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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 (mathoverflow.net/questions/26385/…). 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

1 Answer 1

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: http://arxiv.org/abs/1407.2894. 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 dx.doi.org/10.1016/j.jctb.2010.10.004 for some results and references. –  Michał Kukieła Jun 14 '13 at 18:51

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