On page 338, A User's Guide to Spectral Sequences. 2nd Edition, by John McCleary, Theorem 8.9, there is a Cartan-Leray spectral sequence for homology:
If $X$ is a connected pace on which the group $\pi$ acts freely and properly, then there is a spectral sequence, homological type, with $$ E^2_{p,q}=H_p(\pi,H_q(X)),$$ converging strongly to $H_*(X/\pi)$.
Is there any cohomology version of Cartan-Leray spectral sequence that can get cup product structure of $H^*(X/\pi)$?
Let $k$ be a field. Suppose $H^*(X;k)$ and $H^*(\pi;k)$ are known. How to get the cup product structure of $H^*(X/\pi;k)$?
Best Answer
The Leray-Serre spectral sequence in cohomology is multiplicative, meaning that there is a multiplication $E_{r}^{pq} \otimes E_r^{p'q'} \to E_r^{p+p',q+q'}$ for each $r$, and the multiplication on the $E_{r+1}$ page is induced by that on the $E_r$ page. Note however that this does not actually give you the cup product on $H^\ast(X/\pi)$ (in your case), only on the bigraded algebra $\bigoplus_{p,q} E_\infty^{p,q} = \bigoplus_{p,q} \mathrm{Gr}_L^p H^{p+q}(X/\pi)$. So you're only getting partial information about the cup product on $H^\ast(X/\pi)$; to get the full information you will need some luck (maybe $E_\infty$ is concentrated along a single row, or something) or trickiness.
In general, knowledge of $H^\ast(X)$ and $H^\ast(\pi)$ does not even give you knowledge of the $E_2$ page $E_2^{pq} = H^p(\pi,H^q(X))$, as remarked in the comments. This is only true if $\pi$ acts trivially on the cohomology of the fibers.