$\mathbb{P}^1$ 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

1 Answer

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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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