Which surfaces does a compact orientable surface with boundary cover? Suppose I have a bounded, orientable genus 5 surface with 4 boundary circles. Is there a way to determine what surfaces it covers?
First, I know that there is a covering map from the closed orientable surface of genus 5 to the closed non-orientable surface of genus 6. This covering map "corresponds" to the quotient maps that identifies anti-podal points. With this in mind, I think the the bounded, orientable genus 5 surface with 4 boundary circles covers a bounded non-orientable surface of genus 6 with 2 boundary circles.
Am I understanding this correctly? Can I take this further? Thanks for any responses!
 A: In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.

Denote the closed orientable surface of genus $g$ with $b$ boundary components by $\Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $\chi(\Sigma_{g,b}) = 2 - 2g - b$ and $\chi(S_{g,b}) = 2 - g - b$.
If $p : M \to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{\partial M} : \partial M \to \partial N$ of the same degree. So if $p : \Sigma_{g,b} \to \Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover, 
\begin{align*}
\chi(\Sigma_{g,b}) &= k\chi(\Sigma_{g',b'})\\
2 - 2g - b &= k(2 - 2g' - b')\\
2 - 2g - kb' &= k(2 - 2g' - b')\\
2 - 2g &= k(2 - 2g')\\
\chi(\Sigma_g) &= k\chi(\Sigma_{g'}).
\end{align*} 
The converse is also true. That is, if $\chi(\Sigma_g) = k\chi(\Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $\Sigma_{g,b} \to \Sigma_{g',b'}$. To see this, note that if $\chi(\Sigma_g) = k\chi(\Sigma_{g'})$, then there is a degree $k$ covering map $p : \Sigma_g \to \Sigma_{g'}$; see this answer. If $D \subset \Sigma_{g'}$ is the interior of a closed disc in $\Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $\Sigma_g$. So if $D_1, \dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $\Sigma_{g'}$, then $p^{-1}(\Sigma_{g'}\setminus(D_1\cup\dots\cup D_{b'})$ is $\Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $\Sigma_{g,b}$. Therefore the restriction of $p$ to $\Sigma_{g,b}$ is a degree $k$ covering $\Sigma_{g,b} \to \Sigma_{g',b'}$.
Likewise, $\Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g',b'}$ if and only if $\chi(\Sigma_g) = k\chi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings:


*

*$\Sigma_{5,4} \to \Sigma_{5,4}$ of degree one,

*$\Sigma_{5,4} \to \Sigma_{3,2}$ of degree two,

*$\Sigma_{5,4} \to S_{6,2}$ of degree two,

*$\Sigma_{5,4} \to \Sigma_{2,1}$ of degree four, and

*$\Sigma_{5,4} \to S_{4,1}$ of degree four.

