Does the existence of continuous injections $f: A\rightarrow B$ and $g: B\rightarrow A$ imply the existence of a bicontinuous bijection between A and B (ie topological equivalence)? If not, what is a good counter example?
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That is not a duplicate per se, but actually it has a stronger assumption, and the answer is still no. – tomasz Jun 21 '15 at 02:38
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1Related: http://mathoverflow.net/questions/1058/when-does-cantor-bernstein-hold – Noah Schweber Jun 21 '15 at 03:03
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No. $[0,1]$ injects continuously into its subset $(0,1)$ (say, by means of a linear map), but they're not homeomorphic (e.g., think about connectedness after deleting a point).
Micah
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Another counterexample is the middle-thirds Cantor set $C$ and $X=C\setminus\{0\}$: clearly $X$ embeds in $C$, and it's well-known (and not hard to prove) that $X$ is homeomorphic to $\Bbb N\times C$, where $\Bbb N$ has the discrete topology, so $C$ embeds in $X$. However, $C$ is compact, and $X$ is not, so $C$ and $X$ are not homeomorphic.
Brian M. Scott
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