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 ?



Is a topological fiber-bundle, whose total space admits a retraction onto a fiber, trivial?
The statement is not true. Let $\pi:V\to M$ be a vector bundle over a manifold which is non-trivial as a fiber bundle. Let $U$ be an open neighborhood of $M$ over which $V$ is trivial, fix $x\in U$, and pick a local trivialization
$$
T:\pi^{-1}(U)\xrightarrow{\sim}U\times V\_x.
$$
Let $f:M\to\mathbb{R}$ be a continuous function with support contained in $U$ such that $f(x)=1$. Then we can build a retraction $V\to V\_x$ by
$$
y\mapsto \begin{cases} f(\pi(y))\cdot p\_2(T(y))&:y\in\pi^{-1}(U),\\0&:\text{otherwise}.
\end{cases}
$$
Here $p\_2$ is the projection $U\times V\_x\to V\_x$.


