# Nontriviality of the Möbius bundle without orientability

1. Sep 24, 2008

### quasar987

There's this problem in Lee's Intro to smooth manifold (5-2) that asks to prove that the Möbius bundle is a line bundle over the circle and that it is non trivial.

The Möbius bundle is [0,1]xR/~ 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]xR/~ 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...

2. Sep 25, 2008

### tim_lou

I'm not sure if the following argument works:

on the mobius strip, you can have a loop going very close to the boundary such that it goes in a circle twice without intersecting (that'll be 2 in the fundamental group). If the mobius strip is indeed homeomorphic to the cylinder, you can map that loop on the mobius strip into a loop that goes around a cylinder twice (the fundamental group must be isomorphic), but then it must intersect itself, contradiction, a homeomorphic map must be bijective. (I think you can make this idea precise).

3. Sep 26, 2008

### quasar987

Hey, I think this works! :)