Let $C$ be a smooth projective connected curve of genus $g$ over $\bar{\mathbf{Q}}$. Fix a finite **non-empty** (Edit) set of closed points $S$ in $C$ and let $U$ be the complement of $S$ in $C$.


**Q1.** *(Algebraic formulation)* Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1\_{\bar{\mathbf{Q}}}$ such that $\pi|\_{U}$ is etale?


Equivalently, let $X$ be a compact connected Riemann surface of genus $g$ which can be defined over $\bar{\mathbf{Q}}$ and let $B$ be a finite set of of closed points in $X$ with complement $Y$. 


**Q1.** *(Analytic formulation )* Does there exist a finite topological cover $Y\longrightarrow \mathbf{P}^1(\mathbf{C})-\{0,1,\infty\}$ ?


The equivalence of these two questions follows from the proof of Belyi's theorem and Riemann's existence Theorem.


If the answer to Question 1 is positive, I would be very interested in knowing if the degree of $\pi$ can be bounded effectively. 


**Q2.** Does there exist a finite (surjective) morphism $\pi:C\longrightarrow \mathbf{P}^1$ such that $\pi|\_{U}$ is etale and $\deg \pi \leq c$, where $c$ is a constant depending only on $S$ and $g$?


**Example.** Suppose that $g=0$. Then, following Belyi's proof of his theorem, the answer to Question 1 is yes. The answer to Question 2 is also positive and an explicit upper bound for such a rational function is given by Khadjavi in *An effective version of Belyi's Theorem*.


I don't expect the answer to Question 1 to be easy. In fact, what I'm asking is to prove the existence of a Belyi morphism $\pi:C\longrightarrow \mathbf{P}^1\_{\bar{\mathbf{Q}}}$ with **prescribed ramification**. Now, that's probably very hard but definitely very interesting to find out.


**Trivial Remark.** Suppose that $g>1$. Then the automorphism group of $C$ is finite. Choose a Belyi morphism $\pi:C\longrightarrow \mathbf{P}^1\_{\bar{\mathbf{Q}}}$ and let $U\_0\subset C$ be the complement of the ramification points of $\pi$. Then we see that Question 1 has a positive answer if we take $U$ to be $\sigma(U\_0)$ with $\sigma$ an automorphism of $C$. But that's only finitely many examples.



Can we construct rational functions with prescribed ramification on an algebraic curve over \Qbar
No, it is easy to construct examples where this is not possible (aside from trivial ones with $|S| < 3$). For example, if $g(C)>0$ one can find $S$ arbitrarily large so that the points of $S$ give linearly independent elements in $Pic(C)$. For such an $S$ there can be no map of the kind you want since the elements of $S$ must be mapped to at least $2$ distinct points of $\mathbb{P}^1$ which would give a non-trivial relation on the classes of elements of $S$ in $Pic(C)$.


