The existence of continuous bijections $f: X\rightleftarrows Y: g$ doesn't necessarily imply that $X$ and $Y$ are homeomorphic, if $f$ and $g$ aren't inverses of each other. I'm interested in examples of pairs of spaces that have distinct $\pi_1$ (or even distinct homology groups) where such continuous bijections exist in both direction.
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John Palmieri's answer is terrific, and makes me curious if we could find a path-connected example. – JMM Jun 09 '23 at 21:20
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I think that if we adjoin countably math paths to $X$ to get a space $\tilde{X}= ...---[0, 1)^2 --- [0, 1)^2 --- S^1\times [0, 1) --- (S^1)^2 --- (S^1)^2 ---...$ (and do the same thing to obtain $\tilde{Y}$ from $Y$), then extending the continuous bijections $X \rightleftarrows Y$ to continuous bijections $\tilde{X} \rightleftarrows \tilde{Y}$ in the obvious way gives us a path-connected example. – JMM Jun 09 '23 at 21:31
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The lovely answer at https://mathoverflow.net/a/251028/4194 should work: $X$ is the disjoint union of
- countably many copies of $[0,1) \times [0,1)$
- a single copy of $S^1 \times [0,1)$
- countably many copies of $S^1 \times S^1$
and $Y$ is the disjoint union of
- countably many copies of $[0,1) \times [0,1)$
- countably many copies of $S^1 \times S^1$
The post describes bijections in each direction, the nontrivial parts of which are based on the standard example of the continuous bijection $[0,1) \to S^1$.
There is a choice of basepoint for $X$ for which the fundamental group is $\mathbb{Z}$, but there is no such choice for $Y$.
John Palmieri
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