Left adjoint to the inclusion of Boolean algebras into distributive lattices

Question

Let $\mathbf{Boole}$ be the category of Boolean algebras.

Let $\mathbf{BDL}$ be the category of bounded distributive lattices.

There is a fully faithful functor ${\mathbf{Boole} \rightarrow \mathbf{BDL}}$ and it has a left adjoint because Boolean algebras can be defined as bounded distributive lattices with an additional unary operation satisfying some equations.

I have been unable to give or find an explicit description of the left adjoint.

Can anyone describe it or give a reference? (Please, no Kan extensions.)

Answer 1

Every boolean algebra $B$ can be embedded in a powerset, namely the set of sets of boolean algebra homomorphisms $B \to \{ \bot, \top \}$. (This is a weak version of Stone duality.) We can use the same idea to construct the free boolean algebra on a lattice. (For me, lattices are always bounded.)

For a lattice $A$, let $\textrm{pt} (A)$ be the set of lattice homomorphisms $A \to \{ \bot, \top \}$. There is a natural lattice homomorphism $\eta_A : A \to \mathscr{P} (\textrm{pt} (A))$, defined by the following formula: $$\eta_A (a) = \{ \mu \in \textrm{pt} (A) : \mu (a) = \top \}$$ Let $\mathscr{B} (A)$ be the boolean subalgebra of $\mathscr{P} (\textrm{pt} (A))$ generated by the image of $\eta_A : A \to \mathscr{P} (\textrm{pt} (A))$. Disjunctive normal form tells us that every element of $\mathscr{B} (A)$ is a finitary join of a finitary meet of possibly negated elements of the image of $\eta_A$, but since $\eta_A$ is a lattice homomorphism we can do somewhat better: every element of $\mathscr{B} (A)$ is of the form $$(\eta_A (a_0) \land \lnot \eta_A (a'_0)) \vee \cdots \vee (\eta_A (a_{n-1}) \land \lnot \eta_A (a'_{n-1}))$$ for some $a_0, a'_0, \ldots, a_{n-1}, a'_{n-1}$ in $A$.

Now consider $\eta_A$ as a lattice homomorphism $A \to \mathscr{B} (A)$. I claim it is initial among all lattice homorphisms $\phi : A \to B$ where $B$ is a boolean algebra. Suppose given such a $\phi$. We have a map $\textrm{pt} (\phi) : \textrm{pt} (B) \to \textrm{pt} (A)$ induced by precomposition, and this induces a (complete) boolean algebra homomorphism $\mathscr{P} (\textrm{pt} (A)) \to \mathscr{P} (\textrm{pt} (B))$. What does it do to $\eta_A (a)$? Well, by naturality, it gets mapped to $\eta_B (b)$. So the boolean subalgebra generated by the image of $\eta_A$ is mapped into the boolean subalgebra generated by the image of $\eta_B$, i.e. we get a boolean algebra homomorphism $\mathscr{B} (\phi) : \mathscr{B} (A) \to \mathscr{B} (B)$. But $\eta_B : B \to \mathscr{B} (B)$ is an isomorphism, so we obtain $\phi = \eta_B^{-1} \circ \mathscr{B} (\phi) \circ \eta_A$. This proves existence, and uniqueness is automatic because $\mathscr{B} (A)$ is generated (as a boolean algebra) by the image of $\eta_A$.

Note that I have not restricted to distributive lattices in the above discussion. However, since every boolean algebra is distributive, if $\eta_A : A \to \mathscr{B} (A)$ were an injective map then $A$ would be distributive, so distributivity of $A$ is a necessary condition for injectivity of $\eta_A$. In fact, it is also sufficient.

Suppose $A$ is distributive. We must show that, for any $a_0$ and $a_1$ in $A$, if $a_1 \nleq a_0$, then there is some $\mu : A \to \{ \bot, \top \}$ such that $\mu (a_0) = \bot$ and $\mu (a_1) = \top$. By Zorn's lemma, there is a maximal filter $F \subseteq A$ such that $a_0 \notin F$ and $a_1 \in F$. Define $\mu (a) = \top$ if $a \in F$ and $\mu (a) = \bot$ if $a \notin F$. This is a lattice homomorphism:

• Obviously $\mu (\bot) = \bot$ and $\mu (\top) = \top$.
• Since $F$ is closed under $\land$, if $\mu (a_2) = \mu (a_3) = \top$ then $\mu (a_2 \land a_3) = \top$ too. Since $F$ is upward-closed, if $\mu (a_2 \land a_3) = \top$ then $\mu (a_2) = \mu (a_3) = \top$, hence if $\mu (a_2) = \bot$ then $\mu (a_2 \land a_3) = \bot$. Thus $\mu$ preserves $\land$.
• Since $F$ is upward-closed, if $\mu (a_2) = \top$, then $\mu (a_2 \lor a_3) = \top$. Since $F$ is maximal among filters not containing $a_0$, $a_2 \notin F$ if and only if there is some $a_4 \in F$ such that $a_2 \land a_4 \le a_0$; so if $\mu (a_2) = \mu (a_3) = \bot$ then we have $a_4$ and $a_5$ such that $a_2 \land a_4 \le a_0$ and $a_3 \land a_5 \le a_0$, and since $A$ is distributive, we have $(a_2 \lor a_3) \land (a_4 \land a_5) \le a_0$, hence $\mu (a_2 \lor a_3) = \bot$. Thus $\mu$ preserves $\lor$.

This completes the proof. Incidentally, the construction of $\mathscr{B} (A)$ I outlined is basically what the special adjoint functor theorem would construct, and the description of the elements of $\mathscr{B} (A)$ is what you would get from a generators-and-relations construction.

Answer 2

This is a very good question to ask yourself! The left adjoint called the free Boolean extension of a bounded distributive lattice – this should lead you to the right references. The simplest way to think about it is in terms of the Priestley dual: each bounded distributive lattice is the lattice of clopen upsets of a Priestley space (a certain ordered topological space), and its free Boolean extension is the Boolean algebra of all clopen subsets of this topological space.

If you are not comfortable with Priestley duality, you can restrict the above to the finite case: each finite distributive lattice is the lattice of upsets of a finite poset, and its free Boolean extension is the Boolean algebra of all subsets of this finite poset.