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Let $X$ be a compact connected Riemann surface, and let $S\subset X$ be a finite subset.

Does there exist a morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ which is unramified at the points of $S$?

I'm interested in the case where $X$ is of genus at least $2$. (The genus zero case is trivial: take $f$ to be the identity.)

The answer is trivial when $S$ is empty. (Any morphism $f:X\to \mathbf{P}^1(\mathbf{C})$ will do.)

Let $h:X\to \mathbf{P}^1(\mathbf{C})$ be a morphism with ramification locus $R(h)$. Then, if $S\subset X\backslash R(h)$, the answer is yes.

How effective can our answer be? That is, suppose that there exists such an $f$. Then, can we bound its degree?

The title is a special case of the above question: take $S=\{\textrm{pt}\}$.

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    The answer to your first question is yes by a (a weak version of) Riemann-Roch. Pick a point $P \notin S$ and write $\ell(nP)$ for the dimension of the space of meromorphic functions unramified outside of $P$ and with a pole of order at most $n$ at $P$. Then $\ell(nP) \underset{n\to +\infty}{\longrightarrow} +\infty$. Regarding your second question, the Riemann-Hurwitz formula relates the degree to the local ramification indexes.2011-12-25
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    @YBL: the condition "unramified outside of $P$" is not linear - how does Riemann-Roch say anything about such a space of functions?2011-12-25

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