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 ?


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 $U\times V_x\to V_x$.

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  • $\begingroup$ Thx for this counter-example. But, minor corrections, I think the retraction should be $r : V \to \{x\} \times V_x$ and defined for $e \in \pi^{-1}(U)$ by $r(e) = f(\pi(e))\,T^{-1}\circ \rho \circ T(e)$ with $\rho : U \times V_x \to \{x\} \times V_x$, $(x',v) \mapsto (x,v)$. $\endgroup$ – ychemama Sep 23 '17 at 13:30
  • $\begingroup$ Correction to my correction, the retraction should be $r : V \to \pi^{-1}(x)$, sorry $\endgroup$ – ychemama Sep 23 '17 at 14:11

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.

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  • $\begingroup$ I became confused concerning what is the relationship of this with the answer to another question by the OP, could you comment on that? $\endgroup$ – მამუკა ჯიბლაძე Sep 24 '17 at 6:15
  • $\begingroup$ @მამუკაჯიბლაძე: I don't quite see a relationship. The answer to another question is about point-set questions again, homeomorphisms and so on. Dold's theorem says that a map between fiber bundles which induces equivalences on the fibers would be a fiber homotopy equivalence, but that would not tell us about invertibility, only homotopy invertibility. $\endgroup$ – Matthias Wendt Sep 25 '17 at 8:14
  • $\begingroup$ I see, thanks. In that answer the counterexample was based on a space $X$ with the group $G$ of self-homeomorphisms not a topological group, since the inverse map is not continuous. Then, the "bad" bundle map was the (only) map from the action $G\times X\to X$ to the projection $G\times X\to X$, given by $(g,x)\mapsto(g,gx)$. Presumably there still exists some homotopy inverse map which is continuous? Or even that is not necessary for Dold's theorem? $\endgroup$ – მამუკა ჯიბლაძე Sep 25 '17 at 16:15
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    $\begingroup$ @მამუკაჯიბლაძე: Ok, I think I understand your point now. It's not clear to me at the moment if $X$ and $G$ in the example have the homotopy type of CW-complexes (so that Dold's theorem could be applied). The homeomorphism group of a finite CW-complex is a topological group. $\endgroup$ – Matthias Wendt Sep 26 '17 at 8:07
  • $\begingroup$ Yes I did not notice this - in that answer I linked it is explained that there is no locally connected counterexample, so... $\endgroup$ – მამუკა ჯიბლაძე Sep 26 '17 at 10:16

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