I'm at the end of the proof of uniformization for simply connected manifolds in Farkas-Kra's Riemann Surfaces text. I feel like I'm missing something really obvious here. (The proof in question is on page 203 of the second edition.)
Context: I have a simply-connected Riemann surface $M$, and an 'admissible' function $f: M \rightarrow \Bbb C \cup \{\infty \}$ (i.e. a meromorphic function on $M$), which is to say that $f$ is holomorphic except at some point $P$ where it has $\text{ord}_P f = -1$, and such that $g$ is bounded in every neighborhood outside of $P$.
I want to show that $M$ is compact iff $g$ is surjective. The forward direction follows from using the argument principle to show that a meromorphic function on a compact surface has the same number of zeroes and poles, and this that $g-z_0$ has exactly one zero for every complex $z_0$. However, the reverse direction still evades me. (We also have, at this point, that $g$ is injective; I was hoping to get some sort of topological contradiction, but of course that would assume that our bijective continuous map was FROM a compact space - which is, naturally, a bit problematic.)
$g$ being injective implies it is not constant. A non-constant holomorphic mapping is an open mapping (that holds for the codomain any Riemann surface), hence, $g$ is a homeomorphism between $M$ and $g(M)$.