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.