# Lifting of a proper map in the cover is a proper map

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.

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. $$\square$$