Extending holomorphic function between compact riemann surfaces

Question

I apologize if this is not the right way to go about asking such a question. I am interested in understanding the proof in this question about this theorem.

Let $M$,$N$ be compact connected Riemann surfaces, $X⊂M$ a finite subset. Then every biholomorphic embedding $f:M−X→N$ will extend to a biholomorphic map $M→N$.

I have understood almost all of the first proof given in the answer, however there is one point missing.

$h|D−{x}$ extends to a holomorphic function to the entire disk D. It follows that $f|D−{x}$ also extends to a (necessarily injective) holomorphic function $D→N$

The fact that $h$ extends to a holomorphic function to all of $D$ is clear. What is less clear is how it follows from this that $f$ extends to a holomorphic function to all of $D$. The case where $h(x)$ is not a branch point for $g$ should be clear, because we know that there exists a neighbourhood of $h(x)$ evenly covered by $g$ , and we can suppose $D$ small enough so that $h(D)$ is contained in this neighbourhood, and then we can take the local inverse of $g$ that agrees with $f$ everywhere except for $x$ , and extend $f(x)$ to the value of this local inverse on $h(x)$. By composing $h$ with the local inverse, this extension of $f$ is obviously holomorphic. (Please correct me if I have made any mistakes)

However, in the case where $h(x)$ is a branch point for $g$ , how is one to proceed? I was thinking we could use the fact that there exist local coordinates where $g$ looks like a power, which is probably the only tool I can use, but I can't seem to formalize the arguement.

Answer 1

Attempt at answering, I'm a novice too: I don't think you need a local inverse nor a system of coordinates.

You know that $f(x)$ should be a point in the preimage of $h(x)$ by $g$. Even when $h(x)$ is a branching value, there's a neighborhood $U$ of $h(x)$ such that $g^{-1}(U)$ is composed of a finite number of disjoint connected components $V_i$, and each of them contains only one preimage of $h(x)$ by $g$. This is because points in the same fiber must be isolated for holomorphic maps.

If $D$ is small enough, then $f(D-\lbrace x\rbrace)$ will be contained in only one of the $V_i$ (because continuous maps preserve connectedness), call it $V$. Then you define $f(x)$ to be the only point $y\in V$ with $g(y) = h(x)$.