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Let $p\colon E\to B$ be a fibration with fibres simply connected and homotopy equivalent to a compact CW-complex. Must $p_*\colon H_3(E;\mathbb{Q})\to H_3(B;\mathbb{Q})$ be surjective?

COMMENTS. Yes if (EDIT) $B$ is simply connected, even if the fibre is not compact but just finite-dimensional. In general, finite-dimensionality is not enough: consider the homotopy fibre sequence $\mathbb{R}^3\setminus\mathbb{Z}^3\to T^3\setminus\mathrm{point}\to T^3$.

MOTIVATION. If $p_*\colon H_3(E;\mathbb{Q})\to H_3(B;\mathbb{Q})$ is surjective, then any bundle gerbe over $p\colon E\to B$ is rationally trivial, cf. M. Murray, D. Stevenson, A note on bundle gerbes and infinite-dimensionality (http://arxiv.org/abs/1007.4922).

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  • $\begingroup$ What exactly do you mean for "fibration"? A locally trivial bundle? A Serre fibration? $\endgroup$
    – Daniele Zuddas
    May 3, 2012 at 11:50
  • $\begingroup$ I mean Serre fibration. For locally trivial bundles, the question is equally interesting for me. $\endgroup$
    – Semen Podkorytov
    May 3, 2012 at 12:03
  • $\begingroup$ For locally trivial bundles, it is also natural to ask the fibre to be homeomorphic (not just homotopy equivalent) to a compact CW-complex. $\endgroup$
    – Semen Podkorytov
    May 3, 2012 at 13:37
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    $\begingroup$ @Semen: I tried to prove your claim in the special case when $\pi_1(B) $acts trivially. Are you implicitly using a fact from rational homotopy theory? Namely, the result that appears here: mathoverflow.net/questions/81139/homology-of-loop-space ? $\endgroup$
    – John Klein
    May 4, 2012 at 21:54
  • $\begingroup$ @John: I found a mistake in my proof for trivial $\pi_1$-action (I used only the Serre spectral sequence, no rational homotopy theory). If $B$ is simply connected, the claim follows from Theorem 2 in Gotay-Lashof-Sniatycki-Weinstein "Closed forms on symplectic fibre bundles" (pims.math.ca/~gotay/GLSW.pdf). $\endgroup$
    – Semen Podkorytov
    May 5, 2012 at 13:00

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I'm wondering the extent to which the assumptions can be tweaked. Let's assume $B$ is connected and with basepoint. Let $F$ be the fiber over the basepoint.

However, I won't assume $F$ is homotopy finite (i.e., homotopy equivalent to a finite complex). Nor will I assume anything about the action of $\pi_1(B)$. Rather, I will assume

  • $F$ is $1$-connected (just as Semen does), and

  • $H_2(F;\Bbb Q)$ is trivial.

Assertion: With respect to these assumptions, $H_3(E;\Bbb Q) \to H_3(B;\Bbb Q)$ is surjective.

Proof: By slight abuse of notation, let $E/F$ be the the mapping cone of the inclusion $F\to E$.

Then the Blakers-Massey theorem shows that $E/F \to B$ is 3-connected.

We infer that $E\to B$ is $H_3({-};\Bbb Q)$-surjective if $E \to E/F$ is.

But the long exact homology sequence of $F \to E \to E/F$ and the assumption that $H_2(F;\Bbb Q)$ is trivial implies $H_3(E;\Bbb Q) \to H_3(E/F;\Bbb Q)$ is surjective. $\square$

The above leads to the following question: Is there a relationship between the hypotheses

(1) $F$ is homotopy finite, simply connected and $\pi_1(B)$ acts trivially on $H_*(F;\Bbb Q)$;

(2) $F$ is simply connected and $H_2(F;\Bbb Q)$ is trivial

?

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    $\begingroup$ Your result seems to be clear if you look at the Serre-Leray spectral sequence of the fibration. $\endgroup$
    – Angelo
    May 3, 2012 at 14:45
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    $\begingroup$ Even with a non-trivial $\pi_1$-action? In any case, I tend to be a luddite in these matters: I don't use spectral sequences when I can see the result without using them. $\endgroup$
    – John Klein
    May 3, 2012 at 14:49
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    $\begingroup$ (I'm not surprised though: since the Blakers-Massey theorem is pretty closely related to the Serre exact sequence of the fibration.) $\endgroup$
    – John Klein
    May 3, 2012 at 14:52
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    $\begingroup$ The spectral sequence gives something better: namely, the requested surjectivity holds if and only if the map from the coinvariants in $H_2(F,\mathbb Q)$ to $H_2(E, \mathbb Q)$ is injective. $\endgroup$
    – Angelo
    May 3, 2012 at 16:41
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    $\begingroup$ Angelo: your observation ("something better") can be also seen without the spectral sequence. It uses the observation that there is a fiber sequence ΣF∧ΩB→E/F→B such that the composite ΣF∧ΩB→E/F→ΣF is the Hopf construction of the action of ΩB on F. I had decided not to post this in my original answer, because I wanted to keep things simple. $\endgroup$
    – John Klein
    May 3, 2012 at 18:17

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