28
$\begingroup$

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 $X$ ?

Improve this question
$\endgroup$
2
  • 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
    Commented Jan 31, 2012 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
    Commented Jan 31, 2012 at 15:01

1 Answer 1

Reset to default
5
$\begingroup$

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 $F\to X\to B$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .