Let $(X,\tau), (Y,\sigma)$ be $T\_2$-spaces such that there are injective lattice homomorphisms $f: \tau\to \sigma$ and $g:\sigma\to \tau$.


Does this imply that $(X,\tau)\cong (Y,\sigma)$?



$T_2$-spaces such that the lattices of open sets can be embedded into each other
No, $T\_2$-spaces with this property aren't necessarily homeomorphic.


Let $\mathbb{N} = \{0,1,2,\ldots\}$ be the set of non-negative integers and let $\tau = {\cal P}(\mathbb{N})$ be the discrete topology, and define $$\sigma = {\cal P}(\mathbb{N}\setminus\{0\}) \cup \{U\subseteq \mathbb{N}: 0\in U\text{ and } \mathbb{N}\setminus U\text{ is finite}\}.$$


(Essentially $\sigma$ is the 1-point compactification of the discrete topology on ${\cal P}(\mathbb{N}\setminus \{0\})$.)


Clearly $(\mathbb{N},\tau)$ and $(\mathbb{N},\sigma)$ are not homeomorphic as only one of them is compact, but not the other. Moreover, the inclusion map $\iota: \sigma \to \tau$ is an injective lattice homomorphism, as well as the map $f:\tau \to \sigma$ defined by $A \mapsto A+1:=\{a+1: a\in A\}$.


