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This is from Forster's book "Lectures on Riemann surfaces"

Thorem 8.4

$X$ and $Y$ are Riemann surfaces. $A \subset X $ is a closed discrete subset. $ \pi' : Y' \to X - A $ is a proper unbranched holomorphic covering map.

Then $\pi '$ extends to a branched covering of X, i.e. there exists a Riemann surface $Y$, proper holomorphic map $\pi : Y \to X $ and a fiber preserving biholomorphic map $ \phi : Y - \pi ^{-1}(A) \to Y' $

Proof.

For every $a\in A$ choose a coordinate neighborhood $(U_{a}, z_{a})$ of $a$ on $X$ with the following properties : $z_a(a)=0$ and $z_a(U_a)$ is a unit disk in $\mathbb{C}$ and $U_a \cap U-{a'} = \varnothing $ if $a ≠a'$. Let $U_a ^* = U_a - \{a\}$.

Since $\pi : Y' \to X' $is proper, $ \pi ' ^{-1}(U_a ^*) $ consists of a finite number of connected components $V_{a, \nu} ^*$ $ \nu = 1, ... , n(a) $ For every $\nu$, $ \pi ' |V_{a, \nu} ^* \to U_a ^*$ is a proper unbranched holomorphic covering. ...

Question : I don't understand why $ \pi ' ^{-1}(U_a ^*) $ should have finitely many components. I know that $\pi '$ has finitely many sheets, say $m$ sheets, since it is proper. Thus every $x \in U_a ^* $ has an evenly covered connected neighborhood of which inverse image is disjoint union of $m$ connected components. Do these local inverse images piece together nicely to make $ \pi ' ^{-1}(U_a ^*) $ consist of finite components?

Also I cannot see why for every $\nu$, $ \pi ' |V_{a, \nu} ^* \to U_a ^*$ is a proper unbranched holomorphic covering.

I would appreciate any hint or other references.

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Let $\Omega$ be a connected component of $\pi'^{-1}(U_a^*)$, then the restriction $\pi' : \Omega \to U_a^*$ is a covering, in particular is surjective and so contains at least one point $p \in \pi'^{-1}(b)$ where $b \in U_a^*$. As you said $\pi'$ is proper, so there are only finitely many preimages, say $m$ and by the previous argument the number of connected components of $\pi'^{-1}(U_a^*)$ is less than $m$.

By hypothesis $\pi$ was proper and unbranched, and these properties are stable under restriction of the domain. More precisely if $f : A \to B$ is a proper covering map and $C \subset A$ is any saturated subset then the restriction $f_{|C} : C \to f(C)$ is a proper covering map.

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  • $\begingroup$ The claim that "if $f : A \to B$ is a proper covering map and $C \subset A$ is any subset then the restriction $f_{|C} : C \to f(C)$ is a proper covering map" is false. However, it is true for $C$ which is "saturated", i.e. $C=f^{-1}(f(C))$. $\endgroup$ – Moishe Kohan Jan 24 '18 at 22:47
  • $\begingroup$ @MoisheCohen : you are of course right, I'll edit my answer, thanks for the catch. $\endgroup$ – Nicolas Hemelsoet Jan 25 '18 at 11:07

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