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Suppose I have a manifold M. If E is a fiber bundle over M with non-orientable fiber F, is E necessarily trivial?

I'm imagining bundles over S

But we can continuously deform (with some map we'll call [tex]\Phi[/tex]: [0,1] --> C

(This is not true if the fibers are some vector space V, because local triviality will be violated if you try to move the fibers close to 2pi -- you would have to reverse orientation, which can't be done. In other words, fix a continuous path in V parametrized over [0,2pi) by [tex]\gamma[/tex]

I'm imagining bundles over S

^{1}. For example, if F is the mobius strip, then consider any section S of E. By local triviality, we can think of this section as a continuous function [tex]\gamma[/tex]_{S}: [0,2pi) --> F (the mobius strip), but where [tex]\gamma[/tex]_{S}(0) need not equal [tex]\gamma[/tex]_{S}(2pi).But we can continuously deform (with some map we'll call [tex]\Phi[/tex]: [0,1] --> C

^{0}(S^{1}, F) ) each such [tex]\gamma[/tex]_{S}= [tex]\Phi[/tex](0) into some other continuous function [tex]\widehat{\gamma}[/tex]_{S}= [tex]\Phi[/tex](1) such that [tex]\widehat{\gamma}[/tex]_{S}(0) = [tex]\widehat{\gamma}[/tex]_{S}(2pi) and the local triviality condition is still satisfied for [tex]\Phi[/tex](t) for all t in [0,1].(This is not true if the fibers are some vector space V, because local triviality will be violated if you try to move the fibers close to 2pi -- you would have to reverse orientation, which can't be done. In other words, fix a continuous path in V parametrized over [0,2pi) by [tex]\gamma[/tex]

_{1}. To have local triviality, we need that any other continuous path [tex]\gamma[/tex]_{2}in V over (pi,3pi) remains continuous as [tex]\Phi[/tex] messes with [tex]\gamma[/tex]_{1}. If the fibers near 2pi are considered to have opposite orientation on either side of 2pi in [tex]\gamma[/tex]_{1}but the same orientation on either side of 2pi in [tex]\gamma[/tex]_{2}, this is simply impossible.)
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