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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...

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    So far I'm the onl$y$ person who's up-voted this question.2012-06-27

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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.

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    @Jr. I believe you can appeal to the fact that $f$ is continuous if and only if its coordinate functions are continuous. So it only remains to observe that each coordinate is the composition of continuous functions (outside of $(w,z) = (0,0)$, which is not in $\mathbb{P}^1$).2012-06-27