Suppose $(E,p,B;F)$ is a fiber bundle such that $E$ is homeomorphic to $B\times F$, is it true that the fiber bundle is trivial? A non connected counter example has been provided, so I'll ask for E,B and F to be connected (hopefully low dimensional) manifolds.
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asked Mar 16, 2011 at 20:36
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$\begingroup$ Never mind, this question might not be just the definition of "trivial". Are you assuming p has to be a continuous map, just a fibration? $\endgroup$– Elizabeth S. Q. GoodmanCommented Mar 16, 2011 at 20:38
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1$\begingroup$ Also, it would be nice if you could carefully define ``fiber space'' as I believe that language is a bit old. In particular, do you mean one of the definitions of a fiber bundle, or a map that has the homotopy lifting property for all maps into B--a Hurewicz fibration--or has the homotopy lifting property just for CW complexes into B--a Serre fibration? $\endgroup$– Elizabeth S. Q. GoodmanCommented Mar 16, 2011 at 20:51
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$\begingroup$ Every point in $b\in B$ should a neighborhood $V$ s.t. $p^{-1}(V)$ is homeomorphic to $V\times F$ and, viewed in that 'chart', $p$ is the projection onto the first factor. $\endgroup$– Olivier BégassatCommented Mar 16, 2011 at 21:10
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$\begingroup$ I meant 'fiber bundle' instead of 'fiber space'. $\endgroup$– Olivier BégassatCommented Mar 16, 2011 at 21:12
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$\begingroup$ I think a nontrivial $S^3$ bundle over $S^4$ with vanished Euler class will provide a counter example to this. $\endgroup$– Xiaolei WuCommented Mar 18, 2011 at 2:20
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2 Answers
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Consider the pullback $\xi$ of $TS^2$ via the projection of $S^2\times\mathbb R$ onto the first factor. The bundle $\xi$ is a nontrivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$ because its pullback under the inclusion $S^2\to S^2\times\mathbb R$ is $TS^2$, which is nontrivial. On the other hand, its total space is $\mathbb R\times TS^2$ which is diffeomorphic to $S^2\times\mathbb R^3$, which is the total space of the trivial $\mathbb R^2$-bundle over $S^2\times\mathbb R$.
answered Mar 18, 2011 at 1:21
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$\begingroup$ great, that's just what I wanted! $\endgroup$ Commented Mar 18, 2011 at 3:50
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$\begingroup$ I added some details about this construction here. $\endgroup$ Commented Apr 5, 2023 at 20:35
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No. Let $B=S^1$, $F=\mathbb Z$, $E=S^1\times\mathbb Z$, and let $p$ be a two-fold covering on each component: $p(z,n)=z^2$ where $S^1$ is regarded as the unit circle in $\mathbb C$ (for the purposes of computing $z^2$).
answered Mar 16, 2011 at 21:17
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1$\begingroup$ that's true, what if $E$ is connected? $\endgroup$ Commented Mar 16, 2011 at 21:23