currently I am struggeling with the following issue that indicates my lack of knowlegde about simply connected Riemann surfaces, it goes as follows: Suppose that $\gamma \colon [0,t] \to \mathbb C \cup \{\infty\}$ is some simple curve in the complex plane that is bounded, i.e. infinity is not part of this simple curve.
By other questions on this website I know that $X \subseteq \mathbb C \cup \{\infty\}$ is simply connected iff both $X$ and its complement are connected, which is the case for $\gamma([0,t]) \subset \mathbb C \cup \{\infty\}$.
Question: Does there exist a conformal map $f \colon \mathbb D \to \mathbb C \cup \{\infty\} - \gamma([0,t])$ that is onto?
My thoughts: Since $\hat{\mathbb C} \setminus \gamma([0,t])$ is simply connected, I want to apply the uniformization theorem for simply connected Riemann surfaces. But I am unsure if $\hat{\mathbb C} \setminus \gamma([0,t])$ is a Riemann surface and that the conformal equivalence is given by $\mathbb D$.
My try: It still should be a Riemann surface, since one can take the charts as an atlas
- $U_1 = \mathbb C^{\ast} \setminus \gamma([0,t])$ with $\phi_1(z) = \frac{1}{z}$ and
- $U_2 = \mathbb C \cup \{\infty\} \setminus \gamma([0,t])$ with $\phi_2(z) = z$.
Now suppose the above is true then applying the uniformization theorem we know that $\hat{\mathbb C} \setminus \gamma([0,t])$ is conformally equivalent to either the unit disk, the Riemann sphere or the complex plane. We rule out the latter two by observing that
- I guess the Riemann sphere can be ruled out by invariance of domain, i.e. the punctured Riemann sphere and the Riemann sphere cannot be homeomorphic to each other.
- I guess since the curve cuts out more than two points we can apply an argument directly given by link
Is the above reasoning correct?