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I want to show the following:

Let $F:X\rightarrow Y$ be a non-constant and holomorphic map between compact riemann surfaces with $genus(X)=genus(Y)\geq 2$.

In the above it holds that $F$ is an isomorphism.

I am pretty sure that I should use the formula of riemann-hurwitz which can be found here. But I wasnt able to proof the statement.

Hopefully someone can help me.

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  • $\begingroup$ Hint: try to show that $N=1$ and there is no branched point. $\endgroup$
    – user99914
    Jul 17, 2016 at 12:54
  • $\begingroup$ Well, if there would be no branced points, then we would have $N=1$ and this yields that $F$ is an isomorphism (I already proved this fact). But with so less information I don't know how to prove that there is no branched point. $\endgroup$
    – Marc
    Jul 17, 2016 at 13:13
  • $\begingroup$ Note that $2g-2$ is strictly positive (since $g\ge 2$) and $\sum (e_p - 1)$ is nonnegative. $\endgroup$
    – user99914
    Jul 17, 2016 at 13:15
  • $\begingroup$ Oh okay, quite easy. ^.^. Thanks. $\endgroup$
    – Marc
    Jul 17, 2016 at 13:18
  • $\begingroup$ You may post an answer, @Marc. That finishes the question. $\endgroup$
    – user99914
    Jul 17, 2016 at 14:14

1 Answer 1

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To make this disappear from the list of unanswered questions. The solution to this question is essentially given in the comments already.

Let $F\colon X \rightarrow Y$ be a non-constant holomorphic map between compact Riemann surfaces both of genus $g\geq 2$. The Riemann-Hurwitz formula yields $$(1-\operatorname{deg}(F))(2g-2)=\sum_{p \in X}(\operatorname{mult}_p(F)-1).$$ The right-hand side is non-negative, while, by assumption, the term $2g-2$ is strictly positive. Thus, we have $\operatorname{deg}(F)=1$ and $F$ is an isomorphism.

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