TITLE: Is a topological fiber-bundle, whose total space admits a retraction onto a fiber, trivial?
QUESTION [12 upvotes]: Let $\xi = \pi \colon E \to B$ a topological fiber bundle with connected base $B$, $E_x = \pi^{-1}(x)$ the fiber at $x \in B$, $j \colon E_x \hookrightarrow E$ the canonical injection, and let suppose that there exists a retraction $r \colon E \to E_x$, i.e. $r◦j=Id_{E_x}$. Can we conclude that $\xi$ is trivial ?
I found this proposition somewhere in a book or a pdf, without proof, but can't remember where. I thought I had a proof but it is plain wrong.
The converse is indeed true (i.e. if $\xi$ is trivial, $E \cong B \times F$, there is a retraction of $E$ on any fiber), and this proposition seems intuitively correct... but is it ?

REPLY [11 votes]: It was already pointed out that the statement is not true in the point-set sense. It is, however, true up to homotopy. This is a theorem of Dold and follows from his 

Partitions of unity in the theory of fibrations. Ann. Math. 78 (1963), 223-255.

The following formulations can be found in James "Topology of Stiefel manifolds" (with a couple of added assumptions, purely reformulation): 
Theorem 4.2: Suppose that $B$ is path-connected, that we have fibrations $p:E\to B$ and $p':E'\to B$ such that $E$ and $E'$ have the homotopy type of CW-complexes. Then a fiber-preserving map $f:E\to E'$ is a fiber homotopy equivalence if it induces a homotopy equivalence on fibers. 
Corollary 4.3: Suppose that $X$ is path-connected, that $p:E\to X$ is a fibration with fiber $F$ and that $E$ and $X\times F$ have the homotopy type of CW-complexes. If there exists a homotopy-retraction $E\to F$ then $E$ is trivial in the sense of fiber homotopy theory.