这是一份manchester曼切斯特大学 MATH31052作业代写的成功案例
Suppose $\pi$ is a group and that $B \pi$ is its classifying space. As is usual, the universal cover of $B \pi$ is denoted $E \pi$. ( $E \pi$ is called the universal space for $\pi$.) Given a $\mathbb{Z} \pi$-module $A$, the homology and cohomology groups of $\pi$ with coefficients in $A$ are defined by
$$
H_{}(\pi ; A):=H_{}(B \pi ; A) \quad \text { and } \quad H^{}(\pi ; A):=H^{}(B \pi ; A) .
$$
Let $\varepsilon: C_{0}(E \pi) \rightarrow \mathbb{Z}$ be the augmentation map. Since $E \pi$ is acyclic (it is contractible), the sequence
$$
\longrightarrow C_{k}(E \pi) \longrightarrow \cdots \longrightarrow C_{0}(E \pi) \stackrel{\varepsilon}{\longrightarrow} \mathbb{Z} \longrightarrow 0
$$
is exact. In other words, (F.4) is a resolution of $\mathbb{Z}$ by free $\mathbb{Z} \pi$-modules. (Here and throughout the group $\pi$ acts trivially on $\mathbb{Z}$.)
MATH31052 COURSE NOTES ：
$$
\operatorname{Hom}{G}\left(\mathbb{Z}\left(G / G{\sigma}\right), \operatorname{Hom}\left(F_{}, \mathbb{Z} G\right)\right) \cong \operatorname{Hom}{G{\sigma}}\left(F_{}, \mathbb{Z} G\right) .^{3}
$$
The cohomology of this last complex is just $H^{q}\left(G_{\sigma} ; \mathbb{Z} G\right)$. Taking cohomology first with respect to $q$ we get a spectral sequence whose $E_{1}^{p q}$ term is a sum of terms of the form $H^{q}\left(G_{\sigma} ; \mathbb{Z} G\right)$. Since $\mathbb{Z} G$ is a free $G_{\sigma}$-module and since $G_{\sigma}$ is finite (because the action is proper), these groups vanish for $q>0$. For $q=0$ they are the invariants, $(\mathbb{Z} G)^{G_{\sigma}}$. All that remains is
$$
E_{1}^{p, 0}=\bigoplus(\mathbb{Z} G)^{G_{\sigma}} \cong \operatorname{Hom}{G}\left(C{p}, \mathbb{Z} G\right)=C_{G}^{p}(Y)
$$
So $E_{2}^{p, 0} \cong H_{G}^{p}(Y ; \mathbb{Z} G)$. Since the $G$-action on $Y$ is cocompact, Lemma F.2.1 (i) gives $H_{G}^{p}(Y ; \mathbb{Z} G)=H_{c}^{p}(Y)$, completing the proof.