Covering spaces and cohomology

Question

If $p: B \rightarrow C$ is a finite covering space with covering group $G$. Why (rigorously) $H^{*}(B)^{W} \simeq H^{*}(C)$

Answer 1

In general, $G$ is a group acting continuously on a space $X$ and $M$ is an abelian group, there is a spectral sequence going from $E_2=H^\bullet(G,H^\bullet(B,M))$ to $H^\bullet(B/G,M)$; here $H^\bullet(G,\mathord-)$ is group cohomologgy of the discrete group $G$. This was constructed by Grothendieck in his famous Tohôku paper, for example. (The spaces have to be sufficiently nice for their sheaf cohomology to coincide with whatever cohomology you want; paracompact is enough, and that is not exactly a draconian hypothesis :D )

If $p:B\to C$ is a covering with group $G$, then $B/G=C$. If morevoer the group finite and, say, $M=\mathbb Q$ (or any abelian group in which multiplication by the order of $G$ is an isomorphism) then $H^\bullet(B,M)$ is also a abelian group with that property, and $H^p(G,H^\bullet(B,M))=0$ if $p>0$. Therefore the spectral sequence collapses at $E_2$ and we have an isomorphism $H^0(G,H^\bullet(B,M))\cong H^\bullet(C,M)$. Finally, since $H^0(G,\mathord-)$ is canonically isomorphic to $(\mathord-)^G$, we have your isomorphism.

All this technology, though, is not really needed. Suppose for example the spaces in question are manifolds and we compute cohomology using de Rham. Then the group $G$ acts on the de Rham complex $\Omega^\bullet(B)$, and we can consider its fixed subcomplex, $\Omega^\bullet(B)^G$. It is very easy to show that $\Omega^\bullet(B)^G$ is in fact isomorphic to the de Rham complex $\Omega^\bullet(C)$ of the quotient. On the other hand, it is also easy to show that taking invariants and computing cohomology of these vector spaces commute (it is for this that we want the group finite and real vector spaces), so that $H(\Omega^\bullet(B)^G)\cong H(\Omega^\bullet(B))^G$. The same thing will work wit the complex which computes singular cohomology with coefficients in $\mathbb Q$.