I was reading this paper by Hollands and Yazadjiev, where on page 760, they claim that since $\hat{M}$ is an orientable simply connected analytic $2$-manifold with boundaries and corners, we may map it analytically to the upper complex half plane $\mathbb{H}^2=\{\zeta\in\mathbb{C}\mid\operatorname{Im}\zeta>0\}$ by the Riemann mapping theorem. My guess is that they are really talking about mapping the **interior** of $\hat{M}$ to $\mathbb{H}^2$, and that the Riemann mapping theorem refers to the usual uniformization theorem, which in this case, if I understood their argument correctly, states that $\operatorname{int}\hat{M}$ is conformally equivalent to either $\mathbb{S}^2$, $\mathbb{C}$ or $\mathbb{H}^2$. We can certainly rule out the case $\mathbb{S}^2$ since $\operatorname{int}\hat{M}$ is non-compact, but I'm not sure what rules out the case $\mathbb{C}$. Does this have something to do with the fact that $\hat{M}$ has non-empty boundary? Also, why can we conclude that $\operatorname{int}\hat{M}$ is simply connected from the fact that $\hat{M}$ is?

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