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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 $mathbbRP^2$ 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 $ggeq 1$. I wanted to somehow have a properly discontinuous action of $mathbbZ/2mathbbZ$ on a orientable surface $M_g'$ such that $M_g'/(mathbbZ/2mathbbZ)$ is homeomorphic to $N_g$.

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 $tau: S^1to S^1$. We will need one more observation: The involution $tau$ extends to an involution $tau: M_g,1to M_g,1$. 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 $sigma: M_g,1to M'_g,1$ denote the identification map. Formally speaking, we take $M_g,1times 0,1$ and
$$
sigma(x,i)=(x, 1-i), ~iin 0,1, ~xin M_g,1.$$
Then $M_g,1= M_g,1times 0$, $M'_g,1= M_g,1times 1$.

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 $phi: M_2gto M_2g$ 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 $M_2gto N_p$.

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 $xin M$ is a choice of a generator $mu_x in H_n(M, Msetminus x)$ (in what follows I will write $H_n(Mmid x)$ for this homology group). But recall that $H_n(Mmid x) simeq mathbbZ$ 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 $widetildeM := displaystylecoprod_xin M H_n(Mmid x)^times$ ($H_n(Mmid x)^times$ denotes the set of generators), with the obvious projection map to $M$. We now have to figure out a topology for $widetildeM$.

For any open set $Usubset M$ of the form $phi^-1(mathrmInt(D^n))$ with a homeomorphism $phi : Vto mathbbR^n$, for $V$ an open set, we have $H_n(M, Msetminus U) simeq mathbbZ$ as well and so we can talk about $H_n(M, Msetminus U)^times$ too. Then for any $zin H_n(M, Msetminus U)^times$, put $(U,z) := z_x mid xin U$ where $z_x$ is the image of $z$ under the restriction map $H_n(M,Msetminus U) to H_n(Mmid x)$ (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 $widetildeMto M$ is a cover, that it is double-sheeted, and that a section is precisely an orientation. You can then conclude as I did earlier.