Suppose $f: X \to X'$ is a (holomorphic) covering map between hyperbolic surfaces (that is, their universal covering is the unit disk $\mathbb{D}$). Letting $p: \mathbb{D} \to X$ and $p': \mathbb{D} \to X'$ be the universal coverings, one can obtain a lifting $\tilde{f}: \mathbb{D} \to \mathbb{D}$ after choosing some points. Then apparently this lifting is biholomorphic. I see why $\tilde{f}$ is a holomorphic map, since it is locally a composition of holomorphic maps, but I don't quite see why it must be a bijection. So my question is exactly that, is $\tilde{f}$ a bijection?
Edit: I added the section in which this problem appears, as mentioned in the comments.
Best Answer
Your map $\tilde{f}$ is the unique basepoint-preserving map such that $p'\tilde{f}=fp$, using the lifting property of the covering map $p'$. However, $fp:\mathbb{D}\to X'$ is also a covering map (since it is a composition of covering maps of locally simply connected spaces). So we could also lift $p'$ along $fp$, and get a basepoint-preserving map $\tilde{p}':\mathbb{D}\to\mathbb{D}$ such that $fp\tilde{p}'=p'$.
I now claim that $\tilde{f}$ and $\tilde{p}'$ are inverses, and so in particular $\tilde{f}$ is a biholomorphism since they are both holomorphic. To prove this, note that $p'\tilde{f}\tilde{p}'=fp\tilde{p}'=p'$. So, $\tilde{f}\tilde{p}'$ is a basepoint-preserving lift of $p'$ along itself. But another such lift is just $1_\mathbb{D}$, so by the uniqueness of lifts along covering maps, this means $\tilde{f}\tilde{p}'=1_\mathbb{D}$. Similarly, $\tilde{p}'\tilde{f}=1_\mathbb{D}$ since they are both lifts of $fp$ along itself.