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arxiv: math/0501372 · v1 · submitted 2005-01-22 · 🧮 math.GM

Join-semilattices with two-dimensional congruence amalgamation

classification 🧮 math.GM
keywords everykappalatticeco-brouwerianconditionallyconcexistssemilattice
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We say that a (∨,0)-semilattice S is conditionally co-Brouwerian, if (1) for all nonempty subsets X and Y of S such that X $\leq$ Y (i.e., x $\leq$ y for all (x, y) $\in$ X x Y), there exists z $\in$ S such that X $\leq$ z $\leq$ Y, and (2) for every subset Z of S and all a, b $\in$ S, if a $\leq$ b ∨ z for all z $\in$ Z, then there exists c $\in$ S such that a $\leq$ b ∨ c and c $\leq$ Z. By restricting this definition to subsets X, Y, and Z of less than $\kappa$ elements, for an infinite cardinal $\kappa$, we obtain the definition of a conditionally $\kappa$- co-Brouwerian (∨, 0)-semilattice. We prove that for every conditionally co-Brouwerian lattice S and every partial lattice P, every (∨, 0)-homomorphism $\phi$: Conc P $\to$ S can be lifted to a lattice homomorphism f : P $\to$ L, for some relatively complemented lattice L. Here, Conc P denotes the (∨, 0)-semilattice of compact congruences of P. We also prove a two-dimensional version of this result, and we establish partial converses of our results and various of their consequences in terms of congruence lattice representation problems. Among those consequences, for every infinite regular cardinal $\kappa$ and every conditionally $\kappa$-co-Brouwerian S of size $\kappa$, there exists a relatively complemented lattice L with zero such that Conc L is isomorphic to S.

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