TITLE: Is there a Kan-Thurston theorem for fibrations ? 
QUESTION [28 upvotes]: 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 $X$ ?

REPLY [5 votes]: 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), 165-173.

The idea is to first apply the Kan-Thurston construction to replace $B$ by a classifying space $BQ$. Then pull back the fibration $X\to B$ along $BQ\to B$, and apply the Kan-Thurston 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 Hochschild-Serre spectral sequence for the extension $BN\to BG\to BQ$ should be the Serre spectral sequence associated to the original fiber sequence $F\to X\to B$.