Suppose I have an oriented $F$ bundle over $B$ with total space $E$ (all of the three are closed manifolds) and i have a closed manifold $E'$ which is homotopy equivalent to $E$.Is there any condition in terms of the algebraic topology of $B,F$ or $E$ which would guarantee that $E'$ is also the total space of some fiber bundle?
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5$\begingroup$ I think you want to add more conditions, as $E'$ is trivially a fiber bundle over a point. In particular I think you want the homotopy to interact with your fiber bundle structure. $\endgroup$– Thomas RotCommented Nov 9, 2015 at 15:45
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1$\begingroup$ I guess the question is asking for fibrations over the same base space $B$. For example the exotic 7-spheres are all $S^3$-bundles over $S^4$. In general, however, this seems unplausible: there are many homotopy fibrations that can not be made fibrations and I'd guess that such examples might also be found in the manifolds category, though I don't know of an example. $\endgroup$– ThiKuCommented Nov 9, 2015 at 16:19
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7$\begingroup$ @ThiKu. Of the 28 exotic 7-spheres, only 16 are fiber bundles over $S^4$ with fiber $S^3$. I'm not sure who originally noted this, but it can be deduced from a paper of Itiro Tamura, "Remarks on differentiable structures on spheres" in J. Math. Soc. Japan. 13 (1961), 383-386. $\endgroup$– Allen HatcherCommented Nov 9, 2015 at 17:46
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$\begingroup$ Thank you for the correction. So that example may be given as an answer. $\endgroup$– ThiKuCommented Nov 9, 2015 at 22:41
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