Finitely Many genus-g Quotients of Compact Riemann Surface

I hear there is a semi-famous theorem from my advisor, but he didn’t know the name and I was unable to find it online. Does anybody know of the following?

Let $S$ be a compact Riemann surface. Then for a given genus $g$, up to isomorphism (holomorphism in our case), there are only finitely many compact Riemann surfaces $Y$ of genus $g$ that are images of surjective holomorphisms $\varphi : S \twoheadrightarrow Y$

Has anybody heard of such a result? It may be true only for genus 2 and higher, I have no idea on this one.

It would just be nice to mention in a talk I am giving about the Hurwitz Theorem.

Solutions Collecting From Web of "Finitely Many genus-g Quotients of Compact Riemann Surface"

This can be seen as an application of the Parshin-Arakelov finiteness theorem: There are only finitely many (up to a biholomorphic isomorphism) holomorphic families of Riemann surfaces of the fixed type $(g,n)$ (where $g$ is the genus and $n$ is the number of punctures) with fixed compact base $B$. (See here and here for some references.)

Consider now the special case when $B$ is your Riemann surface $S$. The finiteness theorem then translates to the statement that there are only finitely many nonconstant holomorphic maps $B\to M_{g,n}$ where the target is the moduli space of Riemann surfaces of type $(g,n)$ (I am simplifying a bit here, you should treat the target as an orbifold/stack). Note that $M_{g,1}$ is naturally isomorphic to the universal curve $C_g\to M_g$. (This is where you need $g\ge 2$.) Suppose now that there are infinitely many genus $g$ Riemann surfaces $Y_g$ with nonconstant holomorphic maps $B\to Y_g$. Identify $Y_g$’s with suitable fibers of $C_g$. Then you obtain infinitely many nonconstant holomorphic maps $B\to M_{g,1}$, contradicting the finiteness theorem. In your setting the genus of $Y$ does not exceed the genus of $S$ (this can be derived for instance from the Riemann-Hurwitz formula). qed

As an addendum, here is an easier proof of the easier result (due to de Franchis) that given two compact Riemann surfaces $X, Y$ of genus $\ge 2$, there are only finitely many nonconstant holomorphic maps $f: X\to Y$. This proof is a model of the proofs of the harder finiteness theorem above. By the Schwarz lemma, each map $f: X\to Y$ is 1-Lipschitz with respect to the uniformizing hyperbolic metrics on $X$ and $Y$. (This is one place where genus $\ge 2$ is used.) By Arzela-Ascoli theorem, any sequence of holomorphic maps $f_n: X\to Y$ has to subconverge to a nonconstant holomorphic map $f: X\to Y$. Therefore, after passing to a subsequence, for large $n$ the maps $f_n$ are homotopic to $f$. But then $f_n=f$ for large $n$. (Use the fact that a holomorphic 1-form is uniquely determined by its periods; here again we need the fact that the genus is $\ge 2$.)

Edit. An update. The finiteness theorem is due to Severi. See

Bounds on the number of holomorphic maps,

by M.Tanabe, Proc. of AMS, 2005, for the explicit bounds on the number of pairs $(Y,f)$, given the genus of $X$.