Holomorphic Function constant in $\mathbb{P}^1(\mathbb{C})$ I want to show that a holomorphic function $f: \mathbb{P}^1(\mathbb{C}) \to \mathbb{C}$ is constant. $\mathbb{P}^1(\mathbb{C})$ is the projective line. I'm not very sure how to solve that. I have the idea to start with the Maximum-Principle. For that I need a point $a\in \mathbb{P}^1$, so that \begin{align}
|f(a)| \geq |f(z)| \forall z \in \mathbb{P}^1(\mathbb{C})
\end{align}
Maybe $a= \infty$, but I'm not coming further.  
 A: In the context of compact Riemann surfaces, proving the Liouville theorem is easier.

A function $f : P^1(\mathbb{C}) \to \mathbb{C}$ is holomorphic iff both $ f(z,1),  f(1,z)$ are holomorphic $\mathbb{C} \to \mathbb{C}$.
$\quad$ (for $z \ne 0$ we have $f(z,1) = f(1,1/z)$)
Let $z_0,z_1$ such that $$|f(z_0,1)| = \sup_{|z| \le 1} |f(z,1)|,\qquad |f(1,z_1)| = \sup_{|z| \le 1} |f(1,z)|$$ 
and let $F(z) = f(z,1)$ If $|f(z_0,1)| \ge |f(1,z_1)|$, $F(z)= f(1,z)$ otherwise.
It means that $F$ is holomorphic $\mathbb{C} \to \mathbb{C}$ and $z_0$ or $z_1$ is a local maximum of $|F|$, contradicting the maximum modulus principle if it is non-constant.
A: Preliminary remark : This does not makes sense to talk about metric as the metric of $\Bbb C$ can't be extended to $P^1(\Bbb C)$ as one would have $d(0, \infty) = \infty$ but distance between two points is always finite. On the other hand, it is well known that $S^2 \cong P^1(\Bbb C)$ so it is indeed compact. 
Now, if $f : P^1(\Bbb C) \to \Bbb C$ is not constant and holomorphic, it is an open mapping, therefore the image has to be open and compact, which is not possible. So $f$ has to be constant. 
