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Let $\pi : E \to B$ be a Serre fibration over a CW complex, with circle fibers.

In the orientable case, it is easy to see that $\pi$ is fiber homotopy equivalent to a principal $SO(2)$--bundle.

Given the additional requirement that every fiber of $\pi$ be homeomorphic to a circle, is $\pi$ necessarily a locally trivial fiber bundle?

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  • $\begingroup$ Could you recall the argument for the orientable case? $\endgroup$ Aug 20, 2015 at 14:36
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    $\begingroup$ The paper 131.220.77.52/lueck/data/… could be relevant here. The authors define primary and secondary obstructions for a fibration to be homotopy equivalent to a fibre bundle. These obstructions vanish here, I think, as $Wh(\mathbb{Z})=0$ and any homotopy equivalence of a circle is homotopic to a homeomorphism. But this doesn't quite answer your question (and anyway might be overkill). $\endgroup$
    – Mark Grant
    Aug 20, 2015 at 16:07
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    $\begingroup$ When I say easy, I think I mean easy to someone who works with this stuff, so not me. But I think it's a little more straightforward than the reference Mark gives. The essential point is that the fiber is $K(\mathbf{Z},1)$. Lemma 3.4.2 in More Concise Algebraic Topology tells us that the fibration comes from a map $B \to K(\mathbf{Z},2)$ if and only if $\pi_1(B)$ acts trivially on the fiber. But the orientability should give triviality of the action, and $K(\mathbf{Z},2)$ is the classifying space for principal circle bundles. $\endgroup$ Aug 21, 2015 at 9:11
  • $\begingroup$ I think any argument in the orientable case should be adaptable to the non-orientable one. Let's discuss the orientable case. Why is there a classifying space for Serre fibrations? I know there is a classifying space Hurewicz fibrations over CW complexes (by Stasheff's theorem). $\endgroup$ Aug 21, 2015 at 14:54
  • $\begingroup$ I definitely agree that it should be enough to figure this out in the orientable case. I don't know whether there is a classifying space for Serre fibrations: the result I'm using here is just for the case where the fiber is a $K(G,n)$ space. $\endgroup$ Aug 21, 2015 at 22:19

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