Note that indeed $\Hom{\DELTA}{X}{[n]}\in\Set$, because the collection of morphisms in a category is per definition a set. We do not need to specify the face or degeneracy maps, as we already know that $\mathbf{Hom}$ is a functor (in both arguments).
\todo{sAb: as example do $\Delta[n]$}
\todo{sAb: as example do the free abelian group pointwise}
Note that indeed $\Hom{\DELTA}{X}{[n]}\in\Set$, because the collection of morphisms in a category is per definition a set. We do not need to specify the face or degeneracy maps, as we already know that $\mathbf{Hom}$ is a functor (in both arguments). Still it is useful to write out some cases.
\begin{example}
We will compute how $\Delta[0]$ look like. Note that $[0]$ is an one-element set, so for any set $X$, there is only one function $\ast : X \to[0]$. Hence $\Delta[0]_n =\{\ast\}$ for all $n$. The face and degeneracy maps are now functions from $\{\ast\}$ to $\{\ast\}$. Again there is only one, namely $\id : \{\ast\}\to\{\ast\}$. This gives:
\todo{sAb: insert picture}
\end{example}
\begin{example}
$\Delta[1]$ is a bit more interesting, but still not too hard. We will compute the first three abelian groups $\Delta[1]_0$, $\Delta[1]_1$ and $\Delta[1]_2$. We can use the fact that any monotone increasing map $f: [n]\to[m]$ is a composition of first applying degeneracy maps, and then face maps, ie.: $f: [n]\tot{\sigma^{i_0}\cdots\sigma^{i_M}}[k]\tot{\delta^{j_0}\cdots\delta^{j_N}}[m]$, where $k \leq m, n$.
For $\Delta[1]_0$ we have to consider maps from $[0]$ to $[1]$, we cannot first apply degeneracy maps (there is no object $[-1]$). So this leaves us with the face maps: $\Delta[1]_0=\{\delta_0, \delta_1\}$. For $\Delta[1]_1$ we of course have the identity function and two functions $\delta_0\sigma_0, \delta_1\sigma_0$. Now $\Delta[1]_2$ are the maps from $[2]$ to $[1]$.
We will compute the two face maps $\delta^0$ and $\delta^1$ from $\Delta[1]_1$ to $\Delta[1]_0$. Recall that the $\mathbf{Hom}$-functor in the first argument (the contravariant argument) works with precomposition. So this gives:
Where we in the first calculation used the identity law. In the second and third line we used the third simplicial equation, asserting that $\sigma_0\delta_0=\id$. Similarly we can calculate the face map $\delta^1$:
As we are interested in simplicial abelian group, it would be nice to make these $n$-simplices into simplicial abelian groups. We have seen how to make an abelian group out of any set using the free abelian group. We can use this functor $\Z[-] : \Set\to\Ab$ to induce a functor $\Z^\ast[-] : \sSet\to\sAb$ as shown in the diagram~\ref{fig:diagram_Z}.
\begin{figure}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes]{
\DELTA^{op}&\Set\\
&\Ab\\
};
\path[->]
(m-1-1) edge node[auto] {$ X $} (m-1-2)
(m-1-2) edge node[auto] {$\Z[-]$} (m-2-2)
(m-1-1) edge node[auto] {$ X' $} (m-2-2);
\end{tikzpicture}
\caption{The simplicial set $X$ can be made into a simplicial abelian group $X'$ by postcomposing with $\Z[-]$}
\label{fig:diagram_Z}
\end{figure}
\begin{example}
We can apply this to the standard $n$-simplex $\Delta[1]$. This gives $\Delta[1]_0\iso\Z^2$, since $\Delta[1]_0$ had two elements, and $\Delta[1]_1\iso\Z^3$, where the isomorphisms are taken such that:
\begin{align*}
\delta_0 &\mapstot{\iso} (1, 0) \\
\delta_1 &\mapstot{\iso} (0, 1) \\
\delta_0\sigma_0 &\mapstot{\iso} (1, 0, 0) \\
\id&\mapstot{\iso} (0, 1, 0) \\
\delta_1\sigma_0 &\mapstot{\iso} (0, 0, 1)
\end{align*}
The face maps from $\Delta[1]_1$ to $\Delta[1]_0$ under these isomorphisms are then given by:
Joshua Moerman
11 years ago