Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 6 / a

The Riemann mapping theorem gives a conformal bijection $G\to D$ exactly whengiven the hypothesis that $G$ is a simply connected domain. A domain is nonempty, open, and connected, so no extra nonemptiness qualification is needed.

For necessity, a conformal bijection $\mathbb{C}\to D$ would be a bounded nonconstant entire function, contradicting the Liouville theorem. For sufficiency, $G$ is a proper simply connected domain, exactly the hypothesis of the Riemann mapping theorem.

Use the univalent functions on $D$, namely the univalent functions $F:D\to \mathbb{C}$. Each has nonzero derivative, so its image $F(D)$ is a domain conformally equivalent to $D$, and is therefore simply connected. This image is proper: otherwise $F^{-1}:\mathbb{C}\to D$ would be a bounded nonconstant entire function, contrary to the Liouville theorem.

Conversely, every proper simply connected domain $G\subset \mathbb{C}$ is $F(D)$ for some such $F$, by taking the inverse of a Riemann mapping theorem bijection. Thus these functions parameterize all proper simply connected domains by their images. The complex plane itself is the exceptional simply connected domain and must be listed separately.

The parameterization has redundancy: two univalent functions $F_1,F_2$ have the same image if and only if $F_2=F_1\circ \varphi$ for an automorphism of the unit disk $\varphi$. All such automorphisms have the formIf an interior point $w_0\in G$ is prescribed, the normalization $F(0)=w_0$ and $F^{\prime }(0)>0$ real removes this redundancy. Equivalently one can describe these maps through their convergent Taylor series on $D$, imposing univalence on the resulting holomorphic functions.