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.


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. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.