First I apologize because this is not a research question, but I can't get any answer on MathStackExchange...

Let $\pi \colon E \to B$ and $\pi' \colon E' \to B$ two topological fiber bundles on the same base $B$. A $B$-morphism $f \colon E \to E'$ is an isomorphism of fiber bundles iff for all $b \in B$ the induced morphism between fibers $f_{E_b} \colon E_b \to E'_b$ is a homeomorphism.

Is this true ? If not in general, under which additional conditions on the spaces does it become true ?

The direct implication is obvious, but I have huge doubts on the reverse. It is clear that $f$ is a bijection, continuous by definition, so we just have to show that $f^{-1}$ is also continuous. Continuity being a local property we can suppose that both fibrations are trivial, i.e. $f \colon B \times F \to B \times F'$ is of the form $(b,y) \mapsto (b,f_b(y))$ where $f_b \colon F \to F'$ is a homeomorphism for all $b \in B$.

Define $f^{-1} \colon B \times F' \to B \times F$ by $(b,y') \mapsto (b,(f_b)^{-1}(y'))$. Since $f$ is continuous, the coordinate map $(b,y) \mapsto f_b(y)$ is also continuous, and if we can conclude from this that the coordinate map $(b,y') \mapsto (f_b)^{-1}(y')$ is also continuous, $f^{-1}$ is continuous and we are done.

But why should this last step be correct ?

It is correct for vector bundles (F and F' being vector spaces of the same dimension) because in this case $f_b \in L(F,F')$ forall $b \in B$, and $f_b$ can be represented by a matrix whose coefficients are continuous fonctions of $b$. Then the polynomial formula for the inverse matrix shows that the coefficients of $(f_b)^{-1}$ are also continuous.

It is also trivially true with principal bundles because there any $B$-morphism is an isomorphism of principal bundles.

But quid in the general topological case ? Forall $b \in B$, $f_b \in Homeo(F,F')$. With the compact-open topology on this set, can we say that the application $b \mapsto f_b$ is continuous (I hope so) ? And how can we say anything about the map $b \mapsto (f_b)^{-1}$ and $f^{-1}$ ?


If $F$ is locally connected, locally compact and Hausdorff (or alternatively compact Hausdorff), then the inverse function from $Homeo(F, F)$ to itself is continuous. Moreover, such a space $F$ is exponentiable. This means, in particular, that a map $B\to Homeo(F, F)$ is continuous if and only if the adjoint map $B\times F\to F$ is continuous. So under this assumption the answer to your question is yes.

Presumably other (weaker?) sufficient hypotheses are available. But in general, the inverse function from $Homeo(F, F)$ to itself is not continuous (see here). Moreover, there exist exponentiable spaces with this property, namely locally compact, Hausdorff, but not locally connected spaces (see the example here). So in general the answer is no: Let $F$ be an exponentiable space such that the inverse is not continuous on $Homeo(F, F)$. Then the function from $Homeo(F, F)\times F$ to itself given by $(f, x)\mapsto (f, fx)$ provides a counterexample.

  • $\begingroup$ In principle it could happen that the inverse is not continuous on $\operatorname{{\textit{Homeo}}}(F)$ but the map $\operatorname{{\textit{Homeo}}}(F)\times F\to F$ with $(f,x)\mapsto f^{-1}x$ is still continuous. Is it clear that this cannot be? $\endgroup$ Sep 13 '17 at 10:00
  • 1
    $\begingroup$ I see, yes, you are right - so in any case you do have a counterexample, and this is what counts for the OQ... $\endgroup$ Sep 13 '17 at 10:07

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