Existence of meromorphic function implies biholomorphic map onto the sphere.

Question

Let $M$ be a closed simply connected Riemann surface, and let $f: M \to \overline{\mathbb C}$ be a meromorphic map with a simple pole in a point $p \in M$. Is it true that $f$ is injective? That $f$ is surjective?

Also, one should not use the uniformization for Riemann surfaces to prove it. Surjectivity should follow from injectivity.

Any help would be appreciated.

Answer 1

Since $M$ is a closed surface, every holomorphic function on it has as many zeros as poles (counting multiplicities), by the argument principle. Justification:

1. $M$ can be covered by finitely many patches $U_j$ biholomorphic to domains in $\mathbb C$
2. You can write $M$ as a finite union $\bigcup \overline{V_k}$ where the domains $V_k$ are disjoint, each $V_k$ is contained in some $U_j$, each boundary $\partial V_k$ is piecewise smooth and does not contain the zeroes or poles of $f$.
3. The argument principle, applied to $V_k$, gives the difference between the numbers of zeroes and poles of $f$ in $V_k$ as $\frac{1}{2\pi i}\int_{\partial V_k} (f'/f)$
4. Summing the integrals in 3 over $k$, we get zero because each part of $\partial V_k$ is traveled twice, in opposite directions.

Therefore, for every $w\in\mathbb C$ the function $f-w_0$ has precisely one zero in $M$.