This question may be easy but I could not come up with a proof.


Let $F$ be a hyperbolic surface of finite type (with finitely many boundary and finitely many puncture). Let $\gamma$ be a closed non-simple geodesic. $\gamma$ is not homotopic to a point, a puncture or a boundary. Let $p$ be a self intersection point of $\gamma.$ 


**1) Can there be a simple closed geodesic passing through $p$?**


If not then 


**2) Can there be a closed geodesic passing through $p$ different from $\gamma$?** 



Intersection of closed geodesics in hyperbolic surface
The answer to (1) is yes. 


Take $P$ a hyperbolic surface with one geodesic boundary, called $\delta$, and two punctures. Form $S$, a sphere with four punctures, by doubling $P$ across $\delta$. Note that $S$ has a reflection symmetry $f$ that fixes $\delta$ pointwise. Let $\gamma$ be a figure of eight curve, about two of the punctures of $S$, choosen so that the reflection $f$ fixes $\gamma$ setwise. Deduce that the self-intersection point $p \in \gamma$ lies on $\delta$. 


