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I thought of this question based on my other question here.

I understand that for a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are connected but not necessarily compact, if $F$ is defined to have a ramification point at $p \in X$ if for all neighborhoods $U$ of $p$ in $X$, the restriction $F|_U$ is not injective.

Question 1: So if, on the contrary, $F$ were injective, then no $p$ is a ramification point because we could just pick $U=X$?

Question 2: What is the relationship between Question 1 and the fact that injective holomorphic maps between Riemann surfaces are isomorphisms onto their images (Proposition II.3.9 in Rick Miranda - Algebraic curves and Riemann surfaces; also see here: Whether or not riemann surfaces can be singletons (and consequences and generalisation))?

BCLC
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    Can you clarify your question? It looks like you’re asking about the contrapositive of a statement you know is true. – Michael Burr Sep 04 '20 at 10:01
  • @MichaelBurr Oh right I used contrapositive. Anyway, I edited question. Thanks! – BCLC Sep 04 '20 at 10:15
  • You might think about this in light of the complex analysis exercise that locally a non-constant holomorphic map $\Bbb C\to\Bbb C$ looks like $f(z)=z^k$ for some $k\in\Bbb N$, and a ramification point corresponds precisely to the origin with $k\ge 2$. – Ted Shifrin Sep 10 '20 at 22:30

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