Let $f\colon E\to B$ be a fiber bundle with a connected fiber $F$, $f$ is proper. Let $\underline{\mathbb{C}}_E$ be the constant sheaf on $E$. Let $f_*(\underline{\mathbb{C}}_E)$ denote its direct image in the derived category $D(Sh_B)$ of sheaves of $\mathbb{C}$-vector spaces. Since $F$ is connected, it is clear that $\tau_{\leq 0}(f_*(\underline{\mathbb{C}}_E))=\underline{\mathbb{C}}_B$. Hence there exists an exact triangle in $D(Sh_B)$ $$\underline{\mathbb{C}}_B\to f_*(\underline{\mathbb{C}}_E)\to \mathcal{F},$$ where $\mathcal{F}\in D^{\geq 1}(Sh_B)$.

Question. Is it true that the above exact triangle splits? If not, under what conditions this is true?

Remark. I am aware of a situation when the above triangle splits, but it is too restrictive for my purposes; in fact much more it true in that case. Let $f\colon E\to B$ be a smooth morphism of smooth projective complex algebraic manifolds. Then it is a special case of the decomposition theorem due to Beilinson-Bernstein-Deligne-Gabber that not only the above exact triangle does split, but moreover $f_*(\underline{\mathbb{C}}_E)$ is isomorphic in $D(Sh_B)$ to the direct sum of its cohomology sheaves (which are necessarily shifted local systems).

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    $\begingroup$ Consider the Hopf fibration $S^3 \to S^2$. If your sequence split, then $H^*(S^2)$ would be a summand of $H^*(S^3)$. $\endgroup$ – Vivek Shende Apr 6 '16 at 16:03

No, it is not true. For example, let $E=\mathbb C^2\backslash \{ 0\}$, $B=\mathbb C\mathbb P^1$ (with the obvious map $f$). Then the pushforward as a complex of sheaves on $\mathbb C\mathbb P^1$ has the following cohomology: constant sheaf in degree 0 and constant sheaf in degree +1. If the triangle you are asking about were split, then the cohomology of the pushforward would be $H^*(\mathbb C\mathbb P^1)\oplus H^*(\mathbb C\mathbb P^1)[-1]$. But it the cohomology of the pushforward is the same as $H^*(\mathbb C^2\backslash \{ 0\})$, which has total dimension 2 and not 4.

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    $\begingroup$ Interesting example, thanks. Here the map is not proper, but homotopically your example is equivalent to that in Vivek Shende' comment, where the map is proper. $\endgroup$ – makt Apr 6 '16 at 16:17
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    $\begingroup$ Well, unless you work in an algebraic (or, at least, complex analytic) context, properness is not a sensible condition (almost any nice map is homotopy equivalent to a proper one). $\endgroup$ – Alexander Braverman Apr 6 '16 at 16:19

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