Covering space of a non-orientable surface I have the following problem:
Find the 2-sheeted (orientable) cover of the non-orientable surface of genus g. 
The cases $g=1,2$ are well-known, we have that the cover of $\mathbb{R}P^2$  is $S^2$
and the Klein bottle is covered by the torus. 
My intuition is that the answer is going to be the orientable surface of genus $g-1$ or the $g-1$ torus, and I tried to work with the representation of the non-orientable surface as a  $2g$-gon determined by the word $a_1a_1a_2a_2\dots a_g a_g$... where $a_i$ are the vertices.
Any hints?
 A: There is always a orientation covering(two-sheeted) of non-orientable surface. But I can only give the construction. The construction is as following: Denote OM to be the n-form bundle of the manifold M where n denotes the dimension of M. Now you can check that the group positive reals can act at this manifold. After quotienting this action, you get the covering. And as n-form is one dimensional, so then you got two sheeted covering.
A: Since Euler characteristic is multiplicative with respect to (finite sheeted) coverings, we see the orientable double cover must have twice the Euler characteristic.  This already pins down the homeomorphism type of the cover.
To see it more explicitly, consider the usual embedding of an orientable surface into $\mathbb{R}^3$ (as a long chain-like shape).  This embedding can be translated/rotated around so that the reflections in all 3 coordinate planes maps the surface to itself.
With this embedding, there is a $\mathbb{Z}/2$ action on $\mathbb{R}^3$ given by the antipodal map $x \mapsto -x$.  This action preserves the embedded surface and acts freely on it so the quotient is a manifold.  Since the antipodal map is orientation reversing, the quotient is nonorientable.
