Next week I will be giving a lecture, based on Chapter 2.6 from Jost's book Compact Riemann Surfaces. He states the following theorem:
Theorem 1 (Jost Theorem 2.6.2) Let $S$ and $\Sigma$ be Riemann Surfaces, assume that $\Sigma$ is hyperbolic and complete with respect to the hyperbolic metric. Then any bounded holomorphic map $f: S\backslash\{p\}\to \Sigma$ extends to a holomorphic map $\overline{f}: S \to \Sigma$.
Now, the definition of hyperbolic that Jost is using is not the standard definition, he defines the hyperbolic metric $d_H$ on $\Sigma$ to be
$d_H(p,q) := \inf \{ \sum\limits_{i=1}^n d(z_i,w_i): n \in \mathbb{N},\ p_0,\ p_1,\dots,\ p_n \in \Sigma,\ p_0 = p,\ p_n = q,$ $\phantom{spacespacespace} f_i: D \to \Sigma \ \text{holomorphic}, \ f_i(z_i) = p_{i-1},\ f_i(w_i) = p_i \}$.
For reference purposes this is apparently a special case of the "Kobayashi metric".
Let $D$ be the unit disk in the complex plane. From complex analysis, we have the following theorem (I paraphrase from Rudin's Real and Complex Analysis):
Theorem 2 Suppose $f: D\backslash \{0\} \to \mathbb{C}$ is holomorphic and bounded. Then $f$ has a removable singularity at $0$.
Now, if we consider a coordinate chart $\psi_1$ on $S$ in a neighborhood of $\{p\}$, with $\psi_1(p) = 0$, and an atlas $\{(\phi_n,U_n)\}$ on $\Sigma$, we can write $\phi_i \circ f \circ \psi_1^{-1}: D\backslash \{0\} \to D$.
On immediately seeing the statement of the theorem, I want to claim that this map is bounded. Is this true? And, if so, can we not simply apply Theorem 2 to obtain Theorem 1? My complex analysis intuition says this should be trivial, but Jost produces a much more complicated proof, which lacks intuition.