Two non-homeomorphic spaces which are coverings of each other

Question

Let X and Y be non-homeomorphic spaces. Can X be a cover of Y and vice versa?

Does the answer change if we have a nicer class of spaces eg. path-connected CW-complexes/manifolds?

I posted this question earlier, but it was marked as a duplicate of Can two different topological spaces cover each other?. However, the only answer there seems to have problems, as pointed out in the comments.

Answer 1

Theorem 1. Let $X$ be the 2-sphere with four punctures and $Y$ be the 2-sphere with 3 punctures. Then there exist covering maps $$ Y \stackrel{p}{\to} X\stackrel{q}{\to} Y, $$ where $q$ is a 2-fold covering map and $p$ is an infinite-sheeted covering map.

Proof. The existence of $q$ is clear: Take an involution $\tau$ of $S^2$ fixing two of the punctured points and swapping the other two. Then $X/\tau =Y$.

The proof of the existence of $p$ is more complicated. I will use some reasonably advanced hyperbolic geometry, if you are unfamiliar with it, I cannot help you. My favorite reference for this material is Alan Beardon's book "The Geometry of Discrete Groups."

The surface $X$ admits a complete hyperbolic metric of finite area and, hence, is diffeomorphic to the quotient of the hyperbolic plane (the upper half-plane) ${\mathbb H}^2$ by a discrete group of isometries $\Gamma< PSL(2, {\mathbb R})$ isomorphic to $\pi_1(X)\cong F_3$, the free group of rank 3. The nontrivial elements of $\Gamma$ come in two different types: parabolic and hyperbolic. The latter are characterized by the property they they preserve exactly two points on the circle $S^1= {\mathbb R}\cup \{\infty\}$, the "boundary" of ${\mathbb H}^2$ (the one-point compactification of the real line). Thus, each hyperbolic isometry $g\in \Gamma$ preserves a hyperbolic geodesic $A_g\subset {\mathbb H}^2$ (the "axis" of $g$) in ${\mathbb H}^2$, which is (typically) the semicircle connecting the two fixed points of $g$. For any two hyperbolic elements $g_1, g_2\in \Gamma$ one of two things can happen:

• $A_{g_1}= A_{g_2}$.

• $A_{g_1}\cap A_{g_2}\ne \emptyset$, is a single point in ${\mathbb H}^2$.

• $A_{g_1}\cap A_{g_2}= \emptyset$.

Furthermore, except for the Case 1, the fixed-point sets of $g_1, g_2$ are disjoint.

Lemma 1. There exist elements $g_1, g_2\in \Gamma$ satisfying $A_{g_1}\cap A_{h_2}= \emptyset$.

Proof. Indeed, start with two hyperbolic elements $h_1, g_2\in \Gamma$ such that $A_{h_1}\cap A_{g_2}\ne \emptyset$. Since $\Gamma$ is not cyclic, there exists a third hyperbolic element $g_3\in \Gamma$ whose fixed-point set is disjoint from that of $g_1$ and of $g_2$. Then for every point $z$ on the $x$-axis, apart from one of the fixed points of $g_3$, $$ \lim_{n\to\infty} g_3^n(z)= \xi^+, $$ the attractive fixed point of $g_3$. In particular the sequence $g_3^n$ restricted to $A_1$ converges to $\xi^+$. Thus, for all sufficiently large $n$, $g_3^n(A_{h_1})\cap A_{g_2}= \emptyset$. Set $g_1:= g_3^{n} h_1 g_3^{-n}$. This is a hyperbolic element of $\Gamma$ whose axis is $g_3^n(A_{h_1})$. Hence, $A_{g_1}\cap A_{g_2}= \emptyset$. qed

Now, consider the hyperbolic elements $g_1, g_2\in \Gamma$ as in Lemma 1.

Lemma 2. There exists $n>0$ such that the subgroup $\Lambda=\Lambda_n$ of $\Gamma$ generated by $g_1^n, g^n_2$ is such that ${\mathbb H}^2/\Lambda$ is homeomorphic to the triply-punctured sphere. (Incidentally, if the axes of $g_1, g_2$ cross, then the quotient surface will be homeomorphic to the once-punctured torus.)

Proof. Since the end-point sets $\xi^{\pm}_1, \xi^{\pm}_2$ of the hyperbolic geodesics $A_{g_1}, A_{g_2}$ are pairwise disjoint, we can find small pairwise disjoint Euclidean disk neighborhoods $U_1^\pm, U_2^\pm\subset {\mathbb C}$. I will assume that $\xi^+_i$ is the attractive fixed point of $g_i$, $i=1, 2$. Then for sufficiently large $n$, $$ g_i^n(S^2 - U_i^-)\subset U_i^+, i=1, 2, $$ where $S^2= {\mathbb C}\cup\{\infty\}$. In particular, the open half-disks
$$ U_1^- \cap {\mathbb H}^2, U_2^-\cap {\mathbb H}^2, V_1^+:= {\mathbb H}^2- g_1^n(cl U_1^-), V_2^+:= {\mathbb H}^2- g_2^n(cl U_2^-) $$ are also pairwise disjoint. Let $F$ denote the domain in ${\mathbb H}^2$ equal to the complement to the union of interiors of $U_1^-, U_2^-, V_1^+, V_2^+$. Its boundary in ${\mathbb H}^2$ is the union of four disjoint hyperbolic geodesics, the boundaries of the above half-disks. These boundary geodesics are matched in pairs by $g_1^n$ and $g_2^n$.

The above properties of $F$ imply that $F$ is a fundamental domain of the subgroup $\Lambda=\Lambda_n$ of $\Gamma$ generated by $g_1^n, g^n_2$. In particular, the quotient surface ${\mathbb H}^2/\Lambda$ is homeomorphic to the surface obtained from $F$ by gluing the boundary geodesics via the generators $g_1^n, g^n_2$. Lastly, by looking at the picture of $F$ and the identification pattern, one sees that this quotient-surface is homeomorphic to the triply punctured sphere. qed

Thus, we obtain: ${\mathbb H}^2/\Lambda$ is homeomorphic to $Y$. The inclusion $\Lambda< \Gamma$ induces a covering map $$ p: Y= {\mathbb H}^2/\Lambda \to X= {\mathbb H}^2/\Gamma. $$ This concludes the proof of Theorem 1. qed

Edit. 1. This proof circumvents the issue of tameness of noncompact surfaces with finitely generated fundamental groups which is used (without a proof) in the accepted answer to the linked question. I give a separate proof of tameness in my answer here.

2. One can prove more:

Theorem 2. Suppose that $X, Y$ are connected orientable surfaces (without boundary), such that $Y$ has nonabelian fundamental group and $X$ is noncompact. Then there exists a covering map $X\to Y$.

The special case, when $Y$ is compact, is proven in Theorem 8.1 in

Martin Goldman, An algebraic classification of noncompact 2-manifolds. Trans. Amer. Math. Soc. 156 (1971), 241–258.