Two topologies over the same set which are carried to each other by a lattice isomorphism induce homeomorphic spaces?

Question

This is a question that I've been turning in my head and haven't been able to come up with a way to proceed to either prove it or construct a counter-example:

Let $X$ be a set and $\mathfrak{T}_X$ the set of all topologies over $X$. $\mathfrak{T}_X$ forms a complete lattice when ordered by set-inclusion $\subseteq$. Let $f$ be an automorphism of $\mathfrak{T}_X$ (as in a complete lattice isomorphism). Does it follow that for any topology $\mathcal{T}$ of $X$ that $(X,\mathcal{T})$ and $(X,f(\mathcal{T}))$ are homeomorphic topological spaces?

EDIT

I've added the Category Theory tag, as I feel like someone with more precision might be able to turn this into a question about a directed system in a small category and an endo-functor which preserves all limits.

Answer 1

Automorphisms of the lattice $L={\mathfrak T}(X)$ of topologies on arbitrary set $X$ are classified as follows (see [1], freely available here):

1. If $X$ is infinite or has cardinality $\le 2$, then every automorphism $\phi$ of $L$ is induced by a bijection $f_\phi: X\to X$. In particular, we obtain a natural homeomorphism of topological spaces $$ f: (X,\tau)\to (X, \phi(\tau)), \tau\in {\mathfrak T}(X). $$

2. If $X$ has finite cardinality $\ge 3$, then $Aut(L)$ is the direct product of its subgroup induced by bijections $X\to X$ as above and the order 2 group $Z_2$ whose generator is induced by the automorphism $\theta\in Aut(L)$ swapping subsets of $X$ and their complements. The latter automorphism (and its composition with an automorphism induced by a bijection), of course, will not induce a homeomorphism $(X,\tau)\to (X, \theta(\tau))$ for general $\tau$.

As a specific example, take any finite set $X$ of cardinality $\ge 3$ and its topology $\tau=\{\emptyset, \{x\}, X\}$ where $x$ is a certain element of $X$. Then $\theta(\tau)= \{\emptyset, X\setminus \{x\}, X\}$. It is clear that $(X,\tau)$ and $(X, \theta(\tau))$ are not homeomorphic.

See also [2] (freely available here), for a nice (although dated) survey of properties of the lattice $L$.

[1] Hartmanis, Juris, On the lattice of topologies, Can. J. Math. 10, 547-553 (1958). ZBL0087.37403.

[2] Larson, Roland E.; Andima, Susan J., The lattice of topologies: a survey, Rocky Mt. J. Math. 5, 177-198 (1975). ZBL0296.54003.