Let $(X, \mathcal{T}_X)$ and $(Y, \mathcal{T}_Y)$ be topological spaces, and $f : X \to Y$ and $g : Y \to X$ be surjective local homeomorphisms. Is $\mathcal{T}_X$ homeomorphic to $\mathcal{T}_Y$?
For background, see here.
Since $f$ and $g$ are surjections, we have $|X| = |Y|$, so at least there exists a bijection. I wonder whether König's proof of the Schröder-Bernstein theorem could be modified to prove this.
No, it does not follow. For instance, let $X$ be the 2-dimensional torus with one puncture and $Y$ be the 2-dimensional sphere with three punctures. It is a nice exercise in algebraic topology to verify that $X$ is not homeomorphic to $Y$ but there are covering maps $X\to Y$ and $Y\to X$. One can arrange similar examples where $X$ is not even homotopy-equivalent to $Y$, say, $Y$ is the 2-dimensional sphere with 4 punctures and $X$ is as before.
Edit: There are similar examples of covering maps between compact 3-dimensional manifolds: One can take $X$ to be the total space of the (unique) orientable circle bundle over the torus $T^2$ with the Euler number 1 and $Y$ the total space of he (unique) orientable circle bundle over $T^2$ with the Euler number $2$. This is the example suggested by Yves in his comment to my question here.
