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Let $A$ be a $G$-module.

Let $C_n=\mathbb{Z}G\otimes \cdots \otimes \mathbb{Z}G$ ($n+1$ copies). It is free $\mathbb{Z}$-module with basis $g_0\otimes g_1\otimes \cdots \otimes g_n$, $g_i\in G$.

Denote $1\otimes g_1\otimes \cdots \otimes g_n\in C_n$ by $(g_1,\ldots,g_n)$. These $n$-tuples are basis of free $\mathbb{Z}G$-module $C_n$.

Define $\mathbb{Z}G$-module homomorphisms $d_n:C_n\rightarrow C_{n-1}$ by $$d_n(g_1,\ldots,g_n)=g_1(g_2,\ldots,g_n)+\sum_{1}^{n-1}(-1)^i(g_1,\ldots, g_{i-1}, g_ig_{i+1},g_{i+2},\ldots, g_n)+(-1)^n(g_1,\ldots,g_{n-1}).$$ Let $\varepsilon:\mathbb{Z}G(=C_0)\rightarrow \mathbb{Z}$ be augmentation map.

Claim: The complex $\mathbf{(C)}$: $\cdots \rightarrow C_2\xrightarrow{d_2}C_1\xrightarrow{d_1}C_0\xrightarrow{\varepsilon} \mathbb{Z}\rightarrow 0$ is exact.

Proof in Jacobson's Algebra Vol.2 We define homomorphisms of abelian groups $s_i:C_{i+1}\rightarrow C_i$ such that $$\varepsilon s_{-1}=1_{\mathbb{Z}}; d_1s_0+s_{-1}\varepsilon=1_{C_0}; d_{n+1}s_n + s_{n-1}d_n=1_{C_n}.$$

Question: These conditions say that the maps $s_i$ define a homotopy between identity morphism $(\mathbf{C})\rightarrow (\mathbf{C}$) is homotopic to zero morphism from $(\mathbf{C})\rightarrow (\mathbf{C}$). But I didn't get how this implies the exactness of $(\mathbf{C})$? This could be trivial, but I was unable to see clearly the justification of Jacobson.

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Suppose $d_n(a)=0$ ($a$ is a cycle). Then $$a=(d_{n+1}s_n+s_{n-1}d_n)(a)=(d_{n+1}s_n)(a)=d_{n+1}(b)$$ for $b=s_n(a)$. Thus $a$ is a boundary. All cycles are boundaries, so the chain complex is exact.

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