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.

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    $\begingroup$ All you need is $f : U \to \mathbb{C}$ is meromorphic iff $f : U \to P^1(\mathbb{C})$ is holomorphic iff around each $a \in U$, $f(z)$ or $\frac{1}{f(z)}$ is holomorphic (this is the two charts definition of $P^1(\mathbb{C})$ : one for $|w|< 2, w \mapsto w$ and the other one for $|\frac{1}{w}| < 2, w \mapsto 1/w$). $\endgroup$ – reuns Nov 2 '17 at 1:44
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    $\begingroup$ Yes, by definition of meromorphic around every point $a$, $\ \ f$ is holomorphic or it has a pole, in that case $1/f(z)$ has a zero at $a$ and is holomorphic around $a$ $\endgroup$ – reuns Nov 2 '17 at 1:51
  • $\begingroup$ @reuns I see. Thanks. $\endgroup$ – user45765 Nov 2 '17 at 1:52

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