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Let $F:X \to Y$ be a holomorphic map between Riemann surfaces. $q \in Y$ is a branch point if it is the image of a ramification point. How to prove that the set of branch points is a discrete subset of $Y$. This is from Rick Miranda's Algebraic curves and Riemann surfaces. Thank you.

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  • $\begingroup$ I think it is a mistake. I don't believe branch points is always discrete. However, it's discrete for compact $X$ and $Y$ supposedly. I'm about to make a question about this. In the mean time, see here: math.stackexchange.com/questions/3813927/… $\endgroup$ – John Smith Kyon Oct 31 '20 at 4:03
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    $\begingroup$ Thanks a lot.@JohnSmithKyon $\endgroup$ – yuan Dec 1 '20 at 2:50
  • $\begingroup$ You're welcome yuan. how about upvoting and accepting my answer then? $\endgroup$ – John Smith Kyon Dec 3 '20 at 23:30
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You forgot to say $F$ is non-constant. Then again, I guess $Ram(F)$ is not defined for $F$ non-constant.

In general for any map $F: X \to Y$ of any topological spaces $X$ and $Y$ with $X$ compact and $Y$ Fréchet/T1 and for any closed discrete subspace $A$ of $X$, we have $F(A)$ discrete.

Proof: Closed discrete subspaces $A$ of compact is finite $\implies$ $A$ is finite $\implies$ $F(A)$ is finite $\implies$ $F(A)$ is discrete because finite subspaces of Fréchet/T1 are discrete. QED

Apply this to the case of $A=Ram(F)$ when $F$ is a non-constant holomorphic map between connected Riemann surfaces with $X$ compact (and thus $F$ is surjective, open, closed and proper and $Y$ is compact) to get $F(A)=Branch(F)$ is discrete.

In particular, this means we do not use that $F$ is proper, closed, open, surjective, non-constant or holomorphic or that $X$ is connected or that $Y$ is connected. We can relax this to $X$ compact (and not necessarily Riemann surface) and $Y$ Fréchet/T1 (and not necessarily Riemann surface, Hausdorff/T2 or compact).

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