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 $H^\*(X/\pi;k)$? 



cohomology version of Cartan-Leray spectral sequence that deduces cup product
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.


