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 map, and $\widetilde \Psi,\widetilde \Phi$ are liftings.

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

I am considering this problem 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.

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 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.