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Hello,

I suspect the following is true and easy but I am unable to prove. Suppose (E, B, π, F) is a fiber bundle, where E,B are compact and F is finite, prove that E is a trivial fiber bundle. Any help will be be greatly appreciated! If needed it is OK to assume E,B are locally path connected.

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 what is your definition of a fiber bundle? If it's the usual one, it encompasses covering spaces, and they can very well have finite fiber and be non trivial. – Olivier Bégassat May 16 2011 at 0:59
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 The level of this question is really borderline. Perhaps you should ask at math.stackexchange.com – S. Carnahan♦ May 16 2011 at 3:57

closed as too localized by Ryan Budney, Andrew Stacey, Kevin Lin, S. Carnahan♦ May 16 2011 at 8:22

1 Answer

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This shouldn't be true unless you're using a different definition of fiber bundle than I would expect.

Consider the map $S^1 \to S^1$ which maps $z \mapsto z^2$. This is a fibre bundle whose fiber is a two-point set, but the base and total space are connected, and so the bundle is not trivial.

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