Orientable Surface Covers Non-Orientable Surface I need to describe how a 4-genus orientable surface double covers a genus 5-non-orientable surface. I know that in general every non-orientable compact surface of genus $n\geq 1$ has a two sheeted covering by an orientable one of genus n-1. I tried to use the polygonal representation of these surfaces and try to get one from the other by cutting along some side as is done for the torus as a double cover of the Klein bottle. But It got really confused. 
Thanks!
 A: Another way to proceed, which is a bit different from polygonal presentations, is to think about connect sums.  If $S$ and $T$ are surfaces, then we can form $S \# T$, a new surface, by cutting a disk out of each, and gluing together the resulting circle boundaries.  For example, if $S$ and $T$ are both copies of $T^2$ (torus) then $S \# T$ is a copy of the genus two surface.  Drawing some pictures will help here.
Let's use $S_{g,n,c}$ to denote the connected, compact surface obtained by taking the connect sum of a two-sphere with $g$ copies of $T^2$ (torus), $n$ copies of $D^2$ (disk), and $c$ copies of $P^2$ (projective plane, aka "cross-cap").  One standard notation is $N_c = S_{0,0,c}$ for the non-orientable surface of "genus" $c$. 
Another way to obtain $N_c$ is as follows.  Take the two-sphere $S_{0,0,0}$.  Cut out $c$ disjoint closed disks to get $S_{0,c,0}$.  Identify boundary component with the circle $S^1$.  Finally quotient each boundary component by the antipodal map $x \mapsto -x$.  This gives $N_c$.  (More pictures!)  Thus the orientable double-cover of $N_c$ is obtained by gluing two copies of $S_{0,c,0}$ along their boundaries, to get $S_{c-1,0,0}$.  (Yet more pictures!)
Here are some closely related questions:
Showing the Sum of $n-1$ Tori is a Double Cover of the Sum of $n$ Copies of $\mathbb{RP}^2$
Covering space of a non-orientable surface
Actually, my answer above is a more abstract version of the second half of 
https://math.stackexchange.com/a/279249/1307
