I want to prove the following:
Let $R,S$ be two Riemann surfaces, and suppose Green's function $g_S$ exists for $S$. Let $f: R \to S$ be a nonconstant analytic function. Prove that Green's function $g_R$ exists for $R$, and $g_R(p,q) \leq g_S(f(p),f(q))$ for all $p,q \in R$.
This is an exercise in Gamelin's book, XVI.3.2. The exercise does not forbid $f$ being constant, but it seems evidently necessary.
Some thoughts
Let $(U, z)$ be a coordinate patch of $R$ containing $q$ with $z(q) = 0$, let $v$ be a subharmonic function in $R \setminus\{q\}$ with compact support such that $$\limsup_{x \to q} v(x)+\log |z(x)|< \infty,$$ then by definition $g_R(\cdot,q)$ is the supremum of all such $v$, so it is enough to prove that $v(p) \leq g_S(f(p), f(q))$.
Letting $(V,w)$ be a coordinate chart around $f(q)$ in $S$ with $w(f(q)) = 0$, we know that $$\lim \limits_{x \to q} g_S(f(x),f(q))+\log |w(f(x))|$$ exists, as the function is even harmonic there.
Subtracting, we get that $$\limsup \limits_{x \to q} \;v(x)- g_S(f(x),f(q))+\log \lvert \frac{z(x)}{w(f(x))}\rvert < \infty.$$
If $w \circ f$ is a coordinate function, which is not always true, then $\limsup _{x \to q} \;v(x)- g_S(f(x),f(q)) < \infty$ so for every $\epsilon>0$, the function $\varphi_\epsilon(x) := v(x)- (1+\epsilon)g_S(f(x),f(q))$ extends to be subharmonic at all $R$. It is $\leq 0$ off a compact set, the support of $v$, so the maximum principle forces $\varphi_\epsilon \leq 0$. Letting $\epsilon \to 0$ we are done.
This question is a slightly more quantitative version of the fact, of which I do not know a proof, that an analytic map from a parabolic Riemann surface to a hyperbolic Riemann surface is constant. The main tool in proofs on this topic seems to be the maximum principle; how can it be leveraged here without adding an assumption on $f$?
