Suppose that $G$ is a group, and $\mathbb{Z}[G]$ the group ring. Then $\mathbb Z$ can be considered as a $\mathbb{Z}[G]$-module if every $g (\in G)$ acts trivally, and every $n (\in \mathbb Z)$ acts by multiplication on $\mathbb Z$.

In order to find a projective resolution of the module $\mathbb Z$, thus finding the cohomology module, a book says (translated):

Consider the elements of $G$ as vertexes, then we can define a simplicial complex $K = K(G)$. Suppose that there is a well order in $G$, and $e$ is the first element in this order. So $K_n = \{ (g_0, g_1, \cdots, g_n) | g_i \in G, g_0 <g_1< \cdots <g_n \}$. Let $C_n$ be the free Abelian group in which the simplexes of $K_n$ form a basis. The group $G$ acts on $C_n$ by $g(g_0, \cdots, g_n) = (gg_0, \cdots, gg_n)$. The boundary operator $\partial_n : C_n \rightarrow C_{n-1}$ is $\partial_n(g_0, \cdots, g_n) = \sum_{i=0}^n (-1)^i (g_0, \cdots, \hat{g_i}, \cdots, g_n)$ and the augmentation morphism $\epsilon: C_0 \rightarrow \mathbb Z$ maps every element of $G$ to $1$. The sequence $\cdots \rightarrow C_n \rightarrow C_{n-1} \rightarrow \cdots \rightarrow C_1 \rightarrow C_0 \rightarrow \mathbb Z \rightarrow 0$ is the $G$-module homomorphsim sequence.

In order to prove the exactness of this sequence, the author defines a group homomorphism $s_n: C_n \rightarrow C_{n+1} , (g_0, \cdots, g_n) \mapsto (e,g_0, \cdots, g_n)$ for $n =0,1,2, \cdots$, and use this to prove that the identity map in $C_n$ is null-homotopic. But this definition is unclear to me. I thought one vertex could appear only once in a simplex. So, what will happen if one of the $g_i$ in $\{g_0, \cdots, g_n \}$ is already $e$? Going back to the well-ordering assumption on $G$, is this well-order strict or not? If it is strict, then I think no duplication of vertexes in a simplex would be allowed; otherwise, the chain would be endless even if the group is finite.

Thanks to everyone.

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    $\begingroup$ The action of $G$ on the $K_n$ doesn't seem to work - there was no condition that required that if $g_1 < g_2$ then $gg_1 < gg_2$ $\endgroup$ – Thomas Andrews Feb 27 '12 at 17:01
  • $\begingroup$ @Thomas Andrews: Thanks for the comment. I think the elements in $(gg_0, \cdots, gg_n)$ can be repositioned, i.e., take some $\delta \in S_{n+1}$ such that $\delta(gg_0) < \cdots < \delta(gg_n)$, so $(gg_0, \cdots, gg_n) = (-1)^{\delta} (\delta(gg_0), \cdots, \delta(gg_n))$. $\endgroup$ – ShinyaSakai Feb 28 '12 at 4:58

My understanding of $s_{n}: C_{n}\rightarrow C_{n+1}$ is as follows:

$(g_{0}, g_{1}, \cdots, g_{n}) \mapsto (e, g_{0}, g_{1}, \cdots, g_{n})$ if $g_{0}\neq e$ (in fact $g_{i}\neq e$ for all $0\leqslant i \leqslant n$)

if $g_{0}=e$, then $s_{n}$ will send $(g_{0}, g_{1}, \cdots, g_{n})$ to the zero element in $C_{n+1}$.


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