Past exam of the metric spaces and complex analysis course of the University of Oxford / 2023 / 6 / c

Suppose $F:\mathbb{D}\to \mathbb{C}$ were a holomorphic bijection. By the allowed nonvanishing derivative result, $F^{\prime }(z)\ne 0$ throughout $\mathbb{D}$. The holomorphic inverse function theorem gives a holomorphic inverse locally at every image point. Since $F$ is globally bijective, these local inverses agree with its global inverse $H:\mathbb{C}\to \mathbb{D}$.

Thus $H$ is an entire function with $|H|<1$. The Liouville theorem forces $H$ to be constant, contradicting its being an inverse. ThereforeThis is the analytic obstruction behind the fact that the complex plane and unit disc are homeomorphic but not conformally equivalent.

Yes. An explicit holomorphic surjection is the rational functionIts denominator never vanishes inside the unit disc. To prove surjectivity, take any $w\in \mathbb{C}$ and choose a square root $r$ of $w$ with $\mathrm{Re}r\ge 0$; one of the two opposite roots always has this property. Define $z=r/(1+r)$. Then $1+r\ne 0$ andso $|z|<1$. Also $z/(1-z)=r$, and consequently $F(z)=r^2=w$.

This is a holomorphic surjection from the unit disc onto the complex plane. It does not contradict part (i): it is not injective. For example, $F(i/(1+i))=F(-i/(1-i))=-1$, with two distinct arguments of modulus $1/\sqrt{2}$.