Is every Riemann surface a 2-sheeted covering? Given an algebraic curve $X$ over $\mathbb{C}$, i.e. a Riemann surface and a fixed set of pairs of points $S=\{(p_1,q_1),...,(p_1,q_1)\}$ is there an algebraic curve Y, possibly singular, and a map $f: X \to Y$ such that


*

*$f$ has degree 2.

*$f$ send pairs of points $(p_i,q_i)$ to the same point $y=f(p_i)=f(q_i)$ for all $i$.


We can look at the function field of $X$, say $k(X)$ which if I'm not mistaken looks like a finite extension of $\mathbb{C}(t)$, so $k(X)=\mathbb{C}(t)[X]/f(X)$. If one finds an index two subfield this will yield a map $X\to Y$ using the smoothness of $X$ of degree 2. 
One idea would be to take $\mathbb{C}(t)[X^2]/f(X)$ but I'm not sure this is correct. In any case this does not allow one to describe the image of pairs of points since the map is determined birationally.

Here is an argument for say genus 3 curves. The dimension of the moduli spaceof genus 3 curves is 6 (=$3g-3$). If such an $X$ maps  two to one onto a curve $Y$, then by Riemann-Hurwitz, $g(Y)<3$. If $g(Y)=2$, then again RH says the map has to be etale and thus determined by a genus 2 curve and a 2-torsion line bundle. Since 2-torsion line bundles are finite, the dimension of all these can at most be 4, which is the dimension of the moduli space of genus 2 curves. If $g(Y)=1$, then the map is ramified at 4 points and thus the dimension of such $X$ can be at most the dimension of moduli space of genus 1 curves plus 4, which is 5. Finally, if $g(Y)=0$, then $X$ is hyperelliptic and general curve of genus 3 is not hyperelliptic. Thus a general genus 3 curve can not map 2-1 onto a curve. 
