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. 
 A: As you suggest, a homeomorphism $S^2=\{(x,y,w)\in\mathbb{R}^3 \colon x ^2+ y ^2+ w ^2=1\} \rightarrow \mathbb{CP}^1$ can be constructed in quite a natural manner by factoring through the extended complex plane  $\bar{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$, i.e. the one-point compactification of $\mathbb{C}$. Namely, there are rather natural homeomorphisms $S^2\rightarrow \bar{\mathbb{C}}$ and $\bar{\mathbb{C}}\rightarrow \mathbb{CP}^1$. Composing them yields a homeomorphism $S^2 \rightarrow \mathbb{CP}^1$.
On the one hand, the conjugated stereographic projection, say of the south pole $(0,0,-1)$, extends to a map $$\phi: S^2\rightarrow \bar{\mathbb{C}}; \ (x,y,w)\mapsto \begin{cases}\frac{x}{1+w}-i\frac{y}{1+w} & w\neq-1\\
\infty & w=-1.\end{cases}$$ We could also have used the non-conjugated stereographic projection, but out of habit I chose the conjugated one.*
On the other hand, there is the inverse of a kind of "extended division operator" which Paul Sinclair mentions in the comments:
$$\psi\colon \bar{\mathbb{C}} \rightarrow \mathbb{CP}^1;\ z\mapsto \begin{cases}[z:1] & z\neq \infty\\
[1:0]& z=\infty.\end{cases}$$
Composing $\phi$ and $\psi$, yields a map
$$f\colon S^2\rightarrow \mathbb{CP}^1; \ (x,y,w)\mapsto \begin{cases}[\frac{x}{1+w}-i\frac{y}{1+w}:1] & w\neq-1\\
[1:0] &w=-1. \end{cases}$$

To check that $f$ defines a homeomorphism, consider the function $$g\colon \mathbb{CP}^1 \rightarrow S^2; \ [z:w]\mapsto \frac{1}{\vert w \vert^2+\vert z\vert^2}(2\operatorname{Re}(w\bar{z}),2\operatorname{Im}(w\bar{z}),\vert w\vert^2-\vert z \vert^2).$$ Verify that $g$ is well-defined and that it is the two-sided inverse of $f$. (How do you come up with the map $g$? By composing the inverses of $\phi$ and $\psi$.)
By the universal property of the quotient topology, the map $g$ is continuous if and only if the composition $g\circ q\colon \mathbb{R}^4\setminus\{0\}\cong \mathbb{C}^2\setminus\{(0,0)\}\rightarrow S^2\subseteq\mathbb{R}^3$ is continuous. Here, $q$ denotes the canonical quotient map $\mathbb{C}^2\setminus\{(0,0)\}\rightarrow \mathbb{CP}^1$. Since each component function of $g\circ q$ is the product and sum of continuous maps, the composition $g\circ q$ is continuous. Hence, the canonical projection $q$ is continuous.
Next, recall the following result from topology:

Let $X$ be a compact topological space. Let $Y$ be a Hausdorff space. Then any bijective continuous map $g\colon X \rightarrow Y$ is a homeomorphism.

So far, we have shown that the map $g$ is a continuous bijection. Additionally, the complex projective line $\mathbb{CP}^1$ is compact. Namely, it is the image of the three-sphere in $\mathbb{C}^2$ under the continuous quotient map $q$. The two-sphere is Hausdorff (heck, it's even a manifold). The quoted result now implies that $g$ is a homeomorphism.

* If one wants the two-sphere to form a Riemann surface, where one is chart given by the usual stereographic projection from the north pole, one has to choose the conjugated, instead of the-non-conjugated, stereographic projection from the south pole for the second chart. This is because the two stereographic projections are not holomorphically compatible. 
