Slick proof for uniqueness of the elements in free distributive lattices

Question

Let $D(X)$ be the free distributive bounded lattice on a finite set $X$. Then one can show that every element of $D(X)$ has the form

$$\bigvee_{A \in \mathcal{A}} \bigwedge_{a \in A} a$$

for some antichain $\mathcal{A} \subseteq P(X)$. For example, the antichain $\{\{x\},\{y,z\}\}$ yields the element $x \vee (y \wedge z)$. This is true for every distributive bounded lattice generated by a set $X$. If $X$ is infinite, we just take a finite antichain of finite subsets of $X$.

I would like to show that this antichain is uniquely determined, without constructing $D(X)$ as the set of reduced words (via antichains), which seems to be the usual approach. The reason is that it will be a mess to write down the lattice operations in terms of reduced words, and proving even basic things such as absorption laws require some computational effort that I would like to avoid.

Instead, I would like to follow a similar approach as the one to verify the uniqueness of reduced words in a free group $F(X)$, using its universal property. Here, one writes down a map $X \to \mathrm{Sym}(R)$, where $R$ is the set of reduced words, and then extends it to a homomorphism $F(X) \to \mathrm{Sym}(R)$ via the universal property of $F(X)$. Finally, one verifies that $F(X) \to \mathrm{Sym}(R) \xrightarrow{ev_1} R$ is left inverse to $R \to F(X)$, so that $R \to F(X)$ is indeed injective. This approach works more generally to describe the elements of a coproduct of groups or monoids, see here and there.

So I assume I need to construct a certain lattice $L$ based on the set of antichains on $X$, construct a map $X \to L$ and then extend it to $D(X) \to L$ via the universal property. I tried for $L$ the power set of the set of antichains, but this didn't work out.

Answer 1

There is a construction of the free distributive lattice that follows by (Birkhoff) duality: given the set X, consider $\mathcal{P}(X)$, understood as a powerset under inclusion. This is a partially ordered set. Now take the set of upwards closed subsets of $\mathcal{P}(X)$. Then this forms a distributive lattice (with set-theoretic intersection and union), and indeed is the dual of the free distributive lattice on $X$ many generators. The map from $X$ to $Up(\mathcal{P}(X))$ maps an element $x$ to the principal upset at $\{x\}$.

Your observation follows from this: a given upwards closed subset $U$ is a finite union of principal upsets of the form ${\uparrow}S$ where $S$ is a finite subset of $U$. We think of the element ${\uparrow}S_{0}\cup...\cup {\uparrow}S_{n}$ as $\bigvee_{i=0}^{n}\bigwedge_{s\in S_{i}}s$. This gives an obvious way to lift any map from $X$ to a distributive latice $D$ to a homomorphism, and the lift is unique because it is determined by its image on the singletons.

I can add more details as desired, let me know if this is what you were looking for!