- Does there exist a surjective holomorphic map \(f : D \to \mathbb { C }\) from the unit disc to the whole complex plane? Prove that your answer is correct.
Solution
Yes. One example is \[ \boxed{f(z)=\frac{2z(1+z)}{(1-z)^2},\qquad |z|<1.} \]
1. The disk is biholomorphic to the right half-plane.
The map \(T(z)=(1+z)/(1-z)\) is holomorphic on the unit disk and satisfies \[ \operatorname{Re}T(z)=\frac{1-|z|^2}{|1-z|^2}>0. \] For any \(t\) with \(\operatorname{Re}t>0\), its inverse \(z=(t-1)/(t+1)\) is well defined and satisfies \[ 1-|z|^2=\frac{4\operatorname{Re}t}{|t+1|^2}>0. \] Direct substitution verifies the inverse identities. Thus \(T\) maps the disk onto \(H=\{t:\operatorname{Re}t>0\}\).
2. The polynomial \(p(t)=t^2-t\) maps \(H\) onto \(\mathbb C\).
Given any \(w\in\mathbb C\), the equation \(t^2-t-w=0\) has two complex roots \(t_1,t_2\), counted with multiplicity, and \(t_1+t_2=1\) by the quadratic formula. Their real parts sum to one, so at least one has real part at least \(1/2\). That root belongs to \(H\) and satisfies \(p(t)=w\). Since \(w\) was arbitrary, \(p(H)=\mathbb C\).
3. Composition gives the required surjection.
The composite \(p\circ T\) is holomorphic on the disk and surjective by the preceding two steps. Its formula is \[ T(z)^2-T(z) =\frac{(1+z)^2-(1+z)(1-z)}{(1-z)^2} =\frac{2z(1+z)}{(1-z)^2}. \] The denominator vanishes only at the boundary point one, not in the domain. This proves the assertion for the displayed function.
There is no conflict with the impossibility of a biholomorphism from the disk onto the plane: its inverse would be bounded and entire, hence constant by Liouville’s theorem [@SS03]. The map constructed here is not injective. Indeed, \(t=(1\pm i\sqrt3)/2\) are two distinct points of \(H\) with \(p(t)=-1\), and their distinct inverse images under \(T\) have the same value under \(f\).