Introduction
This question concerns "finite-type" maps from a Riemann surface $X$ to a compact hyperbolic Riemann surface $Y$. (Finite-type maps are roughly like branched covers, but not quite the same. See Question 2 below.)
Definition (following Bishop & Rempe). Let $X,Y$ be Riemann surfaces where $Y$ is compact. A holomorphic function $f:X \rightarrow Y$ is a finite-type map if there is a finite set $S$ such that $$f : X \setminus f^{−1}(S) \rightarrow Y \setminus S$$ is a covering map, and furthermore $f$ has no removable singularities at any punctures of $X$.
Question 1
In [BR21] Bishop & Rempe make the following comment. Suppose there is a finite-type map $f:X \rightarrow Y$ where $Y$ is a compact hyperbolic Riemann surface. Then $X$ is hyperbolic and the boundary of $X$ is uniformly perfect, i.e. the hyperbolic length of any non-contractible closed curve in X is bounded uniformly from below.
The fact that $X$ is hyperbolic is easy. Lifting $f$ to the universal covers gives a map $\tilde{f}: \tilde{X} \rightarrow \mathbb D$. By Liouville’s theorem, we are forced to have $\tilde{X} = \mathbb D$
Question: Why is the boundary of $X$ uniformly bounded?
Attempt at Question 1
It is well known that a holomorphic map between two hyperbolic Riemann surfaces is distance-decreasing, i.e. $d(f(x),f(y)) \leq d(x,y)$. Next let $\delta$ be a small number such that for all $y\in Y$, the neighbourhoods $B(y,\delta) \subset Y$ are simply connected. By compactness, $\delta >0$.
Lastly we try to show every non-contractible closed curve of $X$ has length at least $\delta$. Assume there is a closed curve $\gamma \subset X$ of length less than $\delta$ in $X$. Then, $f(\gamma)\subset Y$ lies in a simply connected open set $U:=B(f(\gamma(0)),\delta)$.
If $f^{-1}(U)$ is simply connected, then we are done because we can contract the curve $\gamma$. But why is this true?
Question 2
What exactly is the difference between finite-type maps and branched covers? The map $\exp: \mathbb{C} \rightarrow S^2$ is a finite-type map but not a branched cover.
Are there examples of branched covers $f:X \rightarrow Y$ which are not finite-type maps? (Such a map should have infinitely many branch points!)
In [BR21] Bishop & Rempe define a branched cover as follows.
Definition (following Bishop & Rempe). An holomorphic map $f:X \rightarrow Y$ is a branched cover if every point $y\in Y$ has a simply connected neighbourhood $U$ such that each connected component $V$ of $f^{−1}(U)$ is simply connected and $f : V \rightarrow U$ is a proper map topologically equivalent to $z \mapsto z^d$ for some $d \geq 1$.
Reference
[BR21] Christopher J. Bishop and Lasse Rempe. Non-compact Riemann surfaces are equilaterally triangulable. arXiv e-prints, page arXiv:2103.16702, March 2021.