Lifted free homotopic closed Curves have same Endpoints at Infinity I'm studying the book of Peter Buser, "Geometry and Spectra of Compact Riemann Surfaces". I want to understand the proof he gives in Lemma 1.6.5.
If $\gamma$ and $c$ are free homotopic closed curves on a hyperbolic surface, you can lift them uniquely (up to choosing a start point) to it's universal (Riemannian) covering, $\tilde{\gamma},\tilde{c}$. That's clear. Then he uses a "standard argument" which I don't understand:
"The cyclic subgroup of the covering transformation group which leaves $\tilde{\gamma}$ invariant also leaves $\tilde{c}$ invariant."
My idea so far: I lift the (free) homotopy $F$ beween $\gamma$ and $c$ and show that it behaves like $f(\tilde{F}(s,t))=\tilde{F}(s,t+\omega)$ under the deck transformation $f$ which leaves $\tilde{\gamma}$ invariant, where $\omega$ is the displacement of the geodesic $\tilde{\gamma}$ under $f$. Is that the right approach? If so, how can i proove that?
From this it should follow that $d(\gamma(t),c(t)) < d$ for all $t \in \mathbb{R}$ and some $d$. Since all deck transformation are isometries and the universal covering is a local isometry aswell, it should work somehow, but i don't get the details here.
Thank you for your help! :)
 A: Let us remark about your title "Lifting of free homotopic closed curves have same endpoints at infinity". For example, if a closed curve is freely-homotopic to a non-simple closed geodesic, say $\gamma$, there is an intersecting point $x_0$ of $\gamma$. If $\tilde{x}_0 \in \pi^{-1}(x_0)$ we can find two different lifting of $\gamma$ passing through $\tilde{x}_0$ having different endpoints at infinity. 
I'm also new in this subject, and I prefer the following statement: 

We start with a lifting $\tilde{c}$ of $c$. Consider the isometry $T_c \in \pi_1(M)$ such that $T_c(\tilde{c}(0))=\tilde{c}(1)$. Define $\tilde{\gamma}$ the only $T_c$-invariant geodesic line. Then, $\tilde{\gamma}$ and $\tilde{c}$ have the same endpoints at infinity.

We want to prove 
$$d(\tilde{x}, \tilde{\gamma}) < \infty \ \forall \tilde{x} \in \tilde{c}.$$ 
Let $\tilde{\gamma}_0$ and $\tilde{c}_0$ be the restrictions of $\tilde{\gamma}$ and $\tilde{c}$ to $[0,1]$. For all $\tilde{x} \in \tilde{c}$, we have  $\tilde{x} = T_{c}^n(\tilde{x}_0)$
for some $n \in \mathbb{Z}$ and $\tilde{x}_0 \in \tilde{c}_0$. We have
$$ d(\tilde{x}, \tilde{\gamma})  <  d(\tilde{x}, T^n_c(\tilde{\gamma}_0)) =  d(T_{c}^n(\tilde{x}_0), T^n_c(\tilde{\gamma}_0)) = d(\tilde{x}_0, \tilde{\gamma}_0) = \text{a constant} < \infty.$$
Reference: Lectures on Hyperbolic Geometry.
A: The equation $f(\tilde F(s,t)) = \tilde F(s,t+\omega)$ should follow from uniqueness of lifting: check that each side is a lift of $F$, and check that they have one common value by plugging in your favorite value of $s$ and of $t$.
For each fixed value of $t \in \mathbb{R}$, as $s \in [0,1]$ varies we obtain a path $p_t(s) = \tilde F(s,t)$. Since $[0,1]$ is compact and $p_t$ is continuous, the image $p_t[0,1]$ is compact and therefore has finite diameter which I will denote $D(t) = \text{diam}\left(p_t[0,1]\right) \in [0,\infty)$. The function $D : \mathbb{R} \to [0,\infty)$ varies continuously with $t$, which is not too hard to check. It follows from compactness that $D(t)$ is bounded for $t \in [0,\omega]$. Also $D(t) = D(t+\omega)$, because the deck translation $f$ is presumed to be an isometry (or, using your word, isotropy) of the hyperbolic plane, and so $D$ is a bounded function, $D(t) \le d$ for some $d > 0$. So 
$$d(\tilde \gamma(t), \tilde c(t)) = d(\tilde F(0,t), \tilde F(1,t)) = d(p_t(0),p_t(1)) \le D(t) \le d
$$
