A non-constant holomorphic map $F$ between riemann-surfaces is an isomorphism

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.

• Hint: try to show that $N=1$ and there is no branched point.
– user99914
Jul 17, 2016 at 12:54
• 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.
– Marc
Jul 17, 2016 at 13:13
• Note that $2g-2$ is strictly positive (since $g\ge 2$) and $\sum (e_p - 1)$ is nonnegative.
– user99914
Jul 17, 2016 at 13:15
• Oh okay, quite easy. ^.^. Thanks.
– Marc
Jul 17, 2016 at 13:18
• You may post an answer, @Marc. That finishes the question.
– user99914
Jul 17, 2016 at 14:14

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.