# Riemann mapping theorem with boundaries and corners

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?