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



Lifting of a proper map in the cover is a proper map
The two squares are pullbacks. This follows from the following more general result:


**Lemma.** *Let $f \colon X \to Y$ be a continuous map of topological spaces that induces an isomorphism $f\_\* \colon \pi\_1(X) \stackrel\sim\to \pi\_1(Y)$. Let $G \subseteq \pi\_1(X)$ be a subgroup with image $H = f\_\*(G)$, and let $X\_G \to X$ and $Y\_H \to Y$ be the covers with Galois group $\pi\_1(X)/G$ and $\pi\_1(Y)/H$. Then any commutative diagram*
$$\begin{array}{ccc}X\_G & \to & Y\_H \\ \downarrow & & \downarrow \\ X & \to & Y\end{array}$$
*lifting $X \to Y$ is a pullback square.*


*Proof.* The map $X\_G \to X \times\_Y Y\_H$ is a map between covering spaces of $X$ that induces an isomorphism on Galois groups, so it is an isomorphism. $\square$


Applying this to $G = \langle \alpha \rangle$ and $G = 0$ shows that the bottom and outer squares are pullbacks, and this implies the result since proper maps are stable under base change. $\square$


