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I am looking for a counterexample to the following claim:

Let $p: E \rightarrow B$ be a fiber bundle, then $p$ is an open map.

This is true when $E \simeq F \times B$ where $F$ is the fiber and $p$ is the projection onto the second coordinate. It is also true when $p$ is a covering map. However, I do not believe it is true without any further assumptions.

A proof of the claim for covering spaces can be found here. It explicitely uses the fact that for some $x \in B$ there is a neighborhood $V$ s.t. $p^{-1}(V)$ is a disjoint union of open sets, each of which homeo to $V$. So I doubt this can be adapted for the general case of a fibre product.

It also does not help that most examples of fiber bundles I work with a covering maps, so there may very well be a very simple example that I just don't see.

Thank you in advance for your help.

G. Chiusole
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    I actually think the statement is true, see https://math.stackexchange.com/questions/1307099/locally-open-maps-are-open-any-reference-and-or-applications – Noel Lundström Apr 08 '20 at 15:11
  • @MoisheKohan It actually does, thank you very much. I thought I had searched the site thoroughly. – G. Chiusole Apr 08 '20 at 18:15
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    @G.Chiusole: Not a problem, I had a similar experience myself: Only after posting a question I found its duplicate by looking at "related" links on the right. – Moishe Kohan Apr 08 '20 at 18:17

1 Answers1

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The claim that fiber bundles are open maps is actually true.

Call $f:X \rightarrow Y$ a locally open map if there exists an open covering $\{U_i \}$ of $X$ such that $f \mid_{U_i}: U_i \rightarrow Y$ is an open map.

Then if $U \subset X$ is open $f(U) = f(\bigcup_i (U_i \cap U)) = \bigcup_i f(U_i \cap U)$ and then obviously all sets $f(U_i \cap U)$ are open and thus $f(U)$ is open. Thus a locally open map is open.

A fiber bundle $\pi: E \rightarrow B$ is locally open since there exists a cover of $B$, $\{ V_i\}$, such that $\pi^{-1}(V_i) \approx V_i \times F$ and such that $\pi$ is the projection onto $V_i$ under this homeomorphism. Thus since the map $V_i \times F \rightarrow V_i$ is open and we take $U_i = \pi^{-1}(V_i)$ to be a cover of $E$ we see that $\pi$ is locally open with respect to $\{U_i \}$ and thus open.

Noel Lundström
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