Geodesics on hyperbolic surfaces Suppose that one has a compact hyperbolic surface realised as a quotient $X=\Gamma\setminus\mathbb{H}$. So in particular, $\Gamma$ consists only of hyperbolic elements of $\mathrm{PSL}(2,\mathbb{R})$ and the identity. Suppose that one takes an element $\gamma\in\Gamma$ non-identity, and for the sake of simplicity suppose is it primitive (so not a non-trivial power of another element in the group).
The axis of this element $\gamma$ in $\mathbb{H}$ will project to a closed geodesic on the surface $X$ through the natural covering map. Suppose now that I consider some other point $z\in\mathbb{H}$ that is not on the axis of $\gamma$ and I consider the projection of the geodesic segment between $z$ and $\gamma(z)$ onto the surface. This will give me a closed curve and will be a geodesic also since we projected a geodesic segment. It won't necessarily however be a closed geodesic since it may not close smoothly at the image of $z$ on the surface.
Question: Will this geodesic loop be freely homotopic to the closed geodesic originating from the projection of the axis? (Of course, in the case where $\gamma$ were non-primitive I would expect that it would be freely homotopic to some power of this projected axis instead).
I believe this to be true but I am not sure how one would go about showing it, any help would be appreciated.
 A: This is true and pretty easy to prove.
Let $\delta(s)$ be the shortest geodesic arc from $z$ to $A = \text{Axis}(\gamma)$, parameterized by $s \in [0,d(z,A)]$ with initial endpoint $\delta(0)=z$, with terminal endpoint $\delta(d(z,a)) = P \in A$, and hitting $A$ at a right angles at the point $P$.
Note that $\gamma \circ \delta(s)$ is, similarly the shortest geodesic arc from $\gamma(z)$ to $A$, hitting $A$ at right angles at the point $\gamma(P)$.
Now take a parameterized family of geodesic arcs $\zeta_t$ from $\delta(t)$ to $\gamma\circ\delta(t)$, defined for $t \in [0,d(z,A)]$. If you parameterize this family by normalized arc length (normalized to have domain $[0,1]$), and if you also normalize the parameter $t$ to vary over $[0,1]$, and then project to the quotient, that will give you the desired free homotopy.
Notice: this proof does not use that $\gamma$ is primitive. The key fact is that for a primitive element $\gamma$, the axis of $\gamma$ is identical to the axis of $\gamma^k$ for all $k \ge 1$, the only distinction being the length of a fundamental domain along the axis.
