## Non-homeomorphic spaces that have continuous bijections between them

What are nice examples of topological spaces $X$ and $Y$ such that $X$ and $Y$ are not homeomorphic but there do exist continuous bijections $f: X \mapsto Y$ and $g: Y \mapsto X$?

-
Here are a couple of examples from Omar Antolín-Camarena: sbseminar.wordpress.com/2007/10/30/…. He's on MO, too: mathoverflow.net/users/644/omar-antolin-camarena. Maybe he'll want to elaborate. – Jonas Meyer Jul 5 2010 at 20:24
As you're looking for a list of examples, this should probably be CW – Charles Siegel Jul 5 2010 at 22:37

Recycling an old (ca. 1998) sci.math post:

": Anyone know an example of two topological spaces X and Y : with continuous bijections f:X-->Y and g:Y-->X such that : f and g are not homeomorphisms?

Let X = Y = Z x {0,1} as sets, where Z is the set of integers. We declare that the following subsets of X are open for each n>0. {(-n,0)} {(-n,1)} {(0,0)} {(0,0),(0,1)} {(n,0),(n,1)} This is a basis for a topology on X.

We declare that the following subsets of Y are open for each n>0. {(-n,0)} {(-n,1)} {(0,0),(0,1)} {(n,0),(n,1)} This is a basis for a toplogy on Y.

Define f:X-->Y and g:Y-->X by f((n,i))=(n,i) and g((n,i))=(n+1,i). Then f and g are continuous bijections, but X and Y are not homeomorphic.

This example is due to G. Paseman.

More generally, take a space X with three successively finer topologies T, T' and T''. Form two spaces which have underlying set ZxX, and "form the infinite sequences" .... T T T T' T'' T'' T'' .... and ... T T T T T'' T'' T'' T'' .... The continuous maps will take a finer topology in one sequence to a rougher topology in the other. You can make them bijective, and show that they are obviously non-homeomorphic for a judicious choice of X, T, T', and T''.

-
 If you want non-compact variations on this, e.g. making the two big spaces normal or connected, then choose X and the T's even more judiciously, and add an extra point or two as needed. And check your work, of course! I leave the compact version(s), or their impossibility, to you. Gerhard "Ask Me About System Design" Paseman, 2010.07.05 – Gerhard Paseman Jul 5 2010 at 23:19

My favorite, which is on the wikipedia page for "homeomorphism", is $\phi:[0,2\pi)\to S^1$, by $\phi(\theta)=(\cos\theta,\sin\theta)$, which is continuous and bijective, but not a homeomorphism.

-
The OP also wants a continuous bijection in the other direction. – Qiaochu Yuan Jul 5 2010 at 22:38
Ahh! Missed that. My apologies. – Charles Siegel Jul 5 2010 at 22:40

Here's a continuum analogue of Gerhard Paseman's answer: Let $X$ and $Y$ be topological spaces whose underlying sets are $\mathbb{R}$. As topological spaces, $X$ is the disjoint union of the open interval $(0,\infty)$ with a discrete space whose points are nonpositive reals, while $Y$ is the disjoint union of $(-1,0)$, $(1,\infty)$, and a discrete space whose points form the complement of those intervals. Translation by adding one is a continuous bijection from $X$ to $Y$, and also a continuous bijection from $Y$ to $X$, but the two spaces are not homeomorphic.

-
I like the example, but I do not see a good proof that they are not homeomorphic. Is e.g. (3,5) an open set in both X and Y, or just a subset of an open set? Gerhard "Ask Me About System Design" Paseman, 2010.07.05 – Gerhard Paseman Jul 6 2010 at 6:02
Look at the one-point compactifications, after noting that both spaces are locally compact Hausdorff. For $X$ we get a disjoint union of a circle and uncountably many points, while for $Y$ we get a disjoint union of a wedge of two circle and uncountably many points. If $X$ and $Y$ were homeomorphic, their one-point compactifications would be as well. – skupers Jul 6 2010 at 9:13
Look at the number of non-singleton connected components: $X$ has one and $Y$ has 2. This is a topological invariant. – Henno Brandsma Jul 7 2010 at 11:08

Here is an example which comes from using the spaces of Charles Siegel's post.

One have a continuous bijection from [0,1) to the circle given by the exponential function ( t-->exp(2ipit) ). The idea is to use this to construct our spaces. Take A to be a wedge of countably many (one for each integer) [0,1) attached at 0. Let A_n be the same wedge but replacing the [0,1) corresponding to the integers from 1 to n by circles.

X is going to be the disjoint union of A_2, A_4, A_6,... and countably many copies of A. Y is going to be the disjoint union of A_1, A_3, A_5,... and countably many copies of A.

We have a continuous bijection from A_n to A_{n+1} given by replacing the copy of [0,1) corresponding to the integer n+1 by a circle (as it is in A_{n+1}). Using this maps we are going to define f and g.

f is going to map one copy of A to A_1, A_2 to A_3, A_4 to A_5,... and so on, and the rest of the copies of A to the rest of the copies of A in Y. g is going to map A_1 to A_2, A_3 to A_4,... and so on, and the copies of A to the copies of A.

-

I don't have my copy of Kelley handy but I think in chapter 1 he gives the example where X is a countable disjoint union of open intervals and a countable discrete set while Y is a countable disjoint union of left-closed, right-open intervals and a countable discrete set.The point is that you can get a half closed interval from an open one by attaching an endpoint and you can build an open interval using a sequence of half closed intervals.

-