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In the Half-plane model, a lobachevski transform is a möbius transform that maps the hyperbolic plane into itself and the infinity line into itself.

I've already proved that given two sets $\{A',B',C'\}$ and $\{A,B,C\}$ of points in the Riemann sphere, there exists exactly two möbius transforms mapping $A \to A'$, $B \to B'$ and $C \to C'$, where if $\phi$ is one of them, then $\rho o \phi$ is the other, where $\rho$ is the reflexion over ther conformal circle through $A',B',C'$. (Proposition 1)

But then I have the following statement: Given two sets $\{A',B',C'\}$ and $\{A,B,C\}$ of points of the infinity line, there is a unique lobachevski transform mapping $A \to A'$, $B \to B'$ and $C \to C'$. And the proof the author gave is: Let $f_1$,$f_2$ be the two Möbius transforms from Proposition 1. One of those, say $f_1$, maps the upper half-plane into the lower half-plane and vice versa. Then, since the other is the composition of $f_1$ with the reflexion over the infinite line, then it is a Lobachevski.

I don't understand why one of the möbius transforms must maps the upper plane into the lower plane and vice versa.

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I think you made a mistake in your proof of proposition 1

I've already proved that given two sets $\{A',B',C'\}$ and $\{A,B,C\}$ of points in the Riemann sphere, there exists exactly two möbius transforms mapping $A \to A'$, $B \to B'$ and $C \to C'$, where if $\phi$ is one of them, then $\rho o \phi$ is the other, where $\rho$ is the reflexion over the conformal circle through $A',B',C'$. (Proposition 1)

The second transformation only works if $\{A',B',C'\}$ are on a hyperbolic line. (an euclidean line orthogonal to the boundary , or a circle with its centre on boundary ) . When the points are not on such a hyperbolic line then $\rho$ does not map the hyperbolic plane into itself.

The second statement you have to proof is I think to show that there is only one transformation (like for 3 points not on an hyperbolic line)

On the boundary some special situations exist:

  • Betweeness does not hold, (like on a circle) you can have transformations like $ A \to A , B \to C \text{ and } C \to B $
  • Fixing $A$ and $B$ does not fix $C$ (2 points on the boundary line are always infinitly far away from eachother) you can have transformations like. $A \to A , B \to B , C \to D \text{ with } C \not= D $.

Hope this helps

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