Tessellation and circle homeomorphism.

If $f$ is an orientation preserving homeomorphism from $S^1$ to $S^1$ and $\tau$ is a tessellation of $\mathbb{D}$ (that is a collection of locally finite geodesics decomposing $\mathbb{D}$ into ideal triangles) then $f(\tau)$ is also a triangulation.

(Locally finite: Any point of $\mathbb{D}$ admits a neighborhood meeting only finitely many geodesics in $\tau$. )

I have read this statement many times but haven't seen any proof of it. Can anyone please give a complete proof of this. In particular I want to know why the image will be locally finite.

Given a geodesic $\gamma$, let $r(\gamma)$ be the Euclidean distance from the center of the disk to $\gamma$.

Claim 1. A collection $\Gamma$ of geodesics is locally finite if and only if for every $\epsilon>0$ the set $\{\gamma\in\Gamma: r(\gamma)\le 1-\epsilon\}$ is finite.

Proof: If $\Gamma$ is locally finite, then only finitely many geodesics meet the closed Euclidean disk $\{z:|z|\le 1-\epsilon\}$, by compactness of the latter. In the converse direction, every point of $\mathbb D$ has a neighborhood contained in some set of the form $\{z:|z|\le 1-\epsilon\}$. $\Box$

Let $d (\gamma)$ be the Euclidean distance between the endpoints of $\gamma$.

Claim 2. There is a continuous strictly decreasing function $\phi:[0,1)\to (0,2]$ such that $d(\gamma)=\phi(r(\gamma))$ for every geodesic. Also, $\phi(1-)=0$.

Proof: either of two quantities $d$ and $r$ determines the geodesic up to Euclidean rotation, and therefore determines the other quantity. A rough upper bound for $\phi$ is $\phi(r)\le 2\sqrt{1-r^2}$, which suffices for $\phi(1-)=0$. This bound follows from the fact that a Euclidean line passing at distance $r$ from the origin cuts off a part of the disk with Euclidean diameter $2\sqrt{1-r^2}$. $\Box$

The function $\phi$ could be written down explicitly, but this is not necessary.

The image of a geodesic $\gamma$ under a map $f:S^1\to S^1$ is understood as the geodesic between the images of the endpoints of $\gamma$.

Claim 3. if $\tau$ is a locally finite collection of geodesics, and $f:S^1\to S^1$ is continuous, then $f(\tau)$ is locally finite.

Proof: Given $\epsilon>0$, we must show that $\{ \gamma \in\tau : r(f(\gamma))\le 1-\epsilon\}$ is finite. By the uniform continuity of $f$, there is $\delta>0$ such that $d(f(\gamma))<\phi(1-\epsilon)$ whenever $d(\gamma)<\delta$. Since $\phi(1-)=0$, there is $\tau>0$ such that $d(\gamma)<\delta$ whenever $r(\gamma)>1-\tau$.

The chain of containments $$\begin{split}\{ \gamma \in\tau : r(f(\gamma))\le 1-\epsilon\} &= \{ \gamma \in\tau : d(f(\gamma)) \ge \phi(1-\epsilon)\} \\ &\subseteq \{ \gamma \in\tau : d( \gamma ) \ge \delta\} \\ &\subseteq \{ \gamma \in\tau : r( \gamma ) \le 1-\tau\} \end{split}$$ proves the claim. $\Box$

This takes care of local finiteness. Another thing to check is that $f(\tau)$ is a tesselation when $\tau$ is a tesselation and $f$ is a homeomorphism. First, the endpoints of geodesics in $f(\tau)$ are dense in $S^1$, because the image of a dense set is dense. Second, the geodesics in $f(\tau)$ do not intersect. Indeed, two geodesics $\gamma_1,\gamma_2$ intersect if and only if the endpoints of $\gamma_1$ separate the endpoints of $\gamma_2$ in $S^1$ (in other words, removing the endpoints of one geodesic leaves the endpoints of the other in different connected components). This property is also invariant under homeomorphisms.