9
$\begingroup$

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 ?

$\endgroup$
14
$\begingroup$

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 $U\times V_x\to V_x$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
11
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.