Are non-constant meromorphic functions on compact Riemann surjective? Let $f$ be meromorphic on compact Riemann surface $X$ (Assume $X$ if connected if need be), i.e. we have $f: X \setminus poles(f) \to \mathbb C$ holomorphic.

*

*This corresponds to holormorphic $F: X \to \mathbb C_{\infty}$, where $F(x):=f(x)$, $x \in Domain(f)$ and $F^{-1}(\infty):=poles(f)$.


*We have $f$ non-constant if and only if $F$ non-constant.


*If $F$ is non-constant, then $F$ is surjective.


*$F$ is surjective if and only if $f$ is surjective.


*Therefore, if $f$ is non-constant, then by (2),(3) and (4), $f$ is surjective.
Question 1: Is this correct?
Question 2: Is this incorrect if $X$ is not connected?

Remark: I believe for the case of $X$ compact and connected: If question 1 is answered affirmatively, then I believe $f$ has a pole because '$f$ has a pole' is equivalent to '$F^{-1}(\infty) \ne \emptyset$'.
 A: The argument is correct for connected $X$. The crucial step is 3, and that works since a non-constant holomorphic mapping on a connected Riemann surface is an open mapping. Thus $F(X)$ is an open subset of the Riemann sphere. Also, since $X$ is compact $F(X)$ is compact, thus closed. And since the Riemann sphere is connected, it follows that $F(X) = \mathbb{C}\cup \{\infty\}$ (using the fact that $X \neq \varnothing$, for otherwise $F$ couldn't be non-constant).
This breaks down if $X$ is not connected. If say $X$ has two connected components $X_1$ and $X_2$ the meromorphic (even holomorphic) function $f(z) = j$ if $z \in X_j, j=1,2$ is non-constant, but not surjective.
We get surjectivity if [and only if] there is (at least) one component of $X$ on which $F$ is non-constant.
A: If $f$ doesn't take the value $a$ then $1/(f-a)$ is holomorphic $X\to \Bbb{C}$.
Holomorphic functions from a compact Riemann surface to $\Bbb{C}$ are constant:
they are continuous, (by compactness of $X$) $1/|f-a|$ takes its maximum at some $p\in X$, locally in some chart $1/(f-a)$ is represented by an analytic function which must be constant because locally a non-constant analytic function is $g(z)=\sum_{n\ge 0} c_n (z-b)^k = c_0+c_k (z-b)^k+O((z-b)^{k+1})$ with $c_k\ne 0$ so $|g(z)|$ is not maximum at $b$.
