# Inverse functor in proof of Dold Kan Correspondence

I´m looking at the proof of the Dold-Kan correspondence. Let $SA$ be the category of simplicial objects in an abelian category $A$ and $CH_{\ge0}(A)$ the category of non-negative chain complexes in $A$. The theorem states that the normalisation functor $N : SA \rightarrow CH_{\ge0}(A)$ is an equivalence of categories.

I am following the proof in Weibel´s book and am trying to construct the inverse functor $K : CH_{\ge0}(A)\rightarrow SA$. For a chain complex $C$ we define $K(C)_n = \oplus_{p \le n} \oplus _{\eta} C_p[\eta ]$ where, for each $p$ $\eta$ ranges over all the surjections $[n]\rightarrow [p]$ and $C_p[\eta ]$ denotes a copy of $C_p$. I am trying to show that this is a simplicial object in our abelian category. I using the definition of a simplical object as a contravariant functor from the ordinal number category $\Delta \rightarrow A$. I understand how it is defined on objects and morphisms however I am stuck trying to show that $K(\alpha \circ \beta) = K(\beta) \circ K(\alpha)$ for $\alpha$, $\beta$ in $\Delta$.

Any help would be much appreciated.

(I'm answering for the sake of leaving no question unanswered; I apologize in advance for resurrecting this 3-year old question.)

This is something you just have to write down I suppose. The key idea here I think is that epi-mono factorizations "compose" in the sense that if the two squares below are epi-mono factorizations of the upper-right path, then so is the outer square.

It also helps to organize the maps. That is, we can represent $\alpha^*$ as the matrix $(f_{\eta,\eta'})$ with respect to the bases $\{\eta\}$ and $\{\eta'\}$ (notation as in diagram), where $$f_{\eta,\eta'} = \begin{cases} \operatorname{id} & \eta' = \eta \alpha \\ d & \delta^p \eta' = \eta \alpha \\ 0 & \text{else}. \end{cases}$$

Similarly, $\beta^*$ and $(\alpha \circ \beta)^*$ are represented respectively by the matrices $(g_{\eta',\eta''})$ and $(h_{\eta,\eta''})$ where $$g_{\eta',\eta''} = \begin{cases} \operatorname{id} & \eta'' = \eta \beta \\ d & \delta^q \eta'' = \eta' \beta \\ 0 & \text{else} \end{cases}$$ $$h_{\eta,\eta''} = \begin{cases} \operatorname{id} & \eta'' = \eta \alpha \beta\\ d & \delta^p \eta'' = \eta \alpha \beta \\ 0 & \text{else}. \end{cases}$$

Fix $\eta, \eta''$. Our goal is now to show that $$h_{\eta,\eta''} = \sum_{\eta'} g_{\eta',\eta''} f_{\eta, \eta'}.$$

The first thing to notice is that most of the terms in the sum are going to be zero. (In fact, it really is a sum of just a single term, since $\eta'$ has to be part of the unique epi-mono factorization of $\eta \alpha$.) There are essentially only three ways $g_{\eta',\eta''} f_{\eta,\eta'}$ can be nonzero.

1. The epi-mono factorization of $\eta \alpha \beta$ has $[r] = [p]$. In this case, we have $\eta'' = \eta \alpha \beta$ and therefore if we construct the same factorization in two steps we find $\eta' = \eta \alpha$ and $\eta'' = \eta' \beta$. This means $f_{\eta,\eta'} = g_{\eta',\eta''} = \operatorname{id}$.
2. The epi-mono factorization of $\eta \alpha \beta$ has $[r] = [p-1]$. In this case, the two step epi-mono factorization may yield $[q] = [p]$ or $[q] = [p-1]$. These are two disjoint cases, and by a similar argument as above we see that exactly one of $\epsilon$, $\zeta$ is the identity and the other is $\delta^p$, so exactly one of $f_{\eta,\eta'}$, $g_{\eta,\eta'}$ is the identity and the other is the differential $d$. Either way, we find that $f_{\eta,\eta'} g_{\eta,\eta'} = d$.
3. The epi-mono factorization of $\eta \alpha \beta$ has $r < p-1$. In this case, at best we have $r = p-2$, $\epsilon = \delta^p$, and $\zeta = \delta^{p-1}$, so $f_{\eta,\eta'} g_{\eta,\eta'} = d^2 = 0$. In all other cases either $p-q$ or $q-r$ is $> 1$, and one of the maps $f_{\eta,\eta'}$ and $g_{\eta', \eta''}$ is zero.

So we have computed all the terms in the sum $\sum_{\eta'} g_{\eta',\eta''} f_{\eta, \eta'}$. Direct comparison shows that we get exactly the same formula as $h_{\eta,\eta''}$. This proves functoriality of $K$.