3
$\begingroup$

Let $\mathcal{O}(P)$ be a finite, completely distributive lattice of all lower sets ordered by set inclusion. Moreover, let $K =\; \mathrel{\{} h(x) \mathrel{|} x \in \mathcal{O}(P) \mathrel{\}}$ be another poset ordered by set inclusion with $h:\mathcal{O}(P)\rightarrow K$ being a surjective function that is order- and join-preserving (which makes $K$ a join-semilattice, right?).

As far as I understand, because of $h$ being order-preserving, there must be a top element $h(\top)$ and a bottom element $h(\bot)$. If understand this question correctly, $K$ must hence be lattice (where meets are defined in terms of joins as $a\wedge b=\bigvee\mathrel{\{} c \mathrel{|} c \subseteq a \text{ and } c \subseteq b \mathrel{\}}$).

What does this mean with respect to $h$? Can I somehow conclude that $h$ is meet-preserving as well (which would leave $h$ as a lattice homomorphism)?

--- Old Question ---

I found this mathoverflow question here.

As part of the answer there is the proposition that "[...] a finite meet-semi-lattice with a maximum element is a lattice". I suppose this holds dually for a join-semilattice with a minimal element.

Is there any citable reference which proves this well-known proposition?

Background: I got a finite, completely distributive lattice (i.e., the set of all lower sets w.r.t a finite poset $P$ ordered by set inclusion; a superalgebraic lattice if I understand correctly), let us denote it as $\mathcal{O}(P)$. On top of that, I got a map $h : \mathcal{O}(P) \rightarrow K$ to another finite (semi-)lattice $K$ whose elements are sets ordered by inclusion as well. I already know that $K$ has a top $h(\top)$, a bottom $h(\bot)$ and that $h$ is join-preserving (which leaves $K$ at least as a join-semilattice, right?).

Now if I understand the aforementioned mathoverflow question correctly, this join-semilattice $K$ must be a lattice since there is a bottom. In particular, I am trying to prove that $h$ is meet-preserving where meets are defined in terms of joins as $a\wedge b=\bigvee\,\{ \, c \;|\; c \subseteq a \textit{ and } c \subseteq b \,\}$.

$\endgroup$
9
  • $\begingroup$ But your "background" paragraph confuses me. In particular, it is unclear there what you are trying to show, versus what you know. Having a poset map into $K$ cannot help you prove that $K$ is a lattice, at least not without more information about the map than you have provided (e.g., that it is surjective). $\endgroup$ Commented Oct 6, 2023 at 14:17
  • $\begingroup$ Anyways, for a reference to the fact you want, see Proposition 3.3.1 of Stanley's textbook "Enumerative Combinatorics," Volume 1. $\endgroup$ Commented Oct 6, 2023 at 14:18
  • $\begingroup$ Thank you for the quick response and the reference. Defining the meet as $a \wedge b = \vee\{c:c\leq a, b\}$ is indeed clear to me. I think my confusion comes from me not knowing enough about $h$ myself yet (as you pointed out as well). The poset $K$ is itself defined using $h$ (i.e., $K=\{ h(x) \mathrel{|} x \in \mathcal{O}(P) \}$) where $h$ is order-preserving and surjective but not injective. $\endgroup$
    – Björn
    Commented Oct 6, 2023 at 14:36
  • $\begingroup$ It's not automatically true that a join semilattice homomorphism of finite lattices preserves meets $\endgroup$ Commented Oct 6, 2023 at 14:37
  • 1
    $\begingroup$ @SamHopkins, my impression from the background of the question and the comments is the op wants to determine if h preserves meets and without more info on h we can't say anything $\endgroup$ Commented Oct 6, 2023 at 14:40

1 Answer 1

3
$\begingroup$

Okay, let me try to summarize.

The fact that a finite join semilattice with a minimum is a lattice is Proposition 3.3.1 of Stanley, "Enumerative Combinatorics," Volume 1.

Now suppose you have a finite lattice $L$ (corresponding to your $\mathcal{O}(P)$, but the fact that it is distributive is irrelevant), and another poset $K$ for which you have an order-preserving, join-preserving, surjective map $h\colon L \to K$. This does imply that $K$ is a join semilattice (for $a,b \in K$, we define the join $a\vee b$ to be $h(a' \vee b')$ where $a', b' \in L$ are pre-images under $h$ of $a,b$). Furthermore, the image of the minimum of $L$ under $h$ must be the minimum of $K$. And clearly $K$ is finite. So indeed, we can conclude that $K$ is a lattice.

EDIT: In response to the further question about whether $h$ must be meet-preserving, very simple examples show this is not the case. For instance, take $L$ to be the rank $2$ Boolean lattice of subsets of $\{1,2\}$, and $K$ to be the two element lattice $\hat{0} < \hat{1}$, with $h(\{1\})=h(\{2\})=h(\{1,2\})=\hat{1}$ and $h(\varnothing)=\hat{0}$. This $h$ is order-preserving, join-preserving, and surjective, but not meet-preserving.

$\endgroup$
3
  • $\begingroup$ Thank you very much for this explanation and your time. On top of $K$ being a lattice, can we conclude anything about $h$? Provided the information we have, is there a way to show that $h$ is meet-preserving? $\endgroup$
    – Björn
    Commented Oct 6, 2023 at 15:27
  • $\begingroup$ @Björn: See my edit. We cannot conclude that $h$ is meet-preserving. $\endgroup$ Commented Oct 6, 2023 at 15:33
  • 2
    $\begingroup$ The (more general) fact that a complete meet semilattice with a maximum (or dually a complete join semilattice with a minimum) is a complete lattice is Theorem 2 in Chapter IV of Birkhoff, Lattice Theory, 2nd edition. $\endgroup$
    – bof
    Commented Oct 7, 2023 at 3:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.