# Non constant meromorphic function of Riemann Sphere

Let $X$ be a compact Riemann surface and $f$ a nonconstant meromorphic function on $X$. Show that $f$ must have a zero on $X$, and must have a pole on $X$.

Suppose that $f$ is a nonconstant meromorphic function on a compact Riemann surface $X$. Then the associated mapping $F : X \to \mathbb{C}_{\infty}$ is a nonconstant holomorphic map. By Liouville's theorem, $F$ is subsequently not bounded. Hence, by the way in which $F$ is defined, $f$ has a pole since $F$ attains $\infty$. For $f$ meromorphic on a compact Riemann surface however, $$\sum_p \text{ord}_p(f) =0.$$ Therefore, $f$ must also have a zero.

This proof is wrong however, because I have clearly misused Liouville's theorem.

• Why have you misused Liouville? Dec 6, 2016 at 1:28
• @TedShifrin $F$ is not entire?
– user319128
Dec 6, 2016 at 1:30
• Fair enough. $F$ has compact image in $\Bbb C_\infty$. If that compact image misses $\infty$, then prove that $f\colon X\to\Bbb C$ must be constant. Dec 6, 2016 at 1:35
• Please don't delete when you get an answer (even if only a hint in comments). Undoubtedly you didn't mean any harm, but it is considered a bit rude, because the time other users spent on thinking about your question then goes to waste. Dec 6, 2016 at 15:19

Let $$f$$ be a non-constant meromorphic function on a compact Riemann surface $$X$$. Consider the associated (non-constant) holomorphic mapping $$F\colon X\rightarrow \mathbb{C}_\infty$$.
By the Open Mapping Theorem (which can be derived from the Local Normal form of a non-constant holomorphic map between Riemann surfaces), the image $$F(X)$$ is open. Additionally, since $$F$$ is continuous and $$X$$ is compact, it follows that $$F(X)$$ is compact. Now, compact subsets of a Hausdorff space are closed. Thus, the non-empty set $$F(X)$$ is clopen. Since $$\mathbb{C}_\infty$$ is connected, $$F$$ is surjective. In particular, $$f$$ has both a zero and a pole.