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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.

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    $\begingroup$ Actually, the proof given in the link is absolutely correct. The claim made by mathreader that there are more than two (connected, oriented) surfaces (without boundary) with the fundamental group isomorphic to $F_2$ is nonsense. (Feel free to ask a separate MSE question on this.) However, the proof given in the link is suboptimal, one can prove that every connected oriented surface with nonabelian fundamental group is covered by the trice punctured sphere. In particular, the 4-times punctured sphere is covered by the trice punctured sphere. $\endgroup$ – Moishe Kohan Dec 21 '19 at 0:06
  • $\begingroup$ @MoisheKohan Just to clarify, do you not need that the surface is compact when you want to make the claim that there are only two such surfaces? $\endgroup$ – siddharth64 Dec 21 '19 at 5:56
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    $\begingroup$ No, the surfaces in question are noncompact. However, one proves that a connected surface with f.g. fundamental group is tame i.e. homeomorphic to the interior of a compact surface with boundary. This is a part of the proof. $\endgroup$ – Moishe Kohan Dec 21 '19 at 13:27
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    $\begingroup$ The degree is infinite. I will write an answer when I have more time. $\endgroup$ – Moishe Kohan Dec 21 '19 at 16:53
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    $\begingroup$ In my answer I did not address the question of tameness of surfaces with f.g. fundamental groups (since I did not need this fact). If you want to see a proof (it again uses hyperbolic geometry!), feel free to ask a separate question. $\endgroup$ – Moishe Kohan Dec 22 '19 at 19:05
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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 $g_2^n$.

enter image description here

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.

  1. 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.

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