Let $M$ be an orientable surface **without** boundary$($**I am not assuming $M$ is compact, it can be non-compact**$)$. Let $\Phi: M\to M$ be a *proper* homotopy-equivalnce$($A proper homotopy-equivalence can be defined analog way as homotopy-equivalence, but here we need to assume all maps, *including homotopies, are proper*$)$.

Let $\alpha\in\pi_1(M)$ and $\beta\in \pi_1(M)$ be two **non-trivial** elements such that $\Phi_*(\alpha)=\beta$, where $\Phi_*$ is the induced map on fundamental group. Consider the lifting problem: 

$\require{AMScd}$
\begin{CD}
\widetilde M @>\displaystyle\widetilde \Phi>> \widetilde M\\
@V  V V @VV  V\\
M_\alpha @>>\displaystyle \widetilde \Psi> M_\beta\\
@V  V V @VV  V\\
M @>>\displaystyle \Phi> M
\end{CD}

Here, $\widetilde M$ is the universal cover of $M$, and $M_\alpha, M_\beta$ are the covers of $M$ corresponding to the subgroups $\langle \alpha\rangle$, and $\langle \beta\rangle$ respectively. All unleveled maps are covering maps, and $\widetilde \Psi,\widetilde \Phi$ are liftings.

> Is it true that $\widetilde \Psi,\widetilde \Phi$ are  *proper*
> maps?

I am considering [this problem][1] and trying to use the fact that the set of all deck-transformations acts properly discontinuously on the universal cover. Note also that both $M_\alpha$ and $M_\beta$ are open annuli (every open connected surface with a finitely-generated fundamental group is homeomorphic to the interior of some compact surface), i.e. $M_\alpha$ as well as $M_\beta$ has exactly two ends, and one has to show $\widetilde \Psi$ induces a bijection on the set of ends.

[1]: https://math.stackexchange.com/questions/1874740/does-lifting-commute-with-the-right-action-of-the-fundamental-group