3

The universal property says the following: for any pair of maps $i: Z \to X$ and $j: Z \to E$ fitting into the following commuting diagram where $p: E \to B$ is a fiber bundle (by the way this is true for any Serre fibration but I am applying it to a fiber bundle) $\require{AMScd}$ \begin{CD} Z @>j>> E \\ @ViVV @VVpV \\ X @>f>> B \end{CD} there exists a unique map $\psi: Z \to X \times_B E$ which factors through $i, j$ as follows (sorry I don't know how to draw the diagonal arrow) $\require{AMScd}$ \begin{CD} Z @>j>> E \\ @ViVV @AAA \\ X @<<< X \times_B E \end{CD} where $X \times_B E \to X$ is defined by $(x,e) \mapsto x$ and $X \times_B E \to E$ is defined by $(x,e) \mapsto e$.

Now suppose $i: Z \to X$ is also a fiber bundle over $X$ thus $\psi: Z \to X \times_B E$ defines a bundle map. $\require{AMScd}$ \begin{CD} Z @>\psi>> X \times_B E \\ @ViVV @VVV \\ X @= X \end{CD} My question is: when is $\psi$ an isomorphism of bundles? i.e. when does $\psi$ have an inverse?

PhysicsMath
  • 1,151
  • I'm not quite sure what you're looking for here. Are you working in the category of topological spaces, and just asking for a concrete description of what a pullback in topological spaces is? (Also, what do you mean by "fibration" exactly? Everything you say makes perfect sense for arbitrary maps...) – Eric Wofsey May 22 '16 at 01:39
  • @Eric: sorry for the confusion! What I mean is fiber bundle. I have revised the question statement to make it clear. – PhysicsMath May 22 '16 at 01:48
  • I think it is when $Z$ also has the universal property, with maps $i, j$ – Henry May 22 '16 at 02:03
  • It's still unclear to me what sort of answer would be helpful to you. Do you know how to concretely describe $X\times_B E$ as a subspace of $X\times E$? The answer to your question is, "when $\psi$ is a homeomorphism", and you can write out exactly what that means concretely in terms of the definition of the underlying set and topology of $X\times_B E$, but that may or may not be illuminating... – Eric Wofsey May 22 '16 at 02:04
  • @Eric: Do you mean the following Eric? If $\psi$ induces the identity map on the base space and a homeomorphism on the fibers then $\psi$ defines a bundle isomorphism? Actually I think this is the answer but I have difficulty in seeing $\psi$ induces a homeomorphism on the fibers. – PhysicsMath May 22 '16 at 02:25
  • I don't think it's true that if $\psi$ induces a homeomorphism on the fibers then it automatically induces a homeomorphism on the total space. That would be saying that for any continuous family of homeomorphisms, the inverses are also a continuous family, which I don't think is true in general. – Eric Wofsey May 22 '16 at 02:56
  • @Eric: I see. Sorry for the confusing. I think I should ask more specific question than this. This question is probably too general to reveal my confusion. – PhysicsMath May 22 '16 at 05:09

0 Answers0