Proving a function covers the unit sphere How can I show $f:\mathbb R^2\to\mathbb R^3$ defined as$$f(u,v)=\left(\frac{2u}{u^2+v^2+1},\frac{2v}{u^2+v^2+1},\frac{u^2+v^2-1}{u^2+v^2+1}\right)$$covers the whole unite sphere except the point $(0,0,1)$? It’s easy to show that the image of $f$ lies in $S^2$ but I can’t show it covers every point except a pole.
 A: The proof is purely geometrical and intuitive.
If you consider another function:
$$
f'=(2u,2v,u^2+v^2-1)
$$
(The same function, just without the denominators) You can see easily that this is a paraboloid poiting towards -Z which ends at $(0,0,-1)$. By normalising the function $f'$, you get your $f$. It projects the cone onto a unit sphere. And you can see easily that all points on the sphere get covered except the point $(0,0,1)$. If you make a line through origin, it intersects the paraboloid twice except the situation when the line leads along the Z axis. The intersections stay on the line after the normalisation. These two intersections always project to two points on the unit sphere. And each line leading through the origin intersects both the sphere and the paraboloid twice (except the one situation where it intersects only once).
A: Take any $(x,y,z)\in S^2$, so $x^2+y^2+z^2=1$. Let $(u,v)=(\frac{x}{1-z},\frac{y}{1-z})$. [This is the inverse of the stereographic projection map.] 
Then $$u^2+v^2+1=\frac{x^2}{(1-z)^2}+\frac{y^2}{(1-z)^2}+1=\frac{x^2+y^2+(1-z)^2}{(1-z)^2}=\frac{2-2z}{(1-z)^2}=\frac{2}{1-z}$$ if $z\ne1$. We also have 
$$u^2+v^2-1=\frac{x^2+y^2-(1-z)^2}{(1-z)^2}=\frac{x^2+y^2-z^2+2z-1}{(1-z)^2}=\frac{-2z^2+2z}{(1-z)^2}=\frac{2z(1-z)}{(1-z)^2}=\frac{2z}{1-z}$$ if $z\ne1$.
Therefore, if $z\ne1$, we have
$$f(u,v)=\left(\frac{2x}{1-z}\cdot\frac{1-z}{2},\frac{2y}{1-z}\cdot\frac{1-z}{2},\frac{1-z}{2}\cdot\frac{2z}{1-z}\right)=(x,y,z).$$
So every $(x,y,z)\in S^2$ except the north pole is in the image.
