I want to show the red line. But from the $(13)$, I can't get the expression of $f^{-1}$ even for $n=2$. I think there's a geometry to it, but I don't see it.
What I try: Assume $n=2$, then $p=(x,y),p_0=(0,-2)$, then $$ p-p_0=(x,y+2),~~~~~~|p-p_0|^2=x^2+(y+2)^2 $$ Therefore, $$ f(p)=\frac{(4x,4-x^2-y^2)}{x^2+(y+2)^2} $$ assume $f(p)=(f_1,f_2)$, then $$ f_1=\frac{4x}{x^2+(y+2)^2},~~~f_2=\frac{4-x^2-y^2}{x^2+(y+2)^2} $$ next, I should calculate $$ x=x(f_1,f_2),~~~y=y(f_1,f_2) \tag{1} $$ if so, I know what is $f^{-1}$.
But, on the one hand, I feel $f$ like inversion, There should be a geometric way. On the other hand, it is complex to get (1) and I'm a little lazy. So I let it go.
PS(2023-12-11): The (13) of below picture imply $$ f(p)+(0,...,0,1)=4\frac{p-p_0}{|p-p_0|^2} $$ So, there is $ |f(p)+(0,...,0,1)|=\frac{4}{|p-p_0|}. $ Namely, $$ |f(p)+(0,...,0,1)|\cdot |p-p_0| =4 $$ Let $q=f(p)$, then $p=f^{-1}(q)$, there is $$ |q+(0,...,0,1)|\cdot |f^{-1}(q)-p_0| =4 \tag{2} $$ Obviously, red line is not completely right. Only spheres center $(0,...,0,-1)$ is mapped to a sphere by $f^{-1}$. And this sphere is not contained in the upper half space.
Pictures below is from 177-178th page of do Carmo's Riemannian Geometry.
