Fibrations with non-simply connected base and rational homology 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).
 A: 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
?
