It is very easy to give examples of topological spaces $A$ and $B$ that are not homeomorphic to each other, with a continuous injection from $A$ to $B$, and a continuous injection from $B$ to $A$. See, for example $(0,1)$ and $[0,1]$, as demonstrated in the answer to a similar question on this forum, or this related question. In other words, there exist topological spaces that do not satisfy the Cantor-Schroeder-Bernstein property, when considering injective maps. I was wondering if this fails when considering surjective maps as well. That is, given two topological spaces $A$ and $B$, if there exists a continuous surjection from $A$ to $B$, and a continuous surjection from $B$ to $A$, must $A$ and $B$ be homeomorphic to each other? I think this shouldn't be true: there should be a simple counterexample, but I just can't find one! It would be great to have a counterexample for locally compact Hausdorff spaces, because I want to eventually understand problems with the Cantor-Schroeder-Bernstein property for $C^\star$ algebras...
Asked
Active
Viewed 150 times
4
-
3$[0,1]$ and $[0,1]^2$ – Alessandro Codenotti Jun 30 '21 at 07:45
-
1$\mathbb N$ and $\mathbb N\times\mathbb Q$ – bof Jun 30 '21 at 07:53
-
1Oops, you wanted locally compact. OK, $\mathbb N$ and $\mathbb N\times X$ where $X$ is a countably infinite compact Hausdorff space. – bof Jun 30 '21 at 07:56
-
1This question on another maths stackexchange site is an almost duplicate. (It's stronger as its asks for two continuous bijections between non-homeomorphic spaces). – Henno Brandsma Jun 30 '21 at 07:57
-
@HennoBrandsma, Aah, I see! The link you provided certainly answers my question, as well as provides a deeper understanding regarding my ulterior motive: CSB for topological spaces (and eventually C* Algebras, I hope...) Thanks a lot! – Ujan Chakraborty Jun 30 '21 at 08:06
-
1Any two spaces which are countable, zero-dimensional, and not compact. – bof Jul 03 '21 at 04:19
2 Answers
6
There is a simple example with $A$ and $B$ both compact!
$[0,2\pi]$ is not homeomorphic to $S^{1}$. The function $t \mapsto (\cos t,\sin t)$ is a continuous surjection from $[0,2\pi]$ onto $S^{1}$, while the function $(x,y) \mapsto \pi (1+x)$ is a continuous surjection from $S^{1}$ onto $[0,2\pi]$.
Greg Martin
- 92,241
Kavi Rama Murthy
- 359,332
3
Take $\Bbb R$ and $[0,\infty)$, with their usual topologies. They are not homeomorphic. But$$\begin{array}{ccc}\Bbb R&\longrightarrow&[0,\infty)\\x&\mapsto&x^2\end{array}$$and$$\begin{array}{ccc}[0,\infty)&\longrightarrow&\Bbb R\\x&\mapsto&x\sin(x)\end{array}$$are continuous and surjective.
José Carlos Santos
- 440,053