Is a map defined on atoms continuous?

Question

We say a map between complete lattices $f: \mathcal{L}\rightarrow\mathcal{J}$ is continuous if, for every $S$ we have that $f(\sup S)=\sup\{f(s)/ s\in S\}$.

Suppose now $\mathcal{L}$ and $\mathcal{J}$ are also atomic, meaning that any element is the supremum of the minimal elements smaller than itself; that is, if $A(\mathcal{L})$ is the set of minimal elements of $\mathcal{L}$ (atoms), we always have that $l=\sup\{k\in A(\mathcal{L})/ k\leq l\}$.

For simplicity, we denote $\{k\in A(\mathcal{L})/ k\leq l\}$ by $A(l)$; now, given a map $f:A(\mathcal{L})\rightarrow A(\mathcal{J})$, I would like to define a map $\overline{f}:\mathcal{L}\rightarrow\mathcal{J}$ by \begin{equation} \overline{f}(l)=\sup\{f(k)/ k\in A(l)\} \end{equation}

So, my question is, is $\overline{f}$ always continuous? If not, are there any reasonable assumptions to make it continuous?

One notices that, for any $S$, $\bigcup_{s\in S}A(s)\subseteq A(\sup S)$, since if $k$ is an atom smaller than some $s\in S$, since $s\leq \sup S$ we have that $k\leq \sup S$, and therefore $k\in A(\sup S)$. It follows from that that \begin{equation} \sup\{\overline{f}(s)/ s\in S\}=\sup \bigcup_{s\in S}\{f(k)/ k\in A(s)\}\leq \sup\{f(k)/ k\in A(\sup S)\}=\overline{f}(\sup S), \end{equation} but I can't work the equality... It looks like I am missing some subtlety, but I can't put my finger on it... It also looks like there is an obvious connection here with category theory and universal algebra, if that helps.

Answer 1

Here is a counterexample. Let $\mathcal{L}=\{0,a,b,c,1\}$ where $a,b,c$ are incomparable, let $\mathcal{J}=\mathcal{P}(\{a,b,c\})$, and let $f(i)=\{i\}$ for $i\in\{a,b,c\}$. Then $\overline{f}(a)=\{a\}$ and $\overline{f}(b)=\{b\}$ so $\overline{f}(a)\vee\overline{f}(b)=\{a,b\}$ but $\overline{f}(a\vee b)=\overline{f}(1)=\{a,b,c\}$.

It is true if you assume $\mathcal{L}$ is a frame, i.e. that it satisfies the infinite distributive law $\bigvee\{a\wedge s:s\in S\}=a\wedge\bigvee S$, since then $A(\bigvee S)=\bigcup_{s\in S} A(s)$ (if $a$ is an atom which is not in $\bigcup A(s)$ then $a\wedge s=0$ for all $s\in S$ and so $a\wedge\bigvee S=0$). Note though that this rather trivializes things, since it implies the map $A:\mathcal{L}\to\mathcal{P}(A(\mathcal{L}))$ is an isomorphism (with inverse taking a set of atoms to their join), so $\mathcal{L}$ is just a power set lattice.