Given a smooth fiber bundle $X \to S^1,$ such that the fiber, $F$, is homotopic to $S^2 \vee S^2.$ Is it true that this is always a trivial fiber bundle?
In general, how to check a fiber bundle is trivial?
Thanks, in advance!!
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5$\begingroup$ No, the holonomy can switch the two wedge summands. In general you have to decide what category you want to live in, for the purposes of saying if the bundle is trivial. But switching the wedge summands is non-trivial even in the fibrewise homotopy category, as the holonomy is non-trivial on $\pi_2$ of the fiber. In general it can be a fairly subtle issue, but the homotopy long exact sequence of the bundle is always a good place to start looking. $\endgroup$– Ryan BudneyCommented Sep 15, 2021 at 23:36
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$\begingroup$ In this case I suppose you want to think of $\pi_2$ of the total space as a module over $\pi_1$, i.e. the Whitehead bracket structure on the "switch the summands" bundle is different than on $S^1 \times (S^2 \vee S^2)$. $\endgroup$– Ryan BudneyCommented Sep 15, 2021 at 23:42
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$\begingroup$ The holonomy could also turn the 2-spheres inside out, an orientation reversing homeomorphism. $\endgroup$– Ben McKayCommented Sep 16, 2021 at 6:29
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1$\begingroup$ For "reasonable" spaces fiber bundles with fiber $F$ and base $B$ are classified by homotopy classes of maps $B→B\operatorname{Homeo}(F)$ where $\operatorname{Homeo}(F)$ is the topological group of self-homeomorphisms of $F$. If $B=S^1$, asking for triviality is more or less equivalent to asking that $π_0\operatorname{Homeo}(F)=0$ (i.e. that the mapping class group is trivial), which is not the case for $F=S^2∨S^2$ as people have noticed in the comments above (e.g. because there's the "switch" homeomorphism) $\endgroup$– Denis NardinCommented Sep 16, 2021 at 7:38
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