$\mathbb{P}^1$ is homeomorphic to $\mathbb{S}^2$ I'd like a hint to prove that this function is a homeomorphism:
$$f[z:w]=\left(\frac{\operatorname{Re}( w \bar{z})}{|w|^2 + |z|^2}, \frac{\operatorname{Im}(w\bar{z})}{|w|^2 + |z|^2},\frac{|w|^2-|z|^2}{|w|^2+|z|^2}\right)$$
of $\mathbb{P}^1$ onto $\mathbb{S}^2$. Thanks.
ADDED(06/27/12):
The previous definition of $f$ was wrong, this new one seems to work...
 A: Sorry, I would write this as a comment, but I don't actually have that privilege currently.  Anyways, as Michael pointed out, your map doesn't seem to be well-defined since it can take a ratio in $\mathbb{P}^1$ to two different values in $\mathbb{R}^3$.  Since you can scale in this way, it makes sense that whatever map you choose, in order to be well-defined, should have its image normalized in some way within $\mathbb{R}^3$.  
Thus, I would first suggest modifying your map to look something like:
$$f[z:w]=\left(\frac{\operatorname{Re}( w \bar{z})}{|w|^2 + |z|^2}, \frac{\operatorname{Im}(w\bar{z})}{|w|^2 + |z|^2},\frac{|w|^2-|z|^2}{|w|^2+|z|^2}\right).$$
Clearly such a homeomorphism exists since you can identify $\mathbb{P}^1$ with the one-point compactification of $\mathbb{C}$, and this looks a lot like stereographic projection, so I would think that this map probably gives it to you.  Now you can proceed as Gerry suggested, checking homeomorphism conditions as you would any map in this instance.
