Riemann mapping theorem

Statement

Let $U$ be a nonempty open subset of $\mathbb{C}$ with $U \neq \mathbb{C}$. Suppose that $U$ is simply connected: $U$ is connected, and each closed path in $U$ can be continuously deformed in $U$ to a constant path. Let $\mathbb{D} = \{ z \in \mathbb{C} : |z| < 1 \}$. Then there is a biholomorphic map $f$ from $U$ onto $\mathbb{D}$, that is, a bijective holomorphic map with a holomorphic inverse. Moreover, for each $z_0 \in U$ there is exactly one such $f$ with $f(z_0) = 0$ and $f'(z_0)$ a positive real number.

Remarks

A biholomorphic map is conformal, so $U$ and $\mathbb{D}$ are conformally equivalent. The case $U = \mathbb{C}$ is excluded: a holomorphic map $\mathbb{C} \to \mathbb{D}$ is bounded, so it is constant by Liouville's theorem. The uniqueness follows from the Schwarz lemma. Riemann stated the theorem in his thesis of 1851, and Osgood gave the first rigorous proof in 1900.

Explanations

L1 human

The theorem says that every simply connected open set in the plane, except the plane itself, looks like the unit disc to complex analysis. The set can have a very irregular boundary, and the map still exists. For the upper half-plane, the Cayley transform is such a map.