# Existence of Riemann surface, holomorphic maps

Say I have compact Riemann surfaces $$X$$, $$Y$$. Is there necessarily a Riemann surface $$Z$$ which maps holomorphically onto both $$X$$, $$Y$$?

Here is some construction. $$X$$ and $$Y$$ are smooth projective algebraic complex curves. So the complex surface $$X \times Y$$ is projective (e.g. by Segre embedding) and so admits a very ample line bundle $$L$$. By Bertini theorem, the vanishing locus of a general section of $$L$$ is a smooth projective complex curve $$Z$$ (so a compact Riemann surface) in $$X \times Y$$. The composition of the inclusion of $$Z$$ in $$X \times Y$$ with the projection on $$X$$ (or $$Y$$) is not constant: if it were constant, $$Z$$ would be contained in some $$Y$$ fiber, having zero intersection with a generic $$Y$$ fiber, and contradicting ampleness of $$L$$.
Choose non-constant meromorphic functions $$f:X\to \mathbb P^1$$ and $$g:Y\to \mathbb P^1$$ and denote by $$Z$$ the normalization of the fiber product $$X\times_{\mathbb P^1} Y=\{(x,y)\in X\times Y; f(x)=g(y)\}$$. It comes equipped with two maps $$\tilde f:Z\to X$$ and $$\tilde g:Z\to Y$$ such that the following diagramm commutes
$$\require{AMScd}$$ $$\begin{CD} Z @>{\tilde f} >> X\\ @V {\tilde g} V V @VV f V\\ Y @>>{ g}> \mathbb P^1 \end{CD}$$
Note that $$Z$$ may be disconnected, but the restriction of $$\tilde f$$ and $$\tilde g$$ to any of its connected components are surjective (because the fibers of those two maps are finite).