# Trivialisation of Möbius strip

I've just started studying Advanced Geometry and I'm in trouble with a (stupid) exercise. It's about finding a trivialisation of the Möbius strip (I'll refer to it as $E$) viewed as a fibre bundle over $S^1$ with fibre any open interval $F\subset\mathbb{R}$.

In order to do that, I have to find a projection $\pi:E\rightarrow S^1$ such that there exists an open cover $\{U_\alpha\}$ of the base space $S^1$ and homeomorphisms $\phi_\alpha:\pi^{-1}(U_\alpha)\rightarrow U_\alpha \times F$ (the trivialisation). First of all, I can't really see the projection, so I don't find the trivialisation of the strip. Should I do the reverse? I mean, looking for homeomorphisms from $\pi^{-1}(U_\alpha)$ to $U_\alpha\times F$ and then try to write the projection? Can someone help me, please?

P.S. My English is not perfect because I'm Italian, so I'm sorry for that! If you see dreadful grammar mistakes, please tell me!

Thanks

## 1 Answer

Think of the Moebius strip $E$ as the quotient $[0,1]^2/\sim$ where $(0,a)\sim(1,1-a)$. Then $\pi:E\to S^1\subseteq \mathbb C$ with $\overline{(x,y)}\mapsto 2\pi x$ gives a well defined fibre bundle. Any open covering of $S^1$ into two open arcs gives you a trivialisation of this bundle.

• Thank you for the answer. You've formally done what I was thinking, but I wasn't able to express my thoughts mathematically! Following your hint, I find that, if I have an open cover of $S^1$ e.g. gived by $U_1$ and $U_2$, then $\pi^{-1}(U_i)=U_i\times[0,1]$, and so the omeomorphism $\phi_i$ that gives the trivialisation is the identity map. Is it correct? – batman Oct 29 '13 at 14:40
• The preimage $\pi^{-1}(U_i)$ lives in the quotient space $[0,1]^2/\sim$ and at least one of these will contain the glued edges, so you have to be a little more careful. – Christoph Oct 29 '13 at 21:44