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This is motivated by an old question of Henno Brandsma.

Two topological spaces $X$ and $Y$ are said to be bijectively related, if there exist continuous bijections $f:X \to Y$ and $g:Y \to X$. Let´s denote by $br(X)$ the number of homeomorphism types in the class of all those $Y$ bijectively related to $X$.

For example $br(\mathbb{R}^n)=1$ and also $br(X)=1$ for any compact $X$. Henno´s question was about nice examples where $br(X)>1$. The list wasn´t too long, but all the examples in there also satisfied $br(X) \geq \aleph_0$. So here is my question:

Is there a topological space $X$ for which $br(X)$ is finite and bigger than $1$?

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I suspect not. It may help to view the induced injective maps (in the opposite direction) on the infinite join-complete lattice of open sets. I think one can produce a countable series of such maps given two such, and from that craft a countable number of homeomorphism types. Gerhard "My View Of Topological Algebra" Paseman, 2012.02.07 – Gerhard Paseman Feb 7 '12 at 20:33
@Gerard: How do you produce such a countable series of maps? on which lattices? – Ramiro de la Vega Feb 8 '12 at 12:08
Let g be a continuous bijective map from X to Y. Let L be the lattice of open sets of X, and M similarly for Y. g induces an injective map from M to L which is not onto when g is not a homeomorphism. The map preserves arbitary joins and possibly arbitrary meets when they exist. Now as a result both lattices are infinite, and f and g induce lattice mappings which are not onto. Consider repeated compositions of these lattice maps. It may inspire you to find a "gap" of some sort that you can repeat. Gerhard "Ask Me About System Design" Paseman, 2012.02.08 – Gerhard Paseman Feb 9 '12 at 4:59
For example, if the maps f and g allow you to produce the examples in my answer to Henno's question, you might see how to multiply the gap and produce a space like (referring to my notation in the other question) ... X X X' X' X' X" X" X" X" ..., and countably many other examples. Also, it may be that this was studied in a framework of category theory, and that the properties of the appropriate category might show that br(X) ie either 1 or infinite. Gerhard "Ask Me About System Design" Paseman, 2012.02.08 – Gerhard Paseman Feb 9 '12 at 5:03
I think the following should have $br(X)=2$. Take Andreas' example of a space which is homeomorphic to itself + two isolated points, but not itself + one point (…). Now take the product of this with $S^1$, and add infinitely many disjoint copies of $[0,1)$. – George Lowther Jun 14 '13 at 15:26

2 Answers 2

Sorry to be coming to the conversation so late (over 3 years late!), but I'm new to MO and just saw it. I've done some research on this very topic, so I couldn't resist sharing.

Let me start by saying that I don't know a complete answer to the question as stated. By I do have a partial answer, some other results that are relevant to the question, and an answer to the question of Doyle and Hocking mentioned in Michal's answer. It's all too long for a comment, so here we go . . .

The following theorem answers the Doyle and Hocking question, and also tells us that any example answering Ramiro's question affirmatively will be at least a little bit exotic.

Theorem: Suppose $X$ is a sequential space. Then either $br(X) = 1$ or $br(X) \geq 2^\mathfrak{c}$.

Proof: Let $Y$ be a space bijectively related to $X$, and let $f: Y \to X$ be a continuous bijection. Without loss of generality, we may assume that $Y$ and $X$ are actually just two topologies on the same set $A$ (because $f$ is a bijection), and $X$ is a refinement of $Y$ (because $f$ is continuous). Let us assume that $Y$ is strictly finer than $X$ (this must be true, in particular, if $Y$ is not homeomorphic to $X$).

Since $X$ is sequential and $Y$ is strictly finer than $X$, there is some sequence $\langle x_n \rangle$ such that $x_n \to x$ in $X$ but $x_n \not\to x$ in $Y$.

In $Y$, since $x_n \not\to x$, there is an infinite $S \subseteq \{x_n : n \in \omega\}$ such that $x \notin \overline{S}$. Thus (passing to a subsequence if necessary) we may assume that $x_n \to x$ in $X$ but $x \notin \overline{\{x_n:n \in \omega\}}$ in $Y$.

Let $\mathcal F$ be any free filter on $\omega$. Define the topology $Z_{\mathcal F}$ on $A$ as follows (we will define the space by defining its closure operator). If $B$ is any subset of $A$, then $\overline{B}$ is the same in $Z_{\mathcal F}$ as in $X$, except possibly for the point $x$. Then we put $x \in \overline{B}$ if and only if either $x \in B$, $x \in \overline{B \setminus \{x_n:n \in \omega\}}$, or $\{n : x_n \in B\} \in \mathcal F^+$.

It's easy to check that this defines a topology on $A$, and that this new topology is finer than $X$ and coarser than $Y$. This means that the identity on $A$ gives continuous bijections $Y \to Z_{\mathcal F} \to X$. So $Z_{\mathcal F}$ is bijectively related to $X$ for any filter $\mathcal F$.

Now let's check that we get $2^\mathfrak{c}$ non-homeomorphic spaces this way. Consider the following topological invariant: $$Filt(Z) = \{\mathcal F : \mathcal F \text{ is a filter and, for some countable set } \{x_n: n \in \omega\},\text{ and some } x, \ \mathcal F = \{\{n : x_n \in U\}: U \text{ open and } x \in U\}\}$$ In other words, $Filt(Z)$ is just a list of all the filters that describe the convergence of some countable set to some point. Since $X$ is sequential, $Filt(X)$ is just the filter of cofinite sets. Given how we've defined $Z_{\mathcal F}$, $Filt(Z_{\mathcal F})$ is equal to either all isomorphs of $\mathcal F$ (if $X$ has one non-isolated point) or the filter of cofinite sets plus all isomorphs of $\mathcal F$ (if $X$ has more than one non-isolated point).

By an "isomorph" of $\mathcal F$ I mean every filter that can be obtained from $\mathcal F$ by a permutation of $\omega$. This class has size $\mathfrak{c}$ for any fixed $\mathcal F$. But the number of filters on $\omega$ is $2^\mathfrak{c}$, so $Filt(Z_{\mathcal F})$ is different for $2^\mathfrak{c}$ different choices of $\mathcal F$. Since it's a topological invariant, we have lots of different spaces.


Next, let's observe that for every infinite $\kappa$ there is some $X$ with $br(X) = \kappa$.

Fix two ultrafilters on $\omega$, say $p$ and $q$. Let $X_p = \omega \cup \{*\}$ be the space obtained by making every point of $\omega$ isolated, and making the neighborhoods of $*$ take the form $A \cup \{*\}$ for $A \in p$. Define $X_q$ and $X_{p \cap q}$ similarly (note that $p \cap q$ is a filter, so this definition still makes sense).

Let $\mathcal X$ be the topological space obtained by taking the disjoint sum of $\aleph_\kappa$ copies of $X_p$, $\aleph_\kappa$ copies of $X_q$, $\aleph_\kappa$ copies of $X_{p \cap q}$, and $\aleph_\kappa$ singletons.

As in the proof of the previous theorem, if $\mathcal Y$ is bijectively related to $\mathcal X$, then we may assume that $\mathcal Y$ is just a refinement of $\mathcal X$. By refining $\mathcal X$, we can obtain any space of the following form (for any $\mu,\lambda,\theta \leq \kappa$): a disjoint sum of $\mu$ copies of $X_p$, $\lambda$ copies of $X_q$, $\theta$ copies of $X_{p \cap q}$, and $\aleph_\kappa$ isolated points.

Using the fact that the only extensions of the filter $p \cap q$ are $p$ and $q$, it is not too hard to check that these are the only spaces possible.

Now suppose we also have $\theta = \aleph_\kappa$. There are $\kappa$ possibilities for choosing $\mu$ and $\lambda$, yielding $\kappa$ spaces of this form. But now we may refine again to get $\mathcal X$ back (leave $\aleph_\kappa$ of the $X_{p \cap q}$ alone, refine $\aleph_\kappa$ of them to $X_p$, and refine another $\aleph_\kappa$ of them to $X_q$).

Now I'll give a really nice example of a space $X$ where $br(X) \geq \mathfrak{c}$, namely the Baire space.

Unfortunately, I don't know a really slick proof of this assertion. I'll have to refer you to a paper of mine for details. Below I'll list a few related results that are also relevant to your question. For most of the results about non-separable spaces, see my other paper, joint with Arnie Miller.

I studied the "bijectively related" relation on (nonempty) perfect completely ultrametrizable spaces (henceforth, PCU spaces).

[Sidebar: At first this may seem like a weirdly specific class to look at. The following folklore result might clear things up a bit: $X$ is a PCU space if and only if there is some pruned, perfect tree whose end space is homeomorphic to $X$. So you can think of these as "tree spaces." The original intent of the two papers above was to explore how the structure of the trees can be leveraged to prove topological results.]

The separable PCU spaces are precisely the perfect, zero-dimensional Polish spaces (see Chapter 2 of Kechris's book for more on this). For this class, we have a pretty good idea of what your relation looks like:

Theorem: The separable PCU spaces are partitioned into exactly three equivalence classes by the relation described in your question: the class of the Cantor space, and the class of the Cantor space minus a point, and the class of the Baire space.

The latter two classes each have $\mathfrak{c}$ members. This result isn't in my paper, but it's a good exercise.

Once you move to non-separable PCU spaces, it is consistent that the situation stays "nice" like it is for Polish spaces. For example,

Theorem: If CH holds, then there are exactly four equivalence classes of PCU spaces of size $\mathfrak{c}$.

More generally,

Theorem: It is consistent with ZFC that $\mathfrak{c} = \aleph_n$ and that there are exactly $n+3$ equivalence classes of size-$\mathfrak{c}$ PCU spaces.

However, if $\mathfrak{c} = \aleph_2$ then the number of equivalence classes is independent of ZFC: it can be $5$ by the above theorem, but it is more if MA holds (I don't know exactly how many).

For cardinals below $\aleph_\omega$, there is a strong connection between these equivalence classes and the question of whether you can partition the Cantor space into $\kappa$ closed sets for $\kappa < \mathfrak{c}$.

I don't know much about what happens after $\aleph_\omega$, and I consider it a very interesting problem.

Two other papers of mine you might want to look at: this one deals with what happens when you refine a topology by just a little bit, and this one (joint with Chris Good, Robin Knight, and Dave McIntyre) deals with finite intervals in the lattice of topologies. (Notice that, by our arguments above, $X$ and $Y$ are bijectively related if and only if there is an interval in the lattice of topologies, with the top and bottom spaces both homeomorphic to $X$, and some space in between homeomorphic to $Y$. But then everything in the interval is also bijectively related to $X$ and $Y$. So one strategy for answering Romiro's question could be to find a finite interval like this).

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Welcome to MathOverflow! I am having trouble directly relating your answer to the question. In particular do you know of an X such that a) br(X) is finite and bigger than 1 for X a PCU space, and b) all members bijectively related to X have to be PCU spaces? – The Masked Avenger Apr 17 at 2:44
Welcome to MO Will. Thank you for your post, but I don´t think it really answers my question since I asked for a finite $br(X)$. Perhaps your post is more relevant for Henno Brandsma´s original question in which mine was inspired. – Ramiro de la Vega Apr 17 at 13:27
You're right -- I don't actually answer the question. Based on the question, I just thought that the asker might be interested in what I wrote. It was too long for a comment (and a brand new user can't comment anyway), so I posted it as an answer. I'll modify my answer above to reflect this. I'll also add a few thoughts that I've had since yesterday afternoon. (p.s.: Great question, Ramiro!) – Will Brian Apr 17 at 15:38
Wow! In spite of my comments above, I never thought of investigating the lattice of topologies of X itself. Is there any reference investigating the notion of equivalence relations on the lattice of topologies which pertain to being bijectively related? More importantly (for my universal algebraic background) are there important lattice congruences (equivalence relations preserving finite meets and joins) that one could use to help investigate this? Gerhard "Lattices Turn My Mind Inside-Out" Paseman, 2015.04.17 – Gerhard Paseman Apr 17 at 17:07
@ Ramiro: Thanks for the suggestion -- I've added an answer to Henno Brandsma's original question that distills the relevant bits of this answer. – Will Brian Apr 17 at 18:31

This problem was stated for manifolds by P. H. Doyle and J. G. Hocking in their 1984 paper "Bijectively related spaces. I. Manifolds" (bottom of page 25). The authors say the problem "seems to be difficult". I don't think a solution has ever been published. However, perhaps one can find more about it in a later survey by Hocking, to which I do not have access. Anyone can comment on this?

EDIT: I've managed to get a copy of the short survey by John Hocking mentioned above. He wrote there that the problem (for manifolds) is still open.

2nd EDIT: There is a recent arXiv preprint by C. Laflamme, M. Pouzet and R.Woodrow on a somewhat related topic: See also the papers it cites (including the one linked to in my comment below this post).

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Thanks! I had read parts of Doyle-Hocking´s paper but I was not aware about page 25´s question. I didn´t restrict my question to manifolds, so it might be easier to find a counterexample out there. I can´t access Hocking´s 2003 survey either. – Ramiro de la Vega Jun 14 '13 at 15:12
Recently a related problem (but not a direct counterpart of your question) has been also considered in the category of graphs. Two graphs are called twins if each of them is isomorphic to a subgraph of the other. Is it true that every graph has (up to isomorphism) either no twins other than itself or infinitely many of them? See for some results and references. – Michał Kukieła Jun 14 '13 at 18:51

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