- Does there exist a holomorphic surjection from the unit disc to \(\mathbb { C } ?\)
[Hint: Move the upper half-plane “down” and then square it to get C.]
Solution
Yes. Let \[ h(z)=i\frac{1+z}{1-z}, \] which maps \(\mathbb D\) conformally onto the upper half-plane \(\mathbb H\). Then \[ g(z)=h(z)-i \] maps \(\mathbb D\) onto the half-plane \(\{w:\Im w>-1\}\).
Define \[ F(z)=g(z)^2. \] This is holomorphic on \(\mathbb D\). To prove surjectivity, let \(\zeta\in\mathbb C\). Choose one square root \(w\) of \(\zeta\) with \(\Im w\ge0\); one of the two square roots always has nonnegative imaginary part. Then \(\Im w>-1\), so \(w=g(z)\) for some \(z\in\mathbb D\). Consequently \[ F(z)=g(z)^2=w^2=\zeta. \] Thus \(F:\mathbb D\to\mathbb C\) is a holomorphic surjection.