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I've been thinking about this for a while, but didn't get very far. Maybe someone here can say something about it.

I know of an example of two spaces $X, Y$ with continuous bijections in both directions. Namely, define $X=Y=\Bbb Z\times\{0,1\}$. We define the basis for $X$ to consist of the sets $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ The basis for $Y$ will be $$\{(-n,0)\},\ \ \{(-n,1)\},\ \ \{(0,0),(0,1)\},\ \ \{(n,0),(n,1)\}$$ Here $n>0$ in both definitions.

Now let $f:X\to Y$ be the map $(n,i)\mapsto(n,i)$. Since the topology on $X$ is finer than that on $Y$, it is continuous. Define $g:Y\to X$ sending $(n,i)$ to $(n+1,i)$. Then both $f$ and $g$ are continuous bijections. However $g$ is not a homotopy equivalence. In fact, the only homotopy inverse of $f$ must map $\{(0,0),(0,1)\}$ to $(0,0)$, so it cannot be bijective. Since such a map exists, $f$ is indeed a homotopy equivalence.

The above example can be generalized: If $(X,\tau)$ is a non-indiscrete space which deformation retracts to a point $a$, $r:X\to\{a\}$ is the retraction and $i:\{a\}\to X$ the inclusion, then the identity map $(X,\tau)\to (X,\tau_{in})$ is a bijective homotopy equivalence with inverse $is$ where $s:(X,\tau_{in})\to\{a\}$.

So the question is

If $f:X\to Y$ is a bijective homotopy equivalence with a bijective homotopy inverse $g$, is $f$ a homeomorphism?

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  • $\begingroup$ This is a great question. Such spaces, if they exist, would have to be pretty bizarre I would think. Most homotopy equivalences which we meet 'in nature' are deformation retracts of one kind or another which certainly don't satisfy your condition. $\endgroup$ – Dan Rust Dec 11 '13 at 17:25
  • $\begingroup$ @DanielRust: I think I might have found a counterexample, but this seems just too easy to be true, so if you could check if this makes sense, that would be nice. What I'm thinking of are the respective cones of the two spaces above, and the maps induces by $f$ and $g$, call them $f'$ and $g'$. Since any two maps into a contractible space are homotopic, we have $g'f'\simeq 1_{CX}$ and $f'g'\simeq 1_{CY}$. So $g'$ and $f'$ are bijective homotopy equivalences inverse to each other. $\endgroup$ – Stefan Hamcke Dec 12 '13 at 21:02
  • $\begingroup$ This looks fine as a counterexample to me (and a clever one at that). My only concern would be in trying to show that $f'$ is not a homeomorphism. I guess if it was a homeomorphism, it would induce a homeomorphism from $X\times\{1\}$ to $f'(X\times\{1\})=Y\times\{1\}$ and the restricted map would be $f\times\mbox{Id}_{\{1\}}$ which is a homeomorphism iff $f$ is a homeomorphism. $\endgroup$ – Dan Rust Dec 12 '13 at 22:47
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After some thinking I found a counterexample. It doesn't need some very strange spaces, actually it is a rather simple construction and I'm surprised that I hadn't thought of this trick before.

Let $X$ and $Y$ be two non-homeomorphic spaces with continuous bijections $f:X\to Y$ and $g:Y\to X$. An example of such spaces is described in my question. The maps $f\times 1:X×I\to Y×I$ and $g×1:Y×I\to X×I$ are continuous bijections, and they induce continuous bijections $\tilde f:CX\to CY$ and $\tilde g:CY\to CX$ between the cones. If $\tilde f$ were a homeomorphism, then so would be its restriction to $X\times\{0\}\cong X$, which is practically just $f$. Hence $\tilde f$ and $\tilde g$ are no homeomorphisms, but they are homotopy inverses to each other as the cone over a space is always contractible and contractible spaces are terminal objects in the homotopy category $\mathbf{hTop}$.

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