Inversion map in higher dimensions (preserving angles' size) Consider the inversion with respect to the sphere $S^n \subset R^{n+1}$, that is the map 
$$ \rho \colon x = (x_1,\dots x_{n+1}) \in R^{n+1}-\mathrm{O} \mapsto \frac{x}{\left \| x \right \|^2} \in R^{n+1}-\mathrm{O}.$$
If $x$, $y$ and $z$ are three nonzero points in $ R^{n+1}$, then the angle between the segments $yx$ and $yz$ is equal (in magnitude) to the angle formed by $\rho(y)\rho(x)$ and $\rho(y)\rho(z)$.
I am looking for the simplest proof of this fact (basically that $\rho$ is anticonformal), possibly a really elementary one in which we don't make use of the transformation's Jacobian. Do you know any?
 A: The right way to do this is to appeal to stereographic projection. First, as joriki mentions, this all happens in an $\mathbf R^4,$ so the stereographic projection is from the North Pole $(0,0,0,1)$ of the standard unit sphere $\mathbf S^3.$  
A complete proof that stereographic projection,  with $\mathbf S^2 \subseteq R^3,$ is conformal is given on pages 248-249 of Geometry and the Imagination by David Hilbert and S. Cohn-Vossen. Only pictures are used, no calculations. I am working on improving that to a picture-only proof for conformality of stereographic projection with $\mathbf S^3 \subseteq R^4.$ It is true, anyway.
Now, see the end of the section 
http://en.wikipedia.org/wiki/Stereographic_projection#Properties 
(just before http://en.wikipedia.org/wiki/Stereographic_projection#Wulff_net )
where they show how inversion is just stereographic projection with a vertical reflection in between, in this case a point $(x,y,z,\omega) \in S^3 \mapsto (x,y,z, -\omega).$ A reflection is  conformal and  an isometry.
Los Links:
http://en.wikipedia.org/wiki/Stereographic_projection
http://en.wikipedia.org/wiki/3-sphere#Stereographic_coordinates
http://en.wikipedia.org/wiki/Hypersphere#Stereographic_projection 
http://en.wikipedia.org/wiki/N-sphere
