This is a part of forster's Riemann surface which I do not totally understand.
1.15 Thm. $X$ is riemann surface and $f\in M(X)$ where $M(X)$ is the set of meromorphic functions on $X$. For each pole $p$ of $f$ define $f(p)=\infty$. Then $f:X\to P^1$ is holomorphic mapping.
Pf. $f\in M(X)$ and $P$ is the set of poles of $f$. Then $f$ induces a mapping $f:X\to P^1$. Suppose $\phi:U\to V$ and $\psi:U'\to V'$ are charts of $X,P^1$ respectively with $f(U)\subset U'$. We have to show $g=\psi\circ f\circ\phi^{-1}:V\to V'$ is holomorphic. Since $f$ is holomorphic on $X-P$, it follows $g$ is holomorphic on $V-\phi(P)$. Hence by Riemann Removable singularities theorem $g$ is holomorphic on all of $V$.
Q: How do I see $g$ is bounded on $V-\phi(P)$? In order to remove singularity, I need to apply boundedness or holomorphic extension.