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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?

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Why do you emphasize "in higher dimensions"? Everything happens in the space spanned by $x$, $y$ and $z$, so this reduces to the three-dimensional case. – joriki Nov 14 '11 at 16:50
I removed the (soft-question) tag. I don't think asking for the simplest proof of something counts as "not admitting a definitive answer". – joriki Nov 14 '11 at 16:52

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 (just before ) 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.

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