0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ $z\mapsto \frac{1}{z}$? $\endgroup$ – carmichael561 Aug 4 '17 at 3:54
0
$\begingroup$

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. ..

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.