Let $M$ be an orientable surface$($without boundary$)$, possibly non-compact. Let $\alpha\in \pi_1(M)$ be a primitive element i.e. $\alpha$ is not a proper power of some other element. Let $f:\Bbb S^1\to M$ be any loop representing $\alpha$. Consider the lifting problem 
Here, $M_\alpha$ be the cover corresponding to the subgroup $\langle\alpha\rangle$ of $\pi_1(M)$ and $\widetilde M$ be the universal cover of $M$. Also, all unleveled maps are covering maps.
Note that $\pi_1(M)$ acts on $\widetilde M$ via covering transformations, and $M_\alpha$ be an open annulus$($any open connected surface with a finitely generated fundamental group is homeomorphic to the interior of a closed surface$)$.
$\textbf{Problem 1:}$ Consider two sets $$\mathscr A_f:=\big\{g\in
\pi_1(M):\text{ the map } \Bbb R\xrightarrow{g\cdot \ell}\widetilde
M\to M_\alpha\text{ runs from one end to other of the open annulus
}M_\alpha\big\},$$ $$\mathscr B_f:=\big\{(z,w)\in \Bbb S^1\times\Bbb
S^1:z\not= w\text{ and }f(z)=f(w)\big\}$$ Is the cardinality of
$\mathscr B$ two-times the cardinality of $\mathscr A$, i.e. $|\mathscr A_f|=\frac{1}{2}|\mathscr B_f|$?
$\textbf{Problem 2:}$ Give a Riemannian metric on $M$ such that
$\alpha$ can be represented by a shortest loop $f^\#:\Bbb S^1\to M$,
i.e. $f^\#$ has the minimum length in its free homotopy class. Is
$|\mathscr A_f|\geq |\mathscr A_{f^\#}|$?
Best Answer
For problem one, the answer is "no" as there may be "bigons". The pair of intersections forming the bigon contribute to $B_f$ but not to $A_f$.
For problem two, the answer is "yes" as the shortest representative has no bigons.
A version of this is discussed in the "Primer on mapping class groups" by Farb and Margalit under the name "the bigon criterion".