# More on a construction on two boolean lattices

Let $\mathfrak{A}$ and $\mathfrak{B}$ be (fixed) boolean lattices (with lattice operations denoted $\sqcup$ and $\sqcap$, bottom element $\bot$ and top element $\top$).

I call a boolean funcoid a pair $(\alpha;\beta)$ of functions $\alpha:\mathfrak{A}\rightarrow\mathfrak{B}$, $\beta:\mathfrak{B}\rightarrow\mathfrak{A}$ such that (for every $X\in\mathfrak{A}$, $Y\in\mathfrak{B}$) $$Y\sqcap^{\mathfrak{B}}\alpha(X)\ne\bot^{\mathfrak{B}} \Leftrightarrow X\sqcap^{\mathfrak{A}}\beta(Y)\ne\bot^{\mathfrak{A}}.$$

(Boolean funcoids are a special case of pointfree funcoids as defined in my free ebook.)

I call a boolean funcoid complete when $\alpha\bigsqcup X=\bigsqcup \alpha[X]$ whenever both suprema $\bigsqcup X$ and $\bigsqcup f[X]$ exist.

Is every boolean funcoid complete?

If not, does it hold for special cases of complete, atomic, atomistic boolean lattices?

## 1 Answer

$$Y \sqcap \alpha \bigsqcup S \neq \emptyset \Leftrightarrow \bigsqcup S \sqcap \beta Y \neq \emptyset \Leftrightarrow \exists X \in S : X \sqcap \beta Y \neq \emptyset \Leftrightarrow \exists X \in S : Y \sqcap \alpha X \neq \emptyset \Leftrightarrow Y \sqcap \bigsqcup \alpha [S] \neq \emptyset$$ for every $Y$. Thus follows $\alpha \bigsqcup S = \bigsqcup \alpha [S]$.