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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!

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    $\begingroup$ Your title says the reverse of your actual question. $\endgroup$ – Andreas Blass Dec 9 '13 at 0:48
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    $\begingroup$ I don't know what you're allowed to use but a nice fact about the Euler characteristic is that if $E \rightarrow B$ is an $n$-sheeted covering, then the Euler characteristic of $E$ is $n$ times the Euler characteristic of $B$. If $E$ is orientable, the Euler characteristic also equals $2-2g$, if it's non-orientable the Euler characteristic is $2-g$ where $g$ is the genus. So... consider the double cover. :) $\endgroup$ – Dylan Wilson Dec 9 '13 at 4:26
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    $\begingroup$ @DylanWilson You comment seems to be detailed enough to be an answer. Have you considered giving an answer for this question? $\endgroup$ – Dan Rust Dec 16 '13 at 7:26
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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

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