Automorphism of Riemann Sphere without Picard's Theorem When I am trying to get familiar with the contributions that Gauss made to the quaternions, the following question comes to my mind:
Question: Is there a known proof that every 1-1 comformal mapping of Riemann Sphere to itself is a Mobius Transformation, without using Picard's Theorem about essential singularities?
 A: There is. It uses the fact that the conformal automorphisms of the complex plane are the nonconstant linear polynomials $z\mapsto az+b,~a\neq0$. These are Möbius transformations.
Now let $\varphi:\overline{\mathbb C}\longrightarrow \overline{\mathbb C}$ be a conformal automorphism of the Riemann sphere. Let $a:=\varphi^{-1}(\infty)$. We consider two cases:
1: $a=\infty$
Then $\varphi\vert_{\mathbb C}$ is a conformal automorphism of the complex plane, so it is a Möbius transformation defined on $\mathbb C$. Since the conformal extension to $\overline{\mathbb C}$ is unique, $\varphi$ itself is also a Möbius transformation.
2: $a\neq\infty$
Then consider the Möbius transformation
$$\mu(z):=\frac{az+1}{z},$$
which sends $\infty$ to $a$. Then $\varphi\circ\mu$ is a conformal automorphism of $\overline{\mathbb C}$ which sends $\infty$ to $\infty$, so this automorphism belongs to case 1, and is thus a Möbius transformation. But then since Möbius transformations are a group, $\varphi\circ\mu\circ\mu^{-1}=\varphi$ is also a Möbius transformation.
The fact from the beginning (automorphisms of the complex plane) can be proved using Casorati-Weierstraß to show that in injective entire function must be a polynomial, and the fundamental theorem of algebra to show that to be injective, the polynomial must also be of degree 1.
A: If $f$ is holomorphic $\Bbb{P^1(C)\to P^1(C)}$ then $f$ is meromorphic $ \Bbb{C\to C}$ and it has finitely many poles, there is a rational function such that $h=f-g$ has no pole.
$h$ is holomorphic $\Bbb{C\to C}$ and $\Bbb{P^1(C)\to P^1(C)}$, if $h(\infty)=\infty$ then  $1/h(1/z)$ is holomorphic and has a zero at of order $n>1$ at $0$ so $1/h(1/z)\sim c z^n$ as $z\to 0$,
$h(z)\sim c^{-1} z^n$ as $z\to \infty$, repeating with $h(z)-c^{-1} z^n$ we'll eventually obtain a polynomial such that $h-p$ is bounded.
So we obtained a polynomial such that $h-p$ is holomorphic $\Bbb{P^1(C)\to P^1(C)}$ and $h-p$ is $\Bbb{C}$-valued and bounded. Since $\Bbb{P^1(C)}$ is compact then $|(h-p)(s)|$ attains its maximum at some $s\in \Bbb{P^1(C)}$ which implies (maximum modulus principle) that $h-p$ is constant.
Whence $f$ was a rational function $u/v$ with $u,v$ coprime polynomials. $f$ is bijective iff $\deg(u)\le 1,\deg(v)\le 1,\deg(u)+\deg(v)\ne 0$ ie. $f$ is a Möbius transformation.
