Whether the fiber of a holomorphic covering of the unit disk over a non-simply-connected domain is infinite or not

Question

Consider a holomorphic covering $f:\mathbb{D}\rightarrow \Omega$. Then for any point $a$ in the domain $\Omega$, consider the fiber $f^{-1}(a)$. If $f$ is non-constant, I know that when $\Omega$ is a simply connected domain, then $f$ is a homeomorphism, thus the fiber consists of only $1$ point. But what if $\Omega$ is not simply connected? Is it true in general that the fiber is infinite?

Answer 1

The fiber has to be infinite. [EDIT: In fact this has nothing to do with holomorphicity. See below.] Top-of-head argument and then one after I opened a book:

Say $\gamma$ is a closed curve in $\Omega$ which is not null-homotopic. Say $\gamma^n$ is "$\gamma$ repeated $n$ times" (sorry, I don't know this stuff). Then $\gamma^n$ is also not null-homotopic for $n>1$.

Say $\Gamma_1$ is a lift of $\gamma$, say starting at $z_0\in\mathbb D$ and ending at $z_1$. Say $\Gamma_2$ is a lift starting at $z_1$ and ending at $z_2$. And so on.

The points $z_0, z_1, \dots$ must all be distinct, hence the fiber is infinite.

They must be distinct. Suppose otoh $z_4=z_2$. Then $\Gamma_3+\Gamma_4$ is a closed curve in the disk, which is null-homotopic. But $\Gamma_j$ is a lift of $\gamma$, hence $\Gamma_3+\Gamma_4$ is a lift of $\gamma^2$; now applying $f$ shows that $\gamma^2$ is null-homotopic in $\Omega$.

Surely there's a better way to put all that in terms of homotopy groups and deck transformations and things. But it seems right...

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Ok, I cheated and opened a book. A slightly more grownup version of more or less the same argument: Since $\Omega$ is not simply connected the "covering group" or group of "deck transformations" is non-trivial. Say $\phi$ is a non-trivial element of that group. On general topological grounds $\phi$ cannot have a fixed point. So $\phi$ is parabolic or hyperbolic; equivalent to either a translation or dilation in the upper half-plane. Hence the orbit of any point under $\phi$ is infinite.

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In fact this has nothing to do with holomorphicity. Say $f:\mathbb D\to X$ is any non-trivial covering map. As before there exists a non-trivial deck transformation $\phi$. Except now $\phi$ is just a homeomorphism of the disk. Being a non-trivial deck transformation, $\phi$ has no fixed point. And the orbit of any point under $\phi$ is infinite: If the orbit of some point is finite then there exists $n$ so that $\phi^n$ has a fixed point. Since $\phi^n$ is a deck transformation this implies that $\phi^n$ is the identity. And I'm told that there is no homeomorphism of the disk with finite order but no fixed point; see Homeomorphisms of the Open Disk .