Given the map: $$f:CP^1 \to S^2\ ,\ f[z:w] = \left(\frac{2\mbox{Re}(w\bar{z})}{|w|^2+|z|^2},\frac{2\mbox{Im}(w\bar{z})}{|w|^2+|z|^2}, \frac{|w|^2-|z|^2}{|w|^2+|z|^2}\right)$$
How would I go about constructing an inverse? I have done something similar in the past when I needed to find the inverse for a stereographic projection, but I used geometrical methods, and here I'm having trouble visualizing things. I have tried to solve this algebraically but I can't seem to separate $z$ and $w$ from their conjugates.
My aim is to eventually prove that this map is a homeomorphism. Thanks
The inverse mapping is
$$g:S^2 \to CP^1: (\xi_1,\xi_2,\xi_3) \mapsto [z:w]=\frac{\xi_1-i\xi_2}{1+\xi_3} \; .$$
All you need is the proportion $[z:w]$. So set $|w|^2+|z|^2=w\bar{w}+z\bar{z}=1$ for the sake of ease. Then also note that
$$\xi_1=2\mbox{Re}(w\bar{z}) = w\bar{z}+\bar{w}z$$
and
$$\xi_2=2\mbox{Im}(w\bar{z}) = -i(w\bar{z}-\bar{w}z)$$
and finally
$$\xi_3=w\bar{w}-z\bar{z} \; .$$
Using the last relation and our condition we see that
$$1+\xi_3 = 2w\bar{w}$$
while combining the relations for $\xi_1$ and $\xi_2$ appropriately we get
$$\xi_1-i\xi_2 = 2\bar{w}z$$
which after division with the previous relation gives us the mapping.