I am presently a learner of Hyperbolic Geometry and am using J. W. Anderson's book $Hyperbolic$ $Geometry$. Now the author presents a sketch proof of why every circle preserving homeomorphism in $\overline{\mathbb{C}}$ is an element of the general Möbius group, which is what I am struggling to understand.
First, a brief outline.
Let $f$ be an element of the set of all circle preserving homeomorphisms which we denote Homeo$^{C}(\overline{\mathbb{C}})$ and let $p$ be a Möbius transormations that maps the triples $(f(0),f(1),f(\infty))$ to $(0,1,\infty)$
Then we see that $p\circ f(0) = 0, p\circ f(1) = 1$ and $p \circ f(\infty)=\infty$, and since $p \circ f(\mathbb{R}) = \mathbb{R}$, either such a composition maps the upper half of the complex plane, $\mathbb{H}$ to itself or the lower half of the complex plane.
If $p \circ f(\mathbb{H}) = \mathbb{H}$, we take $ m =p$, while if $m \circ (\mathbb{H})$ goes to the lower half, we just take $m = W \circ p$, where $W(z) = \overline{z}$
Now here is the thing I don't understand:
- Let $A$ be an euclidean circle in $\mathbb{C}$ with euclidean centre $\frac{1}{2}$ and radius $\frac{1}{2}$. Let $V(0), V(1)$ be the vertical lines through the points $x=0$ and $x=1$.
Can anybody explain why as $V(0)$ and $V(1)$ are vertical tangents to the circle, then these lines under the map $m \circ f(z)$, namely $m \circ f\Big(V(0)\Big)$ and $m \circ f\Big(V(1)\Big)$ are again vertical tangents to the circle $m \circ f(A)$ at $m \circ f(0) = 0$ and $m \circ f(1) = 1$?
I am trying to conclude from here that $m \circ f = Id_z$ , the identity transformation.
Thanks,
Ben