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Mathoverflow.

Are there any two topological spaces $X$ and $Y$ such that they are path connected and such that there exist continuous bijections $X\rightarrow Y$ and $Y\rightarrow X$, but and yet they are not homeomorphic?

Without Path-Connectedness requirement, this is easily fulfilled as the examples in the cited post.

If it indeed implies homeomorphism, how can I prove it?

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    $\begingroup$ My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure. $\endgroup$ – Martin Apr 9 '13 at 14:28
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See the example in my question here. There maps described there are continuous bijections and they induce continuous bijections between the cones of $X$ and $Y$. And cones are path-connected since any point can be joined to the apex.

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    $\begingroup$ Haha, I was wondering why this very similar question to yours appeared at the top of the page. $\endgroup$ – Dan Rust Dec 13 '13 at 13:39
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Yes, there are. But first lets consider criteria for homeomorphisms. let $\psi$$:X\rightarrow Y$ be a continuous bijection where $X$ is compact and $Y$ is Hausdorff. This is a homeomorphism. This means that topological invariants such as path-connectedness are preserved. So if we DO have a continuous bijection, and $X$ is compact and path-connected, and $Y$ is Hausdorff, then $f(X)\subseteq Y$ is path-connected.

Now to answer your question directly. Let $X=\mathbb{R}^2$ and $Y=\mathbb{R}$ with the standard topology. Both of these spaces are path connected. We can construct a bijection between the two. Let $\psi(x,y)=||x+y||-||x-y||$

this map is clearly surjective. It is also clearly injective. So it is bijective, and we know the norm function is continuous. So this is a continuous bijection, both of $\mathbb{R}^2$ and $\mathbb{R}$ are path-connected, but is well known that $X$ and $Y$ are not homeomorphic to each other.

Why don't we have a homemorphism? We have a continuous bijection, $X$ and $Y$ are Hausdorff, but $\mathbb{R}^2$ is not compact. So the assumption of compactness is essential.

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    $\begingroup$ Your function is symmetric in both arguments so it is certainly not bijective $\endgroup$ – ThorbenK Jan 8 '19 at 17:01

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