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quasar987

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The Möbius bundle is [0,1]x

**R**/~ where ~ identifies (0,y) with (1,-y) and the projection map p send [(x,y)] down to exp{(2pi)ix}.

So if the Möbius bundle were trivial, we would have that [0,1]x

**R**/~ is homeomorphic to

**S**^1 x

**R**. (the open Möbius strip homeomorphic to the open cylinder)

The thing is that this problem comes before the chapter on orientation. So either there is a way to prove that the open Möbius strip is not homeomorphic to the open cylinder without invoking orientations, or there is some other way to show that the Möbius fribration is not trivial...