# Which lattices are quotients of finite powerset lattices?

Let $$S$$ be a finite set, and let $$2^S$$ be its powerset, regarded as a lattice. Let $$L$$ be a quotient (in the category of lattices and maps which preserve $$\top,\bot,\wedge,\vee$$) of $$S$$. What can we say about $$L$$?

In fact, what I'd really like to know is: which finite semilattices are retracts (via $$\bot,\vee$$-preserving maps) of finite powerset lattices. I think the two questions are equivalent via a short-but-not-immediate argument.

Note that if $$P$$ is an arbitrary finite poset, then the lattice $$2^P$$ of poset maps $$P \to 2$$ is an example of such a lattice.

• While I do not immediately see that your last claim is true, it implies that the quotients of finite powerset lattices are exactly the finite distributive lattices. On the one hand, distributive lattices form a variety; on the other hand, every finite distributive lattice is of the form $2^P$ for some poset $P$ (see e.g. en.wikipedia.org/wiki/Birkhoff%27s_representation_theorem). – Emil Jeřábek Feb 23 at 8:26
• Except that this does not make any sense. The quotient must be a complemented distributive lattice, i.e., a Boolean algebra. The claim is just false. – Emil Jeřábek Feb 23 at 8:53

The class of finite powerset lattices is closed under quotients, up to isomorphism. That is, the quotients are exactly the lattice reducts of finite Boolean algebras.

In particular, any quotient $$L$$ of $$2^S$$ has to be a bounded distributive lattice, as the class of distributive lattices is a variety. Moreover, if $$x\in L$$ is an image of a set $$A\subseteq S$$, then the image of $$S\smallsetminus A$$ is an element $$y$$ such that $$x\lor y=\top$$ and $$x\land y=\bot$$. Thus, $$L$$ is complemented, i.e., it is a Boolean algebra, and as such it is isomorphic to $$2^{S'}$$ for some set $$S'$$.

Tim-

Emil answered the question you asked, but since you wrote what I'd really like to know is: which finite semilattices are retracts (via $$\bot$$, $$\vee$$-preserving maps) of finite powerset lattices'' let me add to his answer.

The $$2$$-element semilattice is injective in the class of semilattices.

The class of injectives is closed under products and retracts.

Up to isomorphism, the powers of the $$2$$-element semilattice are the power-set semilattices. Hence retracts of power-set semilattices must be injective.

Conversely, since the $$2$$-element semilattice is the only subdirectly irreducible semilattice, every semilattice is embeddable in some power $$2^S$$. And since an injective is a retract of any extension, it follows that every injective arises as a retract of some $$2^S$$,

Thus, the retracts of the power-set semilattices, $$2^S$$, are exactly the injective semilattices.

Theorem 2.8 of

The Category of Semilattices
ALFRED HORN and NAOKI KIMURA
Algebra universalis 1 (1971), 26-38.

proves that a (meet-)semilattice is injective iff it is complete and satisfies an infinite distributive law, namely that the meet distributes over infinite joins.

It follows that a finite semilattice is a retract of a power-set semilattice iff it is the semilattice reduct of a finite distributive lattice. (This shows that the examples in your last struckout paragraph exhaust all examples.)

• Thanks! I had just convinced myself of this using a different argument, but I'm very glad to know that there is an injectivity characterization like this, which makes it apparent. – Tim Campion Feb 23 at 18:23