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Does there exist two topological space $X$ and $Y$ such that $X$ and $Y$ imbed in each other, $X$ is a quotient of $Y$, $Y$ is quotient of $X$, but $X$ and $Y$ are not homeomorphic?

The spaces $[0,1]$ and $(0,1)$ imbed in each other and are not homeomorphic. Also, $[0,1]$ is a quotient of $(0,1)$, but $(0,1)$ is not a quotient of $[0,1]$.

The spaces $[0,1]$ and $S^1$ are quotients of each other and are not homeomorphic. Also, $[0,1]$ imbeds in $S^1$, but $S^1$ does not imbed in $[0,1]$.

I'm wondering if there are two topological spaces which satisfy these properties. I cannot find a pair of examples, but I can't prove the impossibility without some nontrivial relationship between the imbeddings/quotients.

Any help in proving the impossibility or demonstrating a pair of examples is appreciated.

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  • $\begingroup$ Maybe $D^2$ and $D^2\vee S^1$ (wedge at the points $(1,0)$ in either of them). $\endgroup$
    – Stefan Hamcke
    Oct 19, 2014 at 22:26
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    $\begingroup$ What about $X=\{z\in\mathbb C:|z+1|\le1\}\cup\{z\in\mathbb C:|z-1|\le1$ and $Y=\{z\in\mathbb C:|z|\le2\}$? $\endgroup$
    – bof
    Oct 19, 2014 at 22:26
  • $\begingroup$ @StefanHamcke I don't see how $D^2\vee S^1$ is a quotient of $D^2$ $\endgroup$
    – user123641
    Oct 19, 2014 at 22:30
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    $\begingroup$ I think $\{z\in\mathbb C: |z+1/2|\le1/2\}\cup[0,1]\times\{0\}$ is a retract of $D^2$. Then you can wrap the interval around the circle. $\endgroup$
    – Stefan Hamcke
    Oct 19, 2014 at 22:34
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    $\begingroup$ What is nice about these spaces is that they are not even homotopy equivalent :-) $\endgroup$
    – Stefan Hamcke
    Oct 19, 2014 at 22:39

1 Answer 1

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The spaces $X=\{z\in\mathbb C:|z+1|\le1\}\cup\{z\in\mathbb C:|z-1|\le1\}$ and $Y=\{z\in\mathbb C:|z|\le2\}$ are embeddable in each other and are quotients of each other, but are not homeomorphic.

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