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It is well known that the set of branched covers $X\to \mathbf{P}^1(\mathbf{C})$ of bounded degree and given branch locus in $\mathbb P^1(\mathbb C)$ is finite (up to isomorphism).

Let $X$ be a curve of genus at least two. Do there exist an integer $n\geq 2$ and a finite set of finite places $R$ of $X$ such that the set of degree $n$ finite morphisms $X\to \mathbf{P}^1(\mathbf{C})$ whose ramification locus is contained in $R$ is infinite, up to equivalence?

If $f$ and $g$ are finite morphisms from $X$ to $\mathbb P^1$, they are equivalent if $f$ equals $g$ up to an automorphism of $\mathbb P^1$.

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    What exactly is the role of $R$ in your question? You define it but I don't see that you use it. And is $R \subset X$ or is $R \subset \mathbf{P}^1(\mathbf{C})$?2012-12-02
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    Another point: what is the difference between the "well-known" part and your question? Is it that the branch locus is a subset of $\mathbf{P}^1(\mathbf{C})$, while the ramification locus is a subset of $X$?2012-12-02
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    I agree @NilsMatthes, the wording is confusing. Most people who ask "How do I prove this statement is false" want a counterexample, but it seems Harry wants a proof the negation or something? The only way I rearrange it in my head to make sense seems to be equivalent to the "well-known" part.2012-12-02
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    @Harry, to be clear, you want to prove that for all $n, X, R$, there are only finitely many $X\to P^1$ of degree $\le n$ and étale outside $R$ ?2012-12-03

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