If a punctured neighborhood of a Riemann surface is biholomorphic to a punctured disk, is the neighborhood biholomorphic to a disk?

Question

Let $R$ be a Riemann surface, $P_0\in R$ a point, and $D\subset R$ an open neighborhood of $P_0$. Suppose $D - P_0$ is biholomorphic to the punctured unit disk $\{z : 0 < |z| < 1\}\subset\mathbb{C}$. Is $D$ biholomorphic to the unit disk $\{z : |z| < 1\}\subset\mathbb{C}$?

I ask this because Chapter 1, $\S3.2$ in Kurt Strebel's book "Quadratic differentials" seems to make an unnecessarily complicated construction to avoid the failure of the titular question, but I can't think of an example where the question has a negative answer.

Answer 1

For instance, you can argue that $D$ is simply-connected and, hence, by the uniformization theorem, is biholomorphic to the unit disk, or to ${\mathbb C}$ or to $S^2$. Since ${\mathbb C}$ minus a point is not biholomorphic to the punctured unit disk $\Delta^*$ (this follows for instance from the Liouville's theorem: Compose the exponential map ${\mathbb C}\to {\mathbb C}^*$ with the supposed conformal map of ${\mathbb C}^*$ to the punctured unit disk), you rule out the second case. The third case is equally impossible (I trust you can prove this). Hence, $D$ is biholomorphic to the unit disk.

Given this, you can prove more, namely, that the initial conformal map $D-P_0\to \Delta^*$ is bounded, hence, extends to a conformal map $D\to \Delta$ (use Riemann's theorem on removable singularities).