Let $L$ be a complete and completely distributive lattice. An element $x\in L$ is well above $0$, denoted by $x\succ 0$, if for all $S\subseteq L$ with $\bigwedge S =0 $, there exists $s\in S$ with $x\ge s$. Say that $L$ is filtered if for all $x,y\succ 0$ the meet $x\wedge y$ satisfies $x\wedge y \succ 0$.
I'm looking for a reasonable construction that would act like the coproduct of two filtered complete and completely distributive lattices $L_1$ and $L_2$, i.e., a filtered complete and completely distributive lattice $W$ in which both $L_1$ and $L_2$ embed and such that the elements well above $0$ are closely related to the well above $0$ elements from each of $L_1$ and $L_2$. The question can be made more concrete by describing the morphisms to get a category, but at this point I don't want to commit to any particular choice of morphisms. As long as the construction is reasonable, I'm happy.
What if the co-product is simply the product ? That way if you have a lattice $K$ and a lattice $L$, $K \coprod L$ is made of the pairs $(k,l)$ with respective join and meets. You include $k$ to $(k,0)$ and $l$ to $(0,l)$. If you have $f:K \to X$ and $g:L\to X$, then $\theta: K \coprod L \to X: (k,l) \mapsto f(k) \vee g(l)$ is a function that make the coproduct diagram commute. \begin{gather} \theta ( k,0 ) = f(k) \vee g(0) = f(k) \vee 0 = f(k)\\ \theta ( 0,l ) = f(0) \vee g(l) = 0 \vee g(l) = g(l)\\ \end{gather}
If $\xi$ is another morphism $K \coprod L \to X$ such that $\xi(k,0) = f(x)$ and $\xi(0,l) = g(k)$, then \begin{equation} \xi(k,l) = \xi((k,0)\vee(0,l)) = \xi(k,0) \vee \xi(0,l) = f(k) \vee g(l) = \theta(k,l), \end{equation} so $\theta$ is unique.
You could do the same construct using the maximum and the meet instead of the minimun and the join, so I can't really see a canonical one if it were not for your notion of "well above". In this case $0_K$ and $0_L$ are mapped to $0_{K\coprod L}=(0,0)$ and obviously the notion of "well above" translate very nicely into the product.