# The set $\{z\in \mathbb C: |z|\geq 1\}$ with point at infinity is homeomorphic to closed unit disk.

While reading about one way to decompose the extended complex plane (or $\mathbb C\mathbb P^1$), I saw without proof that the set $\{z\in \mathbb C: |z|\geq 1\}$ with a point at infinity added is homeomorphic to the closed unit disk.

I am not sure exactly how this works. I know that the upper half-plane is homeomorphic to the open unit disk, but I am not sure if this helps.

Could someone please explain how we get the desired homomorphism?

• $z\mapsto \frac{1}{z}$? – carmichael561 Aug 4 '17 at 3:54

The set $\mathbb C\cup \{\infty\}$ is homeomorphic to $S^2$, for instance by stereographic projection. Under this homeomorphism, $\{\infty\}$ gets sent to the "north pole". The upper hemisphere corresponds precisely to $\{z\in \mathbb C:|z|\ge1\}$. To do this draw the line through the north pole and each point of the sphere until you hit the complex plane. ..