How does one prove the following fact. I could not find anything in literature.


Let $\pi$ be a subgroup of the symmetric group $S\_p$ and let $W$ be a free $\pi$-complex. Then for any space $X$ there is the isomorphism of homology groups in $\mathbb{F}\_p$-coefficients


$H\_\*(W \otimes\_{\pi} C\_{\ast}(X)^{\otimes p}) \cong H\_{\ast}(W \otimes\_{\pi} H\_{\ast}(X)^{\otimes p})$.


Thank you for your help.



free complex with mod-p coefficients
If $C\_\ast$ is a chain complex of vector spaces over any field and $H\_\ast$ is its homology, then, viewing $H\_\ast$ as a chain complex, there is a chain map $r:C\_\ast\to H\_\ast$ inducing the obvious isomorphism in homology and there is a chain map $i:H\_\ast\to C\_\ast$ inducing its inverse. This results in an equivariant splitting $H\_\ast^{\otimes n}\to C\_\ast^{\otimes n}\to H\_\ast^{\otimes n}$. The other factor contributes nothing to the homology of $W\otimes\_{\pi}-$, because its homology is trivial and $W$ is free.


To see that the resulting isomorphism between $H\_\ast(W\otimes\_{\pi} C\_\ast^{\otimes p})$ and $H\_\ast(W\otimes\_{\pi} H\_\ast^{\otimes p})$ is independent of the chosen splitting $(r,i)$, first note that it depends only on $i$ and that its inverse (therefore also it) depends only on $r$. Now if $(r',i')$ is another such splitting then the splitting $(r',i)$ yields the same isomorphism as $(r,i)$ and also the same isomorphism as $(r',i')$.


By similar reasoning the isomorphism is natural (with respect to chain maps). 


Note that the field did not have to be $\mathbb F\_p$, and $p$ did not have to be prime, and the chain complex did not have to come from a space.


