TITLE: An analytic proof of the De Franchis theorem
QUESTION [9 upvotes]: The De Franchis theorem in its simplest form states that given two compact Riemann surfaces $\Sigma_{g_1},\Sigma_{g_2}$ where $g_1,g_2 > 1$, there are only finitely many non-constant holomorphic mappings from $\Sigma_{g_1}$ into $\Sigma_{g_2}$.
The complete proof can be found in P.Samuel's book Lectures on Old and New Results on Algebraic Curves. This proof uses quite a bit of Algebraic Geometry, in which I have very little background.
Does anyone know how to prove this result using purely complex analytic ideas?

REPLY [2 votes]: We break the proof up into two steps:
1. The set of holomorphic maps is compact in the compact open topology. This is an easy consequence of the hyperbolicity of the Riemann surfaces

The space of holomorphic maps is discrete. Suppose that there is a one parameter family (f_t) of maps between the Riemann surfaces. The derviative (\frac{\partial f}{\partial t}\0 is then a holomorphic section of the pull-back of the tangent bundle of the range over the domain. By hyperbolicity there are no non-zero sections. 

You can find arguments of this type in
M. Kalka, B. Shiffman and B. Wong, Finiteness and Rigidity Theorems for Holomorphic Mappings,
    Michigan Math. J. 28 (1981), 289-295.