Cocompact group action of a deck-transformation Let $\Sigma$ be an orientable surface and $x\in \Sigma$. Let $\alpha\in \pi_1(\Sigma,x)$ be a non-trivial element and $f:(\Bbb S^1,1)\to (\Sigma,x)$ represents $\alpha$. Consider the lift  $\require{AMScd}$
\begin{CD}
 (\Bbb R,0) @>\displaystyle\ell>> (\widetilde \Sigma,\widetilde x)\\
@V  V V @VV  V\\
(\Bbb S^1,1) @>>\displaystyle f> (\Sigma,x)\\
\end{CD}
All unlabeled maps are the universal covers. Let $\varphi$ be a deck-transformation for the universal cover $\widetilde \Sigma\to \Sigma$ of infinite order i.e. for each $n\in \Bbb N$ the $n$-times composition of $\varphi$ with itself, denoted by $\varphi^{(n)}$ is not identity.

Is it true that there is a
$K\subseteq_\text{compact}\text{image}(\varphi\circ \ell)$ such that
$\text{image}(\varphi\circ \ell)=\bigcup_{m\in\Bbb Z}\varphi^{(m)}(K)$ i.e. $\langle \varphi\rangle$ acts cocompactly on $\text{image}(\varphi\circ \ell)$?

$\bullet$ Actually any non-trivial deck-transformation of an orientable surface$(\text{open/closed})$ has infinite order as the fundamental group of such surface does not have torsion element.
$\bullet$ $\ell$ is a proper map, so that image of $\ell$ is closed.
$\bullet$ If $\Sigma$ has a Riemannian metric and $f$ has the shortest length among all elements in its free homotopy class$($of loops$)$, then $\ell$ is an embedding of $\Bbb R$. Then the above problem can be thought of as the standard $\Bbb Z$-action on $\Bbb R$ by the translation which is cocompact as $\Bbb R/\Bbb Z\cong \Bbb S^1$.
I am not too much confident about the last bullet.
 A: This is not true.
Consider $\Sigma = T^2 = S^1\times S^1$ which has universal cover $\tilde{\Sigma}$. The universal covering map $\pi : \mathbb{R}^2 \to S^1\times S^1$ is given by $\pi(x, y) = (e^{2\pi i x}, e^{2\pi i y})$; fix $\tilde{x} = (0, 0)$ and $x = (1, 1)$. The deck transformations of this covering are translations by elements of the integer lattice $\mathbb{Z}^2$, so the group is generated by $\varphi_1$ and $\varphi_2$ where $\varphi_1(x, y) = (x + 1, y)$ and $\varphi_2(x, y) = (x, y + 1)$.
Now note that the fundamental group of $T^2 = S^1\times S^1$ is generated by $\alpha$ and $\beta$ where $f(z) = (z, 1)$ and $g(1, z)$ are representatives of $\alpha$ and $\beta$ respectively. In this case the map $\ell$ is given by $\ell(x) = (x, 0)$; i.e. the image of $\ell$ is the $x$-axis in $\mathbb{R}^2$. Now $(\varphi_2\circ\ell)(x) = (x, 1)$ which has image the line $y = 1$, while for any integer $m$ the image of $\varphi_2^{(m)}\circ \ell$ is the line $y = m$. Let $K \subseteq \operatorname{image}(\varphi_2\circ\ell) = \mathbb{R}\times\{1\}$ be any subset and write $K = K'\times\{1\}$ where $K' \subseteq \mathbb{R}$. Then $\bigcup_{m\in\mathbb{Z}}\varphi_2^{(m)}(K) = K'\times\mathbb{Z} \neq \mathbb{R}\times\{1\} = \operatorname{image}(\varphi_2\circ\ell)$. Note that $K'\times\mathbb{Z}$ is just the union of all copies of $K$ shifted by an integer amount in the vertical direction.
