Take an orientable surface $S_g^s$ of genus $g$ with no boundaries and $s$ points removed and fix a complete hyperbolic metric of finite area (assuming that the Euler characteristic allows an hyperbolic structure on $S$). We know from hyperbolic geometry that there exists a unique geodesic in every non trivial free-homotopy class of curves. I know for sure that there is at least one point on $S_g^s$ through which there is no simple geodesic, but I have no idea about how to prove it. Could you help me with that?
Thank you!
Let $C$ be a horodisk neighborhood of the puncture $p$ so that the horocyclic boundary has length less than $2$. Then no simple, closed geodesic $g$ in $S$ intersects $C$.
Proof: Lift $g$ and $C$ to the upper half plane so that the $p$ is taken to the point at infinity and $C = \{ z \in \mathbb{C} \mid \rm{Im}(z) \geq 1 \}$. Suppose that $g$ meets $C$. Thus each lift of $g$ has horizontal width at least $2$. Thus adjacent lifts of $g$ intersect, a contradiction.