Does every nonorientable surface $S$ have a degree $2$ orientable cover?

Question

As the question title suggests, does every nonorientable surface $S$ have a degree $2$ orientable cover?

Answer 1

I think with this question it depends a lot in which setting you ask it. If you think of surfaces as topological $2$-manifolds, then it is already complicated to define orientability, so it will be hard to avoid a bit of "technology" (like Chapter 3 of Hatcher).

If you are willing to work with smooth surfaces then the concept of orientability becomes much easier and there is a more direct construction. Basically, you choose an atlas $\{(U_i,u_i):i\in I\}$ such that each non-empty intersection $U_i\cap U_j$ is connected. This means that for any such intersection with change of charts $\phi_{ij}$, the expression $\det(D\phi_{ij})$ has constant sign, say $\alpha_{ij}=\pm 1$. Then you take the set of all triples $(i,x,\epsilon)$ where $i\in I$, $x\in U_i$ and $\epsilon=\pm1$. On this you define an equivalence relation by identifying $(i,x,\epsilon)$ with $(j,y,\mu)$ if $x=y$ (and hence lies in $U_i\cap U_j$) and $\epsilon=\alpha_{ij}\mu$. This is immediately seen to be an equivalence relation and you denote by $\tilde S$ the set of equivalence classes. Then from the construction, you get an obvious projection $\tilde S\to S$ and an oriented atlas on $\tilde S$ making this projection a smooth local diffeomorphism and hence a covering. It is easy to see that for connected $S$, $\tilde S$ is connected if and only if $S$ is non-orientable. (Of course, all that generalizes to arbitrary dimensions.)