# Continuous bijections vs. Homeomorphisms

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 $1$?

-
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