Inverse of Multivariable Functions on Manifolds Consider the unit circle, $S=${$(x,y) | x^2 + y^2 =1$} with the stereographic charts $(U_N,\phi_N)$, $(U_S,\phi_S)$
i.e. $U_N=S$ \ {$(0,1)$}, $\phi_N((x,y))=\frac{x}{1-y}$
$U_S=S$ \ {$(0,-1)$}, $\phi_S((x,y))=\frac{x}{1+y}$
How would I find ${\phi_N}^{-1}$ and ${\phi_S}^{-1}$?
These functions are bijective on their domains ($U_N$ and $U_S$ respectively) so their inverses exist.
For ${\phi_N}^{-1}$, I have considered a point $z$ in $\phi_N(U_N)$ and set $\frac{xy}{1-y}=z$. I believe I need to write both x and y in terms of $z$ to get something like ${\phi_N}^{-1}(z)=(f(z), g(z))$ where I need to find $f(z)$ and $g(z)$. And then similarly for $\phi_S$.
How would I do this? Thank you for any help!

Let $\phi_N((x,y))=z=\frac{x}{1-y}$
Then $x=z(1-y)$
Substituting this into $x^2+y^2=1$ we have $z^2{(1-y)}^2+y^2=1$
Then, solving this for $y$, (I used the quadratic formula) $$(1+z^2)y^2-2z^2y+z^2-1=0$$
$$y=\frac{2z^2\pm\sqrt{4z^4-4(1+z^2)(z^2-1)}}{2(1+z^2)}$$
$$y=\frac{2z^2\pm2\sqrt{z^4-(z^4-1)}}{2(1+z^2)}$$
$$y=\frac{z^2\pm1}{1+z^2}$$
Now, if we had $y=\frac{z^2+1}{z^2+1}=1$, then this is not actually in the domain of $\phi_N$, since $(x,1)\notin$ $U_N$ for any x.
So we must have $$y=\frac{z^2-1}{z^2+1}$$
Then, from $x=z(1-y)$, we can just substitute our expression for y in to get
$$x=\frac{2z}{z^2+1}$$
Thus, $${\phi_N}^{-1}(z)=(\frac{2z}{z^2+1}, \frac{z^2-1}{z^2+1})$$
Similar procedure for finding ${\phi_S}^{-1}$
