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 $f^{-1}$ ?



Are topological fiber bundles on the same base with homeomorphic fibers isomorphic?
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](https://ncatlab.org/nlab/show/exponential+law+for+spaces). 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](https://en.wikipedia.org/wiki/Homeomorphism_group)). Moreover, there exist exponentiable spaces with this property, namely locally compact, Hausdorff, but not locally connected spaces ([see the example here](http://www.cs.vu.nl/~dijkstra/research/papers/2005compactopen.pdf)). 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.


