I apologize if my question is trivial. I am a group theorist with a minor knowledge of topology. Suppose $(X, T_X)$ and $(Y, T_Y)$ are two topological spaces and there is an inclusionreversing (inclusion preserving) bijection $\varphi:T_X\to T_Y$. What is the name of this kind of equivalence? What can be said about the relation of these two spaces?

4$\begingroup$ I believe this is the same thing as an isomorphism between the corresponding locales. $\endgroup$– Qiaochu YuanOct 28, 2017 at 16:23

5$\begingroup$ I am a bit confused. Is $\varphi$ inclusionpreserving or inclusionreversing? $\endgroup$– Denis NardinOct 29, 2017 at 12:25

$\begingroup$ @DenisNardin: one of these properties! $\endgroup$– Sh.M1972Oct 30, 2017 at 5:35

$\begingroup$ @QiaochuYuan: May I have some refernces? $\endgroup$– Sh.M1972Oct 30, 2017 at 5:35
2 Answers
As far as I can see there are two possibilities where we can say more:
The bijection is order preserving. As Qiaochu Yuan said in a comment, this is the same as asking that their associated locales are isomorphic. Thanks to the equivalence of sober spaces with spatial locales, this is the same as asking that their soberifications are homeomorphic. In particular, if $X$ and $Y$ are already sober (a very common condition, sober is strictly weaker than Hausdorff), this is the same as asking that $X$ and $Y$ are homeomorphic.
The bijection is order reversing. Then the lattice of opens of both spaces need to have arbitrary meets distribute over finite joins (i.e. each frame of opens must have the property that its opposite lattice is a frame too). This is a very rare situation, that happens mainly for Alexandrov spaces,although as მამუკაჯიბლაძე notes in the comments there might be other examples.

$\begingroup$ Are you sure about the first? I think I had a counterexample here. True, I never actually published it, but... Look at example (4.4) there: $X=(0,1)$ with $T_X:=\{(x,1)\mid0\leqslant x\leqslant1\}$. Now take $Y=X$ with $T_Y:=\{(0,y)\mid0\leqslant y\leqslant1\}$. Clearly both $T_X$ and $T_Y$ are (both isomorphic and anti)isomorphic to $([0,1],\leqslant)$. But they are not Alexandroff, are they? $\endgroup$ Oct 30, 2017 at 9:22

1$\begingroup$ @მამუკაჯიბლაძე Ah! I got mixed up between the poset of opens being closed under intersections and having arbitrary meets... I think this is saying something like "their soberification is Alexandroff", let me think about it $\endgroup$ Oct 30, 2017 at 9:25

$\begingroup$ Mmm not sure, I believe my guys are sober... $\endgroup$ Oct 30, 2017 at 9:27

1$\begingroup$ There are actually several gradually stronger conditions: opens having arbitrary meets (this always holds); arbitrary meets distributing over finite joins (i. e. dual of opens being a frame); dual of opens being a topology (i. e. having enough points, a.k.a. completely primes); Alexandroffness; and (although not relevant here) being selfdual. I believe there are examples separating all of the above. I never took into account nonsober world, though. $\endgroup$ Oct 30, 2017 at 9:30

$\begingroup$ Sorry I took liberty to change the statement accordingly to the comment. Please revise it in case I did something wrong. $\endgroup$ Oct 30, 2017 at 11:31
Edited after realizing that my preceding answer was false.
By elementary tools (i.e., without soberifications) it can be shown that for $T_1$spaces this equivalence coincides with the homeomorphness.
Theorem 1. For $T_1$spaces $X,Y$ the following conditions are equivalent:
$X$ and $Y$ are homeomorphic;
The topologies of the spaces $X,Y$ are order isomophic.
Proof. The implication $(1)\Rightarrow(2)$ is trivial.
To prove that $(2)\Rightarrow(1)$, assume that there exists an order isomorphism $i:\tau_X\to \tau_Y$ between the topologies of the spaces $X$ and $Y$.
Since $X$ is a $T_1$space, for every $x\in X$ the open set $X\setminus\{x\}$ is a maximal element of the poset $\tau_X\setminus\{X\}$. Since $i$ is an isomorphism of the posets $\tau_X$ and $\tau_Y$, the set $i(X\setminus\{x\})$ is a maximal element of the poset $\tau_Y\setminus\{Y\}$ and hence $i(U_x)=Y\setminus \{f(x)\}$ for some point $f(x)\in Y$. The injectivity of $i$ implies that the map $f:X\to Y$ is injective. To see that $f$ is surjective, observe that for every $y\in Y$ the set $i^{1}(Y\setminus \{y\})$ is a maximal element of the poset $\tau_X\setminus\{X\}$ and hence it coincides with the set $X\setminus\{x\}$ for some $x\in X$. It follows that $Y\setminus\{y\}=i(X\setminus\{x\})=Y\setminus \{f(x)\}$ and hence $y=f(x)$.
Therefore the map $f:X\to Y$ is bijective. To show that $f$ is continuous, it suffices to check that for every open set $U\in\tau_Y$ we get $f^{1}[U]=i^{1}(U)$.
To show that $f^{1}[U]\subset i^{1}(U)$, observe that for any $x\in f^{1}[U]$ we get $f(x)\in U$ and hence $(Y\setminus \{f(x)\})\cup U=Y$. Taking into account that order isomorphisms of lattices preserve lattice operations, we conclude that $$(X\setminus\{x\})\cup i^{1}(U)=i^{1}(Y\setminus \{f(x)\})\cup i^{1}(U)=i^{1}(Y)=X$$ and hence $x\in i^{1}(U)$. Therefore $f^{1}[U]\subset i^{1}(U)$.
To see that $i^{1}(U)\subset f^{1}[U]$, observe that for any $x\in i^{1}(U)$ we get $X=(X\setminus \{x\})\cup i^{1}(U)$ and hence $(Y\setminus\{f(x)\})\cup U=i(X\setminus\{x\})\cup U=Y$, which implies $f(x)\in U$ and hence $x\in f^{1}[U]$.
This shows that the preimage $f^{1}[U]=i^{1}(U)$ of the open set $U$ is open and $f$ is continuous. By analogy we can check that the map $f^{1}:Y\to X$ is continuous. So, $f$ is a homeomorphism.
Remark. The theorem does not generalize to $T_0$spaces as the $T_0$space $X=[0,1)$ with topology $\tau_X=\{[0,a):0\le a\le 1\}$ and its subspace $Y=(0,1)$ are not homeomorphic but have order isomorphic topologies.
Also there exist a $T_1$space $X$ and a sober space $Y$ which are not homeomorphic but have orderisomorphic topologies.
Example 1s. Take any infinite set $X$ such that $X\notin X$. On the set $Y= X\cup\{X\}$ consider the topology $\tau_Y=\{\emptyset\}\cup\{Y\setminus F:F\subset X$ is finite$\}$. The space $Y$ is sober but not $T_1$ and its subspace $X$ is $T_1$ but not sober. Nonetheless the spaces $X,Y$ have orderisomorphic topologies.
Added after a comment of მამუკა ჯიბლაძე:
Theorem 1 can be generalized to $T_{Constructible}$spaces. We recall that a set $A$ in a topological space $X$ is constructible (resp. Borel) if it belongs to the smallest algebra (resp. $\sigma$algebra) of subsets of $X$, containing all open subsets of $X$.
A topological space $X$ is defined to be $T_{Constructible}$space (resp. $T_{Borel}$space) if for each $x\in X$ the singleton $\{x\}$ is a constructible (resp. Borel) subset of $X$.
By Theorem 2.1 of this paper, a topological space $X$ is $T_{Borel}$ (resp. $T_{Constructible}$) if and only if each singleton $\{x\}\subset X$ can be written as the intersection $F\cap U$ of a closed set $F\subset X$ and an (open) $G_\delta$set $U$.
This characterization implies that the class of $T_{Constructible}$spaces coincides with the class of $T_D$spaces (i.e. spaces in which each singletion is an intersection of an open and a closed sets).
So, we have the implications
$$T_1\Rightarrow T_D\Leftrightarrow T_{Constructible}\Rightarrow T_{Borel}\Rightarrow T_0.$$
Observe that if the set $X$ in Example 1s is countable, then the spaces $X,Y$ from this example are $T_{Borel}$. This shows that Theorem 1 does not generalize to $T_{Borel}$spaces.
On the other hand, it does generalize to $T_{Constructible}$spaces (and to $T_D$spaces as was suggested by მამუკა ჯიბლაძე).
Theorem 2. Two $T_{Constructible}$spaces are homeomorphic if and only if their topologies are orderisomorphic.
Proof. Let $(X,\tau_X)$ be a $T_{Constructible}$space.
A pair $(V,W)$ of open sets in $X$ is called pointdetermining if
$V\subset W$ and $V\ne W$;
for any open set $U\in\tau_X$ with $V\subset U\subset W$ we get $U=V$ or $U=W$;
for any open set $U\subset X$ with $V\subset U$ and $W\not\subset U$ we get $V=U$.
Using the fact that each $T_{Constructible}$space is a $T_D$space, one can easily check:
Lemma 1. For any point $x\in X$ the pair $(V_x,W_x)$ of the open sets $V_x=X\setminus\overline{\{x\}}$ and $W_x=V_x\cup\{x\}$ is pointdetermining.
Lemma 2. Each pointdetermining pair $(V,W)$ is equal to $(V_x,W_x)$ for a unique point $x\in X$.
Lemma 3. An open set $U\in\tau_X$ is a neighborhood of a point $x\in X$ if and only if $W_x\subset V_x\cup U$.
Now theorem 2 easily follows from Lemmas 13 and the observation that pointdetermining pairs of open sets are preserved by orderisomorphisms of topologies.
Added in an edit. The reconstruction theorems for continuous maps and homeomorphisms on sober and $T_D$spaces can be found in this book, which also contains Example 1s (see Example 3.1 in this book).

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1$\begingroup$ Then it is (itself, rather than only its StoneČech) recoverable from the frame of opens, no? $\endgroup$ Oct 30, 2017 at 11:35

1$\begingroup$ I believe I've seen somewhere that this more generally holds for $T_D$spaces. Actually there is an interesting question here: the property holds for (a) sober spaces; (b) $T_1$spaces, as you show, or, more generally, $T_D$spaces, if I remember correctly. Now none of (a), (b) contains the other, and also the property does not extend to the union of these classes. The question then is whether there are still larger (or any other) classes of spaces for which isomorphism of their topologies implies homeomorphism. $\endgroup$ Nov 1, 2017 at 8:02

1$\begingroup$ @მამუკაჯიბლაძე Thank you for the comment. Indeed, in $T_D$spaces, points can be written as differences of two open sets, so, points can be encoded as pairs $(U,V)$ where $V\subset U$ is a maximal proper subset in an open set $U$. And the $T_0$space in my example is not $T_D$space. So, I will think a bit more. $\endgroup$ Nov 1, 2017 at 12:22

1$\begingroup$ @მამუკაჯიბლაძე I added a generalization to $T_{Constructible}$spaces (which coincide with $T_D$spaces) but have more natural definition. $\endgroup$ Nov 2, 2017 at 7:37