Suppose I have two smooth fibre bundles $E$ and $E'$ over a base space $B$ and a smooth bundle map $H : E \to E'$ such that for each $b \in B$ the restriction of $H$ to $E_b$ is diffeomorphism $H_b \colon E_b \to E_b'$ on the fibres. I would like to extend this to obtain a bundle isomorphism $H : E \to E'$. My idea is to show that this family varies smoothly over $B$, but how would one do that? I'm not sure what's the precise definition of a family of diffeomorphism to vary smoothly and can't quite find a standard definition either. Or is there some other approach?
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Without more assumptions it’s false. You should edit the question accordingly to assume that the glued map is smooth, then you can ask whether it is a diffeomorphism. – peek-a-boo Mar 02 '25 at 19:39
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Okay, I've included the precise statement – A.Z Mar 02 '25 at 19:43
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2I think this result is what you want. – peek-a-boo Mar 02 '25 at 19:45
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@peek-a-boo That question is phrased in terms of vector bundles. This question is about smooth fiber bundles. The argument generalizes verbatim, just pointing this out for clarity. – Thorgott Mar 02 '25 at 20:15
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So is the idea to use the inverse function theorem? If so, then I suppose this boils down to using my hypothesis some how to conclude that the differential $d H_b$ is bijective for each $b \in B$. I am trying to solve this problem on my own without referring to an exact solution, so hints are appreciated. – A.Z Mar 02 '25 at 20:26
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1@Thorgott yes, but the theorem I stated and proved is about fiber bundles :) – peek-a-boo Mar 02 '25 at 21:19
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1@A.Z roughly speaking, along the fiber direction you have a linear isomorphism, and along the base direction you have the identity, so $dH_b$ is pretty much block triangular. This is an easy argument so there’s a fine line between a hint and a solution, so I suggest just reading what I wrote there. – peek-a-boo Mar 02 '25 at 21:21