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.



Finite covers of punctured Riemann surfaces
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). 


