Is a bijective homotopy equivalence with bijective homotopy inverse a homeomorphism?

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?

Solution 1:

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 $\mathbf{hTop}$.