# $\mathbb{P}^1$ 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...

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Hint: prove that it's continuous, one-one, onto, and has a continuous inverse. In case you already knew that, which of those four parts can you do, and which one(s) give you trouble? – Gerry Myerson Jun 27 '12 at 3:16
If you multiply each of $w$ and $z$ by $100$, you still have the same point in $\mathbb{P}^1$, and the first two components of this triple become $100^2$ times as big, but the third one stays the same, so this can't be right. – Michael Hardy Jun 27 '12 at 4:51
@MichaelHardy this is indeed true, this is Problem I.2 C from Miranda's book on Riemann surfaces and algebraic curves, as you said, something isn't right... – Jr. Jun 27 '12 at 15:38
@GerryMyerson I don't know how to prove the continuity, of $f$ and its inverse(I'm still trying to find its inverse) – Jr. Jun 27 '12 at 15:45
I see you've now put in the denominators in the first two components, so maybe now it works. – Michael Hardy Jun 27 '12 at 18:47

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 $\mathbb{R}^3$.
$$f[z:w]=\left(\frac{\operatorname{Re}( w \bar{z})}{|w|^2 + |z|^2}, \frac{\operatorname{Im}(w\bar{z})}{|w|^2 + |z|^2},\frac{|w|^2-|z|^2}{|w|^2+|z|^2}\right).$$
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.
Alternatively, think of $\mathbb{P}^1$ as the quotient of $S^3$ by unit length complex numbers. In this case, the OPs formula is valid (though, one has $|w|^2+|z|^2 = 1$, so it was superfluous to write the denominator.) – Jason DeVito Jun 27 '12 at 14:42
@MGNewman How does one check that $f$ is continouos? – Jr. Jun 27 '12 at 18:29
@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 $\mathbb{P}^1$). – MGN Jun 27 '12 at 18:46