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Let $X$ be a compact Riemann surface, i.e. compact smooth complex analytic (hence automatically algebraic) curve. Let $A\subset X$ be a finite subset, and $X_0:=X\backslash A$.

Let $Y_0$ be a smooth complex analytic curve (necessarily non-compact) with a holomorphic map $f_0\colon Y_0\to X_0$ which is a finite covering.

QUESTION. Do there exist a compact smooth complex analytic (hence algebraic) curve $Y$, a finite subset $B\subset Y$, and a holomorphic (hence in fact algebraic) map $f\colon Y\to X$ such that $Y_0=Y\backslash B$, and $f|_{Y_0}=f_0$?

If yes, a reference would be helpful.

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    $\begingroup$ Yes, it's true. Although I don't have a reference right now. $\endgroup$ – Donu Arapura Aug 2 '16 at 12:27
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I think you can just define $Y$ by gluing disks to $Y_0$ since you know how finite covers of small punctured disks look like: namely $\mathbb C\setminus \{0\} \ni z\mapsto z^k\in \mathbb C\setminus \{0\}$.

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A nice reference is chapter 2 of Narasimhan s Riemann surface book when X is the Riemann sphere. Clearly, the techniques (adding algebraic boundary points) extend to the general case.

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The positive answer to the above question (even in a more general form) in explicitly contained in Theorem 8.4 in the book "Lectures on Riemann surfaces" by Otto Forster (1981).

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