}}$ for some $a_{i,j}\geq0$.  We want to show $a_{i,i}=a_{i,i+1}=1$ and all other $a_{i,j}=0$. 
It's easy to exhibit that $a_{i,i}\geq1$ and $a_{i,i+1}\geq1$.  Plugging the direct sum decomposition into $\hom(F(m),F(n))$ gives 
$$
\binom{n+m+1}{m+1} = \sum_{i,j} \binom{m}{i}\binom{n}{j} a_{i,j}.
$$
But you can use standard combinatorial identities to show this is already an equality when we set $a_{i,i}=a_{i,i+1}=1$ and other $a_{i,j}=0$.  So that must be the answer.