How to prove that a sphere without two points is homeomorphic to a hyperbolic cylinder in $\mathbb{C}^2$?

Question

If I understood correctly, then a sphere without two points is the same as a standard cylinder. how then to prove that a hyperbolic cylinder is homeomorphic to a standard one? I have no idea how to correctly construct a mapping, especially from a hyperbolic cylinder to a standard one

Answer 1

I have no idea what your background or tools are. The easiest way for me to see this is to think about the closure of the algebraic curve $y^2=x(x-1)$ in $\Bbb CP^2$. By projecting on the $x$-axis, you get a branched $2$-fold cover of $\Bbb CP^1$, branched at the two points $x=0$ and $x=1$. (The point at infinity has two preimages.) Such a branched cover (by, for example, the Riemann-Hurwitz formula) is another copy $X$ of $\Bbb CP^1$ [think of the usual branched cover $z\rightsquigarrow z^2$, branched over $0$ and $\infty$]. But now I must remove two points from $X$, namely the two points lying over $\infty$. So $X=\Bbb CP^1 - \{p,q\} \cong \Bbb C - \{q\}$, and the punctured plane is indeed homeomorphic to a cylinder (think polar coordinates).

EDIT: Here is a more concrete approach. By considering the pencil of lines through the origin (slope $t$), we parametrize the curve by $$x=\frac1{1-t^2},\quad y=\frac t{1-t^2}, \quad t\in\Bbb CP^1-\{\pm1\}.$$ This parametrization gives the explicit homeomorphism to $\Bbb C^*$. If you so wish, you can compose with an automorphism (linear fractional transformation) of $\Bbb CP^1$ carrying $0,\infty$ to $\pm 1$, and then the parametrization will be explicitly by $\Bbb C-\{0\}$.