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 $a\wedge b=\bigvee\,\{ \, c \;|\; c \subseteq a \textit{ and } c \subseteq b \,\}$.

  • $\begingroup$ 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
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    $\begingroup$ @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


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.

  • $\begingroup$ 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
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    $\begingroup$ 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

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