Let $X$ and $Y$ be connected Riemann surfaces. Let $F: X \to Y$ be holomorphic non-constant (and thus open and continuous). Denote the set of ramification points $Ram(F) :=\{p \in X \ | \ mult_pF \ge 2\}$ and the set of branch points $Branch(F) := F(Ram(F))$.
Question 1: Just to double check, are the following correct?
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$Ram(F)$ is discrete, and I think $Ram(F)$ is also closed for the same reason $Ram(F)$ is discrete.
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Apparently, for any discrete subspace $A$ of $X$, $F(A)$ is not necessarily discrete.
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$F$ is 'discrete' in the sense that its fibres are discrete (i.e. each $F^{-1}(y)$ is discrete, for any $y \in image(F)$, or even $y \in Y$) but not necessarily in the sense that $F$ maps discrete sets to discrete sets.
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For $X$ compact, we have that $F$'s image is compact, and $F$ is surjective. Thus, $Y$ is compact.
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For $X$ compact, $F$ is a proper and closed map because $Y$ is Hausdorff and because $F$ is continuous.
Question 2: For $X$ compact, is $Branch(F)$ discrete? (possibly answered Question 3)
Question 3: More generally, for $X$ compact and for any closed discrete subspace of $A$ of $X$, is $F(A)$ discrete?
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Question 3.1: I think yes: closed discrete subspace 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. Is this correct?
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Question 3.2: 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. Is this correct?
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In particular, I think 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).
Question 4: (I was about to ask alternatives to that $X$ is compact to say $F(Ram(F))$ or even $F(A)$ is discrete, but I think this is another story already.)
Related:
Set of branch points isn't discrete, but branch points are isolated?
About branch points of a holomorphic map
Best Answer
I believe yes to questions 1,2,3,3.1 and 3.2.