Why is every holomorphic bijection of the Riemann sphere a Möbius transformation? Just based on some reading, I know that every Möbius transformation is a bijection from the Riemann sphere to itself. 
I'm curious about the converse. For any holomorphic bijection on the sphere, why is it necessarily a Möbius transformation? Is there a proof or reference of why this converse is true? Thanks.
(I would appreciate an explanation at the level of someone whose just self-studying complex analysis for the first time.)
 A: Suppose $f$ is an holomorphic bijection of the sphere to itself. There is a Moebius transformation $g$ which maps $f(\infty)$ to $\infty$. Let $h=g\circ f$, which is again an holomorphic bijective map of the sphere to itself, and which maps $\infty$ to $\infty$. It follows that $h(\mathbb C)\subseteq\mathbb C$, because of injectivity, and the restriction $h|_{\mathbb C}:\mathbb C\to\mathbb C$ is a injective entire function.
Now, the big theorem of Picard, or several others, imply that $h|_\mathbb C$ is necessarily linear. It follows that $h$ itself is linear, and then $f=g^{-1}\circ h$ is a Moebius transformation.
A: Here is another proof using results from the theory of Riemann surfaces.
Any holomorphic automorphism $F\colon\mathbb{P}^1\rightarrow \mathbb{P}^1$ of the Riemann sphere $\mathbb{P}^1$ can be viewed as a meromorphic function $f\colon\mathbb{P}^1\rightarrow \mathbb{C}$ in the obvious way. This works since, by the Identity Theorem, the fiber $F^{-1}(\{\infty\})$ is discrete. (Since $\mathbb{P}^1$ is compact, the fiber is even finite.)
Now, every meromorphic function on $\mathbb{P}^1$ is rational, i.e. there are complex univariate polynomials $g,h$ with $f=\frac{g}{h}$. A proof can be found after Corollary 2.9. of Forster's Lectures on Riemann Surfaces. For completeness, let me recall Forster's proof:

Note that, by the Identity Theorem, $f$ has only finitely many poles and zeroes. Without loss of generality, assume that $\infty$ is not a pole of $f$. Otherwise, consider $\frac{1}{f}$. Let $a_1,\ldots,a_n\in\mathbb{C}$ denote the poles of $f$. For each $\nu=1,\ldots,n$, consider
$$h_\nu(z)=\sum_{j=-k_\nu}^{-1}c_{\nu j}(z-a_\nu)^j$$
the principal part of the Laurent expansion of $f$ around the pole $a_\nu$. Then the function $f-(h_1+\ldots +h_n)$ is holomorphic on $\mathbb{P}^1$. Now, assume for a contradiction that the function  $f-(h_1+\ldots +h_n)$ were non-constant. Then it would be open by the Open Mapping Theorem. Since $\mathbb{P}^1$ is compact, the function would hence be surjective. Consequently, $\mathbb{C}$ would be compact which is absurd. It follows that $f$ is rational.

Since a rational function $\frac{g}{h}$ is bijective only if $g$ and $h$ are linear (see here), we have that $f$, and thus $F$, is of the form $z \mapsto \frac{az+b}{cz+d}$ with $a,b,c,d\in \mathbb{C}$. Such a mapping is bijective if and only if $ad-bc\neq 0$.
