# Möbius bundle no global trivialization

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?