Left adjoint to the inclusion of Boolean algebras into distributive lattices

boolean-algebracategory-theorylattice-ordersuniversal-algebra

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.)

Best Answer

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.

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