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.

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.
