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Given a fibration $F \to X \to B$ with all spaces path-connected. 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 $X$ ?

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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. –  Tom Goodwillie Jan 31 '12 at 14:43
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 :) –  Ralph Jan 31 '12 at 15:01

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