Free distributive lattice

Remarks

It exists for every $S$ and is unique up to a unique isomorphism compatible with $\iota$. For nonempty $S$, every element of $F$ is a finite join of finite nonempty meets of generators $\iota(s)$. For finite $S$ with $n \ge 1$ elements, $F$ is isomorphic to the lattice of non-constant monotone Boolean functions of $n$ variables, so it has $M(n) - 2$ elements, with $M(n)$ the Dedekind number. If one also asks for $0$ and $1$ and homomorphisms that preserve them, the free bounded distributive lattice on $n$ generators has $M(n)$ elements.

Example: on $S = \set{x, y}$, the lattice $\set{x \wedge y, x, y, x \vee y}$ with $x \wedge y < x, y < x \vee y$ is free. A map $g$ into $L$ extends by $x \wedge y \mapsto g(x) \wedge g(y)$ and $x \vee y \mapsto g(x) \vee g(y)$, and this is forced.

Non-example: the two-element chain $0 < 1$ with $\iota(x) = 0$ and $\iota(y) = 1$ is distributive and generated by the image of $\iota$, but it is not free. The map $g(x) = 1$, $g(y) = 0$ into the same chain would need a homomorphism sending $0$ to $1$ and $1$ to $0$, which does not preserve the order.