Prove that every proper holomorphic map \[ f:\Delta\to\Delta \] of degree two can be written \[ f=A\circ S\circ B, \] where \(A,B\in\operatorname{Aut}(\Delta)\) and \(S(z)=z^2\).
Solution
1. The map \(f\) has a critical point \(c\in\Delta\).
Suppose instead that \(f'\) never vanished. Then \(f\) would be a covering map. Since the target \(\Delta\) is simply connected, such a covering from the connected disk would be an isomorphism. Its degree would then be \(1\), contrary to the hypothesis that \(f\) has degree \(2\). Hence \(f'(c)=0\) for some \(c\in\Delta\).
2. The point \(c\) is the unique point of \(f^{-1}(f(c))\), counted withmultiplicity two.
Because \(f'(c)=0\), the local degree of \(f\) at \(c\) is at least \(2\). The degree of a proper holomorphic map is the number of points in any fiber counted with multiplicity, and the total degree of \(f\) is \(2\). Therefore the local degree at \(c\) is exactly \(2\) and there can be no other point in the fiber over \(f(c)\).
3. Choose automorphisms \(B_0,A_0\in\operatorname{Aut}(\Delta)\) with\[ B_0(0)=c, \qquad A_0(f(c))=0, \]and set\[ g=A_0\circ f\circ B_0. \]Then \(g\) is a proper degree-two disk map whose only zero is \(0\), withmultiplicity two.
Pre- and post-composition by automorphisms preserve properness and degree. Moreover \[ g(0)=A_0(f(c))=0. \] By 2, the fiber of \(f\) over \(f(c)\) consists only of \(c\) with multiplicity two. Applying \(B_0^{-1}\) in the domain and \(A_0\) in the range therefore shows that the zero divisor of \(g\) is exactly \(2[0]\).
4. There is a real number \(\theta\) such that\[ g(z)=e^{i\theta}z^2. \]
Every proper holomorphic map \(\Delta\to\Delta\) is a finite Blaschke product. A degree-two Blaschke product has two zeros counted with multiplicity. By 3 both zeros of \(g\) are at \(0\), so its Blaschke-product representation is \[ g(z)=e^{i\theta} \left(\frac{z-0}{1-0\cdot z}\right)^2 =e^{i\theta}z^2. \]
5. Therefore\[ f=A\circ S\circ B \]for suitable \(A,B\in\operatorname{Aut}(\Delta)\) and \(S(z)=z^2\).
From 4, \[ A_0\circ f\circ B_0=R_\theta\circ S, \qquad R_\theta(z)=e^{i\theta}z. \] Hence \[ f =A_0^{-1}\circ R_\theta\circ S\circ B_0^{-1}. \] Both \(A_0^{-1}\circ R_\theta\) and \(B_0^{-1}\) are automorphisms of the disk. Taking \[ A=A_0^{-1}\circ R_\theta, \qquad B=B_0^{-1} \] gives the required form.