# Can two different topological spaces cover each other?

I.e. do there exist non-homeomorphic $X$, $Y$ and covering maps $f:X\rightarrow Y$, $g: Y\rightarrow X$?

I have a basic understanding of covering space theory as its taught in school.

I was inspired by this question Two covering spaces covering each other are equivalent?

I have done some of the things you would do, looking at what happens to fundamental groups. And I've tried to find counter examples by drawing graphs.

• Have you tried looking at Hatcher or another algebraic topology text to see what conditions this would place on $X$ and $Y$? This would be useful information to know. If on the other hand, you haven't because you only just started learning about covering spaces and don't know the relationship to the fundamental group, that would also be useful information for a potential answerer to know. Aug 5, 2016 at 11:13
• Maybe you were inspired by this question and your answer. Might be worth mentioning. Aug 5, 2016 at 11:19
• @MichaelAlbanese there are many examples of two different groups $G_1$ and $G_2$ for which there exist embeddings $G_1\to G_2$ and $G_2\to G_1$. maybe, you can give an example two spaces with corresponding covering properties, or prove that such an example does not exist? Aug 5, 2016 at 13:56
• @AndreyRyabichev: Thanks for your comment, but my questions were to the OP to demonstrate what could be added to provide context to the question. Aug 5, 2016 at 13:57
• @MichaelAlbanese I have not looked at Hatcher. But I do know the basic facts about covering spaces contained in Hatcher. And yes, I was inspired by the question you reference. Aug 5, 2016 at 15:46

The answer is yes and X and Y can even be surfaces.

Let $X_1$ be the twice punctured torus, $X_2$ the 4 punctured sphere, $Y_1$ the once punctured torus, and $Y_2$ the thrice punctured sphere.

Here is a proof that either $X_1$ and $Y_1$ mutually cover each other, or $X_2$ and $Y_2$ mutually cover each other. Note that there is a natural 2-sheeted covering from $X_1$ to $Y_1$ and likewise for $X_2$ and $Y_2$ so we only need to establish the other direction.

Now $\pi_1(X_i) = F(3)$ and $\pi_1(Y_i) = F(2)$, where $F(n)$ is the free group on $n$ generators. Note that $F(2)$ is a natural subgroup of $F(3)$ so the correspondence between subgroups of the fundamental group and covering spaces guarantees that there is a covering space of each $X_i$ which have fundamental group $F(2)$, and since $X_i$ is a surface the covering space must also be a surface.

It is a fairly straightforward exercise from an introductory course on the fundamental group to show that the fundamental group of a genus $g$ surface with $m$ punctures is $F(m+2g-1)$. A less obvious observation is that the fundamental group of a non-punctured surface is not free. Here is a link with quite a few proofs of this fact, most aren't that elementary unfortunately.

This tells us that $Y_1$ and $Y_2$ are the only surfaces which have $\pi_1 = F(2)$. This means that either $Y_1$ or $Y_2$ covers $X_1$ and the same is true with $X_2$. Obviously if either $Y_1$ covers $X_1$ or $Y_2$ covers $X_2$ we are done, since they mutually cover each other and are not homeomorphic. So assume for the sake of contradiction that neither of those coverings exist.

This implies that $Y_1$ must cover $X_2$ and $Y_2$ must cover $X_1$. But the composition of covers of $Y_1$ to $X_2$, $X_2$ to $Y_2$, and $Y_2$ to $X_1$ gives a covering map from $Y_1$ to $X_1$ which is a contradiction.

So either the once and twice punctured tori cover each other or the 3 and 4 punctured spheres do. (My money is on the spheres, but I could be wrong).

Sadly this proof isn't strictly constructive as it only gives you 2 possible pairs of spaces. If anyone has a simpler proof, or a more constructive example (or if they can rule out one of the pair, or prove that it's both) that would be awesome.

• I would only add that $Y_1$ and $Y_2$ are the only oriented surfaces with $\pi_1=F(2)$. That's not a problem for your proof because a covering of an oriented manifold has to be oriented. Aug 10, 2016 at 2:08
• Also, the wikipedia article on surfaces points out that not all orientable surfaces are subsets of compact surfaces, but the only examples they give have huge fundamental groups... Aug 10, 2016 at 2:33
• @Devin, your reasoning has a flaw. You assume that the only surfaces with fund. group $F(2)$ are $Y_1$ and $Y_2$. However, there are many more such surfaces, having infinitely many ends. Your reasoning also works if we replace all punctures by small circular boundary components, which makes all our surfaces compact. Since the fundamental group of $Y_i$ has infinite index in the fund. group of $X_2$, say, the covering space cannot be compact. Think of the images of boundary circles under the covering map. 3 circles of $Y_2$ should cover 4 circles of $X_2$ in a $\infty$-to-one fashion. Jun 5, 2018 at 21:50
• @Devin (cont.) To build the cover of $X_2$ corresponding to $F(2)$ subgroup realize $X_2$ as a euclidean neighborhood of a bouquet of 3 loops a,b,c in the plane. The covering corresponding to the subgroup F(a,b) can be built as a fattened graph corresponding to the covering of planar graphs: a figure 8 on loops a,b with two infinite 6-valent trees attached. The fattening happens in the plane in such a way that the links of all vertices are as in $X_2$. We observe that there are infinitely many boundary components each isomorphic to $\mathbb R$, infinitely covering boundary components of $X_2$. Jun 5, 2018 at 22:02
• In the context of this answer, "surface" implies empty boundary, and it is true that every surface with fundamental group $F(2)$ is homeomorphic to $Y_1$ or to $Y_2$. In general, any noncompact surface (with empty boundary) with finitely generated fundamental group has finitely many ends, and is homeomorphic to a closed surface minus a finite set, i.e. it is a "punctured surface" in the language of this answer. @mathreader Dec 21, 2019 at 22:32