How to construct a two-sheeted cover of a non-orientable surface?

Question

Let $S$ be a non-orientable surface. Then there exists a two-sheeted covering map $p:S'\to S$ with $S'$ an orientable surface. I want to know how to construct $p$.

I know that $\mathbb{R}P^2$ is 2-sheeted covered by $S^2$ and $K$ is 2-sheeted covered by the torus. This makes me think that I must double the Euler characteristic for such a covering space $S'$. I want to use a similar construction to show the desired result. So I wanted to started as follows

Every non-orientable surface is homeomorphic to $N_g$ for some $g\geq 1$. I wanted to somehow have a properly discontinuous action of $\mathbb{Z}/2\mathbb{Z}$ on a orientable surface $M_{g'}$ such that $M_{g'}/(\mathbb{Z}/2\mathbb{Z})$ is homeomorphic to $N_g$.

Answer 1

When talking about orientability of a connected $n$-dimensional manifold $M$, there is a very natural cover that appears, and it's called the orientation double cover of $M$.

As its name suggests, it is a double-sheeted covering of $M$. What makes it interesting is that an orientation of $M$ is equivalent to a section of that cover : in particular the cover is connected if and only if $M$ is non orientable (if it is connected then there can be no section, hence no orientation; if $M$ is orientable, there is a section, and so it can't be connected.)

Thus when $M$ is a connected non-orientable manifold, this automatically gives you a connected double-sheeted covering of $M$.

Now to describe this covering. Recall that a local orienration of $M$ at $x\in M$ is a choice of a generator $\mu_x \in H_n(M, M\setminus \{x\})$ (in what follows I will write $H_n(M\mid x)$ for this homology group). But recall that $H_n(M\mid x) \simeq \mathbb{Z}$ so there are only two choices for such a generator.

A global orientation of $M$ is a family of local orientations that satisfies a certain compatibility condition. We can already see our covering popping up : a section should be an orientation, thus a choice of local orientations, with compatibility conditions. This means that the fiber over $x$ should be the local orientations at $x$ and the topology of the space should be such that compatibility of local orientations coincides with continuity of the section.

So let $\widetilde{M} := \displaystyle\coprod_{x\in M} H_n(M\mid x)^\times$ ($H_n(M\mid x)^\times$ denotes the set of generators), with the obvious projection map to $M$. We now have to figure out a topology for $\widetilde{M}$.

For any open set $U\subset M$ of the form $\phi^{-1}(\mathrm{Int}(D^n))$ with a homeomorphism $\phi : V\to \mathbb{R}^n$, for $V$ an open set, we have $H_n(M, M\setminus U) \simeq \mathbb{Z}$ as well and so we can talk about $H_n(M, M\setminus U)^\times$ too. Then for any $z\in H_n(M, M\setminus U)^\times$, put $(U,z) := \{z_x \mid x\in U\}$ where $z_x$ is the image of $z$ under the restriction map $H_n(M,M\setminus U) \to H_n(M\mid x)$ (which is an isomorphism of groups, hence it sends a generator to a generator)

One easily checks that the $(U,z)$ form a basis for a topology if we make $U$ and $z$ move as described above.

Now it's mostly a matter of unwrapping the definitions but you can see quite easily that the projection $\widetilde{M}\to M$ is a cover, that it is double-sheeted, and that a section is precisely an orientation. You can then conclude as I did earlier.

Answer 2

Here is a hands-on construction of the orientation covering for compact connected surfaces (without boundary). Let $N_p$ be a genus $p$ nonorientable surface, i.e. the connected sum of $p$ projective planes. I will describe the orientation covering when $p$ is odd and leave you the even case to work out by analogy.

If $p$ is odd then $N_p$ is homeomorphic to the connected sum of $M_g$ and $RP^2$, where $M_g$ is the compact oriented surface of genus $g$ satisfying $p=2g+1$.

Another way to describe this is to say that $N_p$ is obtained from the surface $M_{g,1}$ (compact oriented of genus $g$ with one boundary component) by "adding a cross-cup", i.e. identifying the boundary circle with itself via the antipodal map $\tau: S^1\to S^1$. We will need one more observation: The involution $\tau$ extends to an involution $\tau: M_{g,1}\to M_{g,1}$. Let me know if you do not know how to find such an extension. Now, given this, let $M'_{g,1}$ be another copy of $M_{g,1}$, let $\sigma: M_{g,1}\to M'_{g,1}$ denote the identification map. Formally speaking, we take $M_{g,1}\times \{0,1\}$ and $$ \sigma(x,i)=(x, 1-i), ~i\in \{0,1\}, ~x\in M_{g,1}.$$ Then $M_{g,1}= M_{g,1}\times \{0\}$, $M'_{g,1}= M_{g,1}\times \{1\}$.

I will glue the surfaces $M_{g,1}$ and $M'_{g,1}$ along their boundary $S^1$ using the map $\sigma$. The result is the oriented surface $M_{2g}$ containing a copy of $S^1$ which is a topological circle I will denote $C$. Define an involution $\phi: M_{2g}\to M_{2g}$ as the composition of $\sigma$ (fixing the common circle $C$ pointwise and swapping $M_{g,1}$ with $M'_{g,1}$) and the involution $\tau$. I will leave you to check that $\phi$ is a fixed-point free involution; its restriction to $C$ is the antipodal map $\tau$. From this, you see that the quotient $M_{2g}/\phi$ is the original surface $N_p$. This is your orientation covering map $M_{2g}\to N_p$.