Does two lifts of a non-simple closed geodesic on a hyperbolic surface always intersect. Let $S$ be a hyperbolic surface of finite type, i.e. $S$ is a surface of genus $g$, with $b$ boundary components and $n$ punctures with $2-2g-b-n<0$. 
Fix a hyperbolic metric on $S$. Let $x$ be a non-simple closed geodesic in $S$, i.e. $x$ have self-intersections. 
If $A$ and $B$ are any two geodesic lifts of $x$ to the upper half plane $\mathbb{H}$, then does $A$ and $B$ always intersect?
 A: The answer is no: there will always be two geodesic lifts of $x$ which do not cross. 
If you understand the deck transformation action of $\pi_1 S$ on $\mathbb{H}$, the proof is relatively straightforward. Here's some details of how this is done.
Let $\{X_i\}_{i \in I}$ be the set of all lifts of $x$ to $\mathbb{H}$, each of which is a geodesic line in $\mathbb{H}$; I'll call these simply lines. This set of lines in $\mathbb{H}$ is invariant under the deck transformation action of the group $\pi_1 S$ on $\mathbb{H}$. 
Here are some examples of how such a set of lines might appear:


*

*http://tilings.org/images/T433.gif, 

*http://www.geom.uiuc.edu/apps/teich-nav/report/_28991_figure29.gif
You can see in these examples that although there are many lines that do cross, there are also many lines that do not cross. In fact there must always be a pair of lines that does not cross. 
To prove it, choose a line $X_i$ in the set. Let $E_i$ be the unordered pair of endpoints of $X_i$.
We may always choose a nontrivial $g \in \pi_1 S$ with corresponding deck transformation $G : \mathbb{H} \to \mathbb{H}$ so that $G$ is a "translation" having two fixed points $p^-,p^+$ on the circle at infinity, and so that 
$$E_i \cap \{p^-,p^+\} = \emptyset
$$
(I'll show below why $g$ and $G$ can always be chosen in this manner).
A key property of $G$ is that $p^+$ is an attractor and $p^-$ is a repeller, meaning that for each point $q$ in the circle at infinity, if $q \ne p^-,p^+$ then 
$$\lim_{n\to\infty}G^n(q)=p^+ \quad\text{and}\quad \lim_{n\to-\infty} G^n(q)=p^-
$$ 
Apply this to the two points $q \in E_i$, and it follows that for large $n>0$ the endpoints $E_j = G^n(E_i)$ of the line $X_j = G^n(X_i)$ are close to $p^+$ and the endpoints $E_k = G^{-n}(E_i)$ of the line $X_k = G^{-n}(X_i)$ are close to $p^-$. If they are close enough, i.e. if $n$ is big enough, it follows that $X_j$ and $X_k$ do not cross. 
To choose $G$, start with three closed geodesics $c_1,c_2,c_3$ on $S$ which are pairwise transverse. Let $g_1, g_2, g_3 \in \pi_1(S)$ be elements representing the conjugacy classes associated to $c_1,c_2,c_3$. Let $G_1,G_2,G_3$ be the corresponding deck transformations. It follows that the attractor repeller pairs $\{p^+_1,p^-_1\}$, $\{p^+_2,p^-_2\}$, $\{p^+_3,p^-_3\}$ of $G_1,G_2,G_3$ are pairwise disjoint. The two-point set $E_i$ can intersect at most two of these attractor repeller pairs, and so is disjoint from at least one of those pairs.
