6
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

(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.

  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 $K$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.