Lemma: Given two sets of three points of Riemann Sphere, there is exactly two Möbius transforms mapping $(A,B,C) \mapsto (A',B',C')$. If $\phi_{1}$ is one of them, then the other is $r \circ \phi_{2}$, where $r$ is the reflexion through the conformal circle defined by $A',B',C'$.
The proof of this lemma is alright, I understood it.
Theorem: Given two sets of improper points (points of $r_{\infty}$) $\{A,B,C\},\{A',B',C'\}$, there is exactly one Möbius transform $f$ mapping $(A,B,C) \mapsto (A',B',C')$, such that $f(\mathbb{H}^2)=\mathbb{H}^2$ and $f(r_{\infty}) = r_{\infty}$.
Proof: Let $\phi_{1},\phi_{2}$ the two Möbius given by Lemma. One of those Möbius, say $\phi_{1}$, maps the upper half-plane into the lower half plane and vice-versa, so $\phi_{2}$ will be a composition $r\circ \phi_{1}$, where $r$ is the refletion in $r_{\infty}$, hence $f(\mathbb{H}^2)=\mathbb{H}^2$ and $f(r_{\infty}) = r_{\infty}$.
My problem is exactly with the bold part. How can the author assume that it will take the upper half-plane into the lower and vice-versa? Since all it is said about $\phi_{1}$ is that it is a Möbius, it can be something like a reflexion through a euclidean line orthogonal to $r_{\infty}$ (Whose upper ray is a hyperbolic line), and then won't take the upper half-plane in the lower... Can someone please explain this to me? Thanks.