Show that \[ \Aut(\CP^1) =\left\{z\longmapsto {az+b\over cz+d}:a,b,c,d\in\CC,\ ad-bc\neq0\right\}. \]
Solution
Let \[ A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\GL_2(\CC). \] The linear automorphism \(A:\CC^2\to\CC^2\) descends to a holomorphic automorphism of projective space \[ [z_0:z_1]\longmapsto[az_0+bz_1:cz_0+dz_1]. \] On the affine chart \(z=[z:1]\), this is \[ z\longmapsto\frac{az+b}{cz+d}. \] Its inverse is induced by \(A^{-1}\), so every fractional linear transformation with \(ad-bc\neq0\) lies in \(\Aut(\CP^1)\).
Conversely, let \(f\in\Aut(\CP^1)\). By the classification of meromorphic functions on \(\CP^1\), write \[ f(z)=\frac{p(z)}{q(z)} \] with coprime polynomials \(p,q\), and put \[ n=\max\{\deg p,\deg q\}. \] For every finite value \(w\) except possibly the ratio of the leading coefficients when \(\deg p=\deg q=n\), the equation \[ f(z)=w \] is equivalent to \[ p(z)-wq(z)=0, \] a polynomial equation of degree \(n\). Hence it has \(n\) solutions counted with multiplicity.
Because \(f\) is an automorphism, its inverse is holomorphic. Therefore \(f\) has no critical points: in local coordinates, \[ (f^{-1})'(f(z))f'(z)=1. \] Thus every point in every fiber has multiplicity one. Since \(f\) is also injective, a generic fiber contains exactly one point. Consequently \(n=1\).
Therefore \(p(z)=az+b\) and \(q(z)=cz+d\) for some \(a,b,c,d\in\CC\). The determinant \(ad-bc\) is nonzero, since otherwise the two linear polynomials are proportional and \(f\) is constant. Hence \(f\) is fractional linear.