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