Given a fibration $F \to X \to B$ with all spaces pathconnected. Is there a (discrete) group $G$ with normal subgroup $H$ such that $$H^\ast(BG;\mathcal{A}) = H^\ast(X;\mathcal{A})$$ $$H^\ast(BH;\mathcal{A}) = H^\ast(F;\mathcal{A})$$ $$H^\ast(B(G/H);\mathcal{A}) = H^\ast(B;\mathcal{A})$$ for every local coefficient system $\mathcal{A}$ on $X$ ?

2$\begingroup$ To be more precise, I think you want there to be homomorphisms from these groups to the fundamental groups, therefore maps of cohomology groups (from that of BG to that of X, for example), and the question is whether these data can be chosen such that those cohomology maps are always isomorphisms. Also, for the third equation you want a coeff system on B, not on X. $\endgroup$ – Tom Goodwillie Jan 31 '12 at 14:43

$\begingroup$ Yes, that's right. Actually I wonder if it's always possible to replace a Serre spectral seq. of a fibration by the spectral seq. of a group extension (therefore my missing caution for the coefficients on B :) $\endgroup$ – Ralph Jan 31 '12 at 15:01
I think the positive answer to this question is given in the proof of Proposition 1.5 of
 A.J. Berrick and B. Hartley: Perfect radicals and homology of group extensions. Topology and its Applications 25 (1987), 165173.
The idea is to first apply the KanThurston construction to replace $B$ by a classifying space $BQ$. Then pull back the fibration $X\to B$ along $BQ\to B$, and apply the KanThurston construction again to replace the total space $X\times_BBQ$ by a classifying space $BG$. Then consider the classifying space $BN$ for $N$ the kernel of $G\to Q$. The result is a map from the fiber sequence $BN\to BG\to BQ$ to the fiber sequence $F\to X\to B$ such that all the maps (on base, total space, fiber) are acyclic. Then the HochschildSerre spectral sequence for the extension $BN\to BG\to BQ$ should be the Serre spectral sequence associated to the original fiber sequence $F\to X\to B$.