Path Connectedness and continuous bijections

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?

• My mistake: I overlooked that you wanted a map in both directions. Please ignore my vote for closure. – Martin Apr 9 '13 at 14:28

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.

• Haha, I was wondering why this very similar question to yours appeared at the top of the page. – Dan Rust Dec 13 '13 at 13:39

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.

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