Looking for a homeomorphism $\mathbb{C}P^1 \cong S^2$ I want to show $\mathbb{C}P^1 \cong S^2$ by explicit construction. Everything I tried so far did not work out unfortunately :(
Any hints? 
 A: Let $\widehat{\mathbb{C}}$ be the one-point compactification. It is standard to see that $\widehat{\mathbb{C}}\cong S^2$ via the stereographic projection. Explicitely you have a map $S^2-{(0,0,1)} \rightarrow \mathbb{R}^2\cong \mathbb{C}$ given by $(x_1, x_2,x_3) \mapsto \frac{1}{1-x_3}(x_1, x_2)$, with inverse given by $(y_1, y_2) \mapsto \frac{1}{1+ \Vert y \Vert ^2}(2y_1, 2y_2, \Vert y \Vert ^2 -1)$. Now use the fact that a homeomorphism between local compact Hausdorff spaces can be extended to an homeomorphism to their one-point compactifications.
I assume that $\mathbb{C}P^1=(\mathbb{C}^2-{(0,0)})/\sim$ where the relation is : $$ (z_0,z_1) \sim (z_0',z_1')\mbox { iff } \exists \lambda \in \mathbb{C}^*: z_i=\lambda z_i'.$$
Now define a map $f: \mathbb{C}^2-{(0,0)}\rightarrow \widehat{\mathbb{C}}$ by $(z_0,z_1)\mapsto \frac{z_0}{z_1} $ if $z_1 \neq 0$, and $(z_0, 0)\mapsto \infty$ otherwise. The universal property of the quotient gives you a unique map $\widehat{f}:\mathbb{C}P^1\rightarrow \widehat{\mathbb{C}}$ from a compact space to a Hausdorff space. So it only remains to prove that $f$ is continuous in order to show that $\widehat{f}$ is your desired homeomorphism.
