Find all holomorphic diffeomorphisms $f:\mathbb{CP}^1\to\mathbb{CP}^1$ The complex projective line $\mathbb{CP}^1$ is the complex manifold defined by the quotient of $\mathbb{C}^2-\{(0,0)\}$ by the relation $z\sim w$ if $z=\lambda w$ for $\lambda\in\mathbb{C}-\{0\}$. I am trying to show that a map 
$$f:\mathbb{CP}^1\to\mathbb{CP}^1$$
is a (holomorphic) diffeomorphism if and only if $f$ is obtained from an invertible matrix $M\in\mathrm{GL}(2,\mathbb{C})$ by quotienting
$$M:\mathbb{C}^2-\{0\}\to\mathbb{C}^2-\{0\}.$$
I was able to show that every such $M$ indeed gives a diffeomorphism, but I am not able to prove the other direction. How show that every diffeomorphism arise in this way?
 A: Note that $GL(2,\mathbb{C})$ acts transitively on $\mathbb{C}P^1$. Take a diffeomorphism $\phi$. Let $\phi ( \infty ) = \alpha$. Take $g\in GL(2,\mathbb{C})$ with $g(\alpha) = \infty$ 
( for instance $g(z) = \frac{1}{z-\alpha}$). Consider the diffeomorphism $\psi = g \circ \phi$. 
We have $\psi( \infty) = \infty$, $\psi(\mathbb{C}) \subset \mathbb{C}$ and $\lim_{z\to \infty}\psi(z) = \psi (\infty)= \infty$. Therefore, $\psi_{\mid \mathbb{C}}$ is entire with limit $\infty$ at $\infty$. Consider the power series expansion of $\psi$
$$\psi(z) = \sum_{n\ge 0} a_n z^n$$
Now introduce the coordinate $t= \frac{1}{z}$ around $\infty$ for the subset $\mathbb{C}\backslash \{0\}$ of the domain  $\mathbb{C}$. Then we have
the Laurent expansion on $\mathbb{C} \backslash\{0\}$:
$$\chi(t) \colon = \psi(\frac{1}{t}) = \sum_{n\ge 0} a_n t^{-n}$$.
Moreover, since $\lim_{t \to 0} \chi(t) = \infty$, $\chi$ cannot have an essential singularity at $0$  (see http://en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_theorem) and so $0$ must be a pole so the Laurent expansion of $\chi $ terminates at some degree $-d$
$$\chi(t) = \sum_{n=0 }^{d} a_n t^{-n}$$
 That means that the power series expansion  of $\psi$ is finite and therefore $\psi$ itself is a polynomial of some degree $d$. 
with $a_d\ne 0$ 
$$\psi(z) = \sum_{n = 0}^d a_n z^n$$
Now for every $u \in \mathbb{C}$ the equation $\psi(z) = u$ will have $d$ roots in $\mathbb{C}$ ( with multiplicity) and for all but finitely many values $u$ it will have exactly $d$ roots ( we have to eliminate the $u$'s the values of the polynomial at the roots of its derivative). Since $\psi$ is also known to be injective we conclude that $d=1$ and therefore 
$$\psi(z) = a_0 + a_1 z$$
with $a_1 \ne 0$. 
We showed that the  diffeomorphism $\psi$ of $\mathbb{C}P^1$ is given by $\psi(z) = a z + b $ for some $a$, $b \in \mathbb{C}$. Therefore  $\psi= g \circ \phi $ is given by a transformation from $GL(2,\mathbb{C})$, and therefore $\phi$ is. 
Obs: We reduced the problem to the one of  finding the diffeomorphisms of $\mathbb{C}$.  In fact we can show that any holomorphic and injective map $\psi \colon \mathbb{C}\to \mathbb{C}$ is of the form $\psi(z) = a z + b$. Let's prove first that $\psi$ is surjective. Note that any injective holomorphic map is open. Therefore, $\psi \colon \mathbb{C} \to \psi (\mathbb{C})$ is a homeomorphisms, and therefore the open subset $\psi (\mathbb{C})$ is simply connected. Assume that $\psi (\mathbb{C})$ is not $\mathbb{C}$. Then there exists a holomorphic diffeomorphism  $u$ from $\psi (\mathbb{C})$ to the unit disk. In fact, it is enough that the diffeomorphism maps $u$ $\psi (\mathbb{C})$ to a subset of the unit disk. But then $u \circ \psi$ is a non-constant and bounded holomorphic function, and therfore (http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29) the function $u \circ \psi $ is constant, contradiction.
Therefore, an injective map $\psi$ from $\mathbb{C}$ to itself is a diffeomorphism. In particular, $\lim_{z \to \infty} \psi(z) = \infty$. We conclude like before that $\psi$ is a polynomial of degree $1$. 
