# Moebius strip and orientability

Lemma: Let $M=U\cup V$ be a smooth manifold with $U$ and $V$ open, connected, oriented and such that $U\cap V$ has two connected components $W_1$ and $W_2$. Then $M$ is orientable $\Leftrightarrow$ for any 2 chosen orientations for $U$ and $V$, either they coincide in $W_1, W_2$ simultaneously or differ in $W_1, W_2$ simultaneously.

Exercise: Use the lemma to prove that the Moebius strip is not orientable.

I know how to prove the lemma, but I'm having trouble on applying it to the case of the Moebius strip. How am I supposed to define $U$ and $V$?

Thanks!

## 1 Answer

Think of a cylinder $S^{1} \times I$ as a product of a circle and an interval, and let $u$ and $v$ be distinct points of $S^{1}$; let $U$ be the complement of $\{u\} \times I$ and $V$ the complement of $\{v\} \times I$.

Analogously, the Möbius strip may be viewed as a bundle of intervals over a circle. Define $U$ and $V$ in similar fashion, taking each to be the complement of an interval.

• Making this CW because 1. I don't think it deserves a bounty, and 2. It may be appropriate for someone to add details (though this description is intended to suffice for OP's purposes). – Andrew D. Hwang Oct 20 '16 at 3:20