Given a poset $(P,\leq)$ the *interval topology* $\tau\_{\text{int}}(P)$ on $P$ is generated by
$$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$
where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.


Let $P, Q$ be posets and $e:P\to Q$ be order-preserving and surjective. Assume that $(P,\tau\_{\text{int}}(P))$ is Hausdorff. Does $(Q,\tau\_{\text{int}}(Q))$ have to be Hausdorff?



Image of poset with Hausdorff interval topology
The answer is no, not necessarily. 


For a counterexample, let $Q$ be any atomless complete Boolean algebra, and let us view it via Stone's theorem as a field of sets, so that $Q$ is a subalgebra for some set $X$ of the power set algebra $P=P(X)$, which is atomic. 


By [Proposition 5 in this paper of Northam](http://www.ams.org/journals/proc/1953-004-05/S0002-9939-1953-0057534-2/S0002-9939-1953-0057534-2.pdf#page=2) (due originally to Katetov 1951), a Boolean algebra is Hausdorff in the interval topology just in case every non-zero element sits over an atom. So $P$ is Hausdorff, but $Q$ is not. 


But meanwhile, we have a surjective order-preserving map $f:P\to Q$, defined by $f(x)=$ the join in $Q$ of the elements of $Q$ that are below $x$ in $P$. We use the completeness of $Q$ in order to know that this join exists in $Q$. 


