Hi:


When I read the book "An introduction to Homological algebra" by Weibel, the page 206, line 9 says that


"Shapiro's Lemma tell us that 


$H\_q(S\_n(X)\otimes\_{Z}A)$ is zero if $q\neq 0$ and is isomorphic to $S\_n(X/G)\otimes\_ZA$ if $q=0$" ,


here $Z$ is the ring of integers, $G$ is a group, $X$ is a path-connected topological space and $S\_n(X)$ is the sequence of singular complex of $X$, and $A$ is an abelian group.


But I think is strange, and I can not prove this, also modify it. 


I really want to know the correct statement and a proof for this.


Thanks!



A computation by the Shapiro Lemma
It seems to me that "$H\_q(S\_n(X)\otimes\_{\mathbb{Z}}A)$" should read "$H\_q(G;S\_n(X)\otimes\_{\mathbb{Z}}A)$".


Then 
$$H\_0(G;S\_n(X)\otimes\_{\mathbb{Z}}A)=(S\_n(X)\otimes\_{\mathbb{Z}}A)\_G = S\_n(X/G)\otimes\_{\mathbb{Z}}A$$
by 6.10.2, and Shapiro's lemma gives $H\_q(G; S\_n(X)\otimes\_{\mathbb{Z}}A)= 0$ for $q>0$ since $$S\_n(X)\otimes\_{\mathbb{Z}}A=\bigoplus\operatorname{Ind}^G\_1 A$$ is induced from a module over the trivial group (sum is over the singular $n$-simplices of $X$). 


Compare Brown's "Cohomology of groups", sections III.6 and VII.7. 


