Riemann surfaces

Question

It is known that $\mathbb{H}$ (upper hyperplane), $\mathbb{C}$, $\mathbb{C} \cup \infty$ (riemann sphere) are riemann surfaces. $\mathbb{H} \cup i\infty$ is not, but what is the principal difference between this case and $\mathbb{C} \cup \infty$?

Answer 1

To answer this, one must understand that there is no isomorphism between an annulus $A_a=\{a<\vert z\vert <1\}$ and the punctured disk $\{0<\vert z\vert <1\}$.

Let us think about $\bf H$ as a disk ${\bf D}=\{ \vert z\vert <1\}$, and consider the annulus $A= A_{1/2}$ ; assume that there exists a complex structure on ${\bf D}\cup \infty$ which restricts to the given structure on $A$. Then $A\cup \infty$ would be a simply connected Riemann surface, hence either $\bf C$ of $\bf D$.

It cannot be $\bf C$ as $\bf C\setminus \{1pt\}$ is not isomorphic to a punctured disk (the universal cover of $\bf C\setminus \{1pt\}$ is $\bf C$). So it must be $\bf D$, and (as $:bf D$ is homogeneous) removing a point would provide an isomorphism between the punctered disk and $A_{1/2}$.

This proof is perhaps to complicated (it uses Poincaré Koebe uniformisation) hopefully there is a simpler one using just the Riemann mapping theorem?