# Complex Projective Line is homeomorphic to the 2-sphere

I was wondering if someone could help me out with proving the fact that the complex projective line is homeomorphic to the 2-sphere.

I've defined the complex projective line $$\mathbb{CP}^1 = (\mathbb{C}^2 \backslash \textbf{0}) \backslash \sim$$, where $$\sim$$ is the equivalence relation such that $$(u,v) \sim (z,w) \iff (u,v) = (\lambda z, \lambda w)$$ for some $$\lambda \in \mathbb{C}$$. Hence it has the quotient topology induced by the natural projection. My attempts have included realising that $$\mathbb{CP}^1 = \{[z:1] : z \in \mathbb{C} \} \cup [1:0]$$, i.e. effectively the Riemann sphere (a copy of $$\mathbb{C}$$ plus an extra point), and of course the Riemann sphere is homeomorphic to the 2-sphere via stereographic projection, but I'm having trouble rigorously showing all of these homeomorphisms. I know that if I can find a map that is a continuous bijection and is also open or closed, then it is a homeomorphism, but proving each of these statements is what I'm having trouble with. Particularly the open/closed part I think.

If anyone could help me out I would greatly appreciate it!

Thanks.

• The map you are looking is for is $$f: \overline{\Bbb C} \to \Bbb C\Bbb P^1 : z \mapsto \begin{cases} [z:1] & z \ne \infty\\ [1:0] & z = \infty\end{cases}$$ whose inverse is $$g : \Bbb C\Bbb P^1 \to \overline{\Bbb C} : [u:v] \mapsto \begin{cases} \frac uv & v\ne 0\\ \infty & v = 0\end{cases}$$ Composing this with the natural projection from $\Bbb C^2 \setminus \{(0,0)\}$ is just the division operator extended to dividing by zero. – Paul Sinclair Feb 24 at 16:41
• @PaulSinclair Ah okay, thanks. The maps make sense and I had been playing around with something of the sort, but how would I go about proving that they're continuous? Particularly proving $f$ is continuous and $g$ is at the point $\infty$ (obviously apart from this point it is, as it's just a rational map). – jamesmbcn Feb 27 at 2:56