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.
 A: (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.
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
[l] & \ra{\beta} & [m] & \ra{\alpha} & [n] \\
\da{\eta''} & & \da{\eta'} & & \da{\eta} \\
[r] & \ra{\zeta} & [q] & \ra{\epsilon} & [p] \\
\end{array}
$$
Here, the vertical arrows are surjections and the bottom horizontal arrows are monomorphisms. I will try to use the terminology and notation from Weibel, and also refer to the diagram above constantly.  
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.  


*

*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 $f_{\eta,\eta'} = g_{\eta',\eta''} = \operatorname{id}$.  

*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 $f_{\eta,\eta'} g_{\eta,\eta'} = d$.  

*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 $K$.  
