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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.

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    $\begingroup$ I take it that $r_\infty = \Bbb R\cup\{\infty\}$. The two mappings from the lemma map the real numbers $A,B,C$ to other real numbers $A',B',C'$, and therefore each one maps the (conformal) circle defined by $A,B,C$ to the circle defined by $A',B',C'$; in other words, each one maps $r_\infty$ to $r_\infty$. Therefore each one maps the upper half-plane (is that what $\Bbb H^2$ is?) either to the upper half-plane or the lower half-plane. $\endgroup$ – Greg Martin Sep 14 '16 at 23:26
  • $\begingroup$ why mapping $r_{\infty}$ to itself would mean mapping upper half-plane to lower half-plane? $\endgroup$ – user286485 Sep 14 '16 at 23:58
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    $\begingroup$ @Alnitak: As Greg said, sending $r_\infty$ to itself does not imply that $H^2$ maps to the lower half-plane. But if the image is the upper half-plane, you are done. $\endgroup$ – Moishe Kohan Sep 15 '16 at 2:10
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When they say that $A, B, C$ is mapped to $A', B', C'$ this means the order of the points is respected, i.e. if $\phi$ is the Mobius map they are talking about, then $\phi(A) = A', \, \phi(B) = B', \, \phi(C) = C'$ and not $\phi(A) = B'$ or some permutation of that. The order is important. Now, a Mobius transformation maps circles to circles, circles to lines, lines to circles or lines to lines. So if $A,B,C \in r_{\infty}$ and $A',B',C' \in r_{\infty}$ too, then the three points $A,B,C$ define the straight line $r_{\infty}$ and since $\phi_1$ is Mobius, $\phi_1(r_{\infty})$ is either a straight line or a circle. But the images of $A,B,C$ under $\phi_1$ are $A',B',C'$ and so the image of the straight line through $A,B,C$, which is $r_{\infty}$, is the circle or line passing through $A',B',C'$, which is the line $r_{\infty}$ itself. Thus $\phi_1(r_{\infty}) = r_{\infty}$. As the points $A,B,C$ are already mapped to $A',B',C'$, then the Mobius involution $r$ should keep them fixed, i.e. $r(A') = A', r(B')=B', r(C') = C'$ which means it should fix the whole line $r_{\infty}$ pointwise, i.e. $r$ acts as identity on $r_{\infty}$. Clearly, $r$ cannot be any of your suggestions, such as reflections in lines orthogonal to $r_{\infty}$ will permute points on $r_{\infty}$. The only option is a reflection in $r_{\infty}$.

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