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.


1 Answer 1


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. $\square$


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