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I am working on a problem (problem 10.19 (d)) in John M. Lee's Introduction to Smooth Manifold.

Assume that $\pi$: $E$ $\rightarrow$ $M$ is a fiber bundle with model fiber $F$, I need to prove if $E$ is compact, then so are $M$ and $F$. Clearly, $M$ is compact and if we assume that $M$ is Hausdorff, then it follows easily from part (c) of this problem.
($\pi$: $E$ $\rightarrow$ $M$ is proper iff $F$ is compact.)

(Edited: I think it suffices to assume that $M$ is $T_1$, i.e., singletons are closed, then it follows from the continuity that the fibers are compact.)

However, I have no idea how to proceed in the general case. Any hints are appreciated.

Yunmath
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1 Answers1

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Since the author chimed in in the comments, let me record the full answer here.

This is still true without separation assumptions on the base. Indeed, by compactness of $M$ there exists a finite cover $\{U_i \subseteq M\}_{i \in I}$ over which $E$ is trivialised. Without loss of generality we may assume that no strict subset of $\{U_i \subseteq M\}$ covers $M$. Then $$Z = M \setminus \bigcup_{j \neq i} U_j \subseteq U_i$$ is a nonempty closed subset over which $E$ is trivial, i.e. $E|_Z \cong F \times Z$. Since $E|_Z \subseteq E$ is closed, it is compact, hence so is $F$ since it is the image of the first projection $F \times Z \to F$. $\square$

Remy
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  • Thank you very much! – Yunmath Feb 04 '21 at 01:03
  • Sorry, why the WLOG part is true? – C.F.G Feb 04 '21 at 08:19
  • @C.F.G since $I$ is finite, there is an inclusionwise minimal subset $J \subseteq I$ such that the $U_j$ for $j \in J$ cover. This is only used to show that $Z$ is nonempty, for which there might be other methods. – Remy Feb 04 '21 at 15:16
  • @Remy: You meant "that no strict subset of $U_i$ (i.e. $V_i\subset U_i$) covers $M$" Or "for at least a $k$, $U_i$'s do not cover $M$ for $i\neq k$"? – C.F.G Feb 04 '21 at 18:10
  • @C.F.G apparently my previous wording could be interpreted in two ways, so I have edited for clarification. – Remy Feb 04 '21 at 18:58
  • Thanks. Sorry again, I don't want to annoy you, but why $E|_Z$ is closed? – C.F.G Feb 04 '21 at 19:17
  • @C.F.G $\pi$ is continuous – Remy Feb 04 '21 at 19:23
  • So your argument works for every sufficiently small closed subset of $M$ and not just $Z$? – C.F.G Feb 04 '21 at 19:36
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    That's the point: you don't need a closed point; any closed subset over which $E$ is trivialised will do. Just construct one. – Remy Feb 04 '21 at 19:40
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    Nice proof. It might even be what I had in mind when I wrote the problem; but since I didn't see it at first, I've decided to add a hint to my correction list. Thanks for chiming in. – Jack Lee Feb 06 '21 at 23:53