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I there! I am trying to write down why the Möbius bundle has no global trivialization. I just read that there is none but I want to see a written prove for this. I am not even sure which definition of the Möbius strip would be the best.

First let me define this relation: $(a,b) \tilde{} (c,d)$ if $b = 1-d$ and $a = 0, c = 1$ or $a=1, c = 0$. Now let me define the bundle: Let ($E$, $\pi$, $S^1$) be a vector bundle where $E = [0,1] \times \mathbb{R} / \tilde{}$, $S^1$ is the unit sphere and $\pi([x,y]) = e^{2\pi i x}$.

So far I have the following idea. Assume there is a global trivialization. This mean there exists a diffeomorphism $t: E \to S^1 \times \mathbb{R}$ which is equal to the infinite cylinder. This implies that $t$ is a continuous mapping between the Möbius band and the cylinder.

How can I proof that there is no such mapping?

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If the interval bundle were trivial then the boundary would be a disjoint union of a pair of circles (corresponding to the pair of endpoints of the interval) but the boundary of the Moebius bundle is a single circle ("twice as long"). Therefore the bundle is not trivial.

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  • $\begingroup$ Thanks! What you described is clear to me. I was more looking for a strict proof. $\endgroup$ – JDoe Jul 4 '16 at 20:34
  • $\begingroup$ @JDoe: As soon as you fill in the details here (which is rather straightforward), you will have a "strict proof." $\endgroup$ – Pete L. Clark Jul 4 '16 at 21:11
  • $\begingroup$ @PeteL.Clark Can you give me a hint how to do this? This is exactly the part I don't know how to start with. Should I use a parametrization of the boundaries? $\endgroup$ – JDoe Jul 5 '16 at 7:13
  • $\begingroup$ Just show that in the presentation you gave, the boundary is connected. $\endgroup$ – Mikhail Katz Jul 5 '16 at 7:17

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