A+ CATEGORY SCIENTIFIC UNIT

Join-semilattices with two-dimensional congruence amalgamation

Volume 93 / 2002

Friedrich Wehrung Colloquium Mathematicum 93 (2002), 209-235 MSC: 06B10, 06B23, 06D15 DOI: 10.4064/cm93-2-2

Abstract

We say that a $\langle\vee,0\rangle$-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 $\langle{x,y}\rangle\in X\times Y$), 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 $a\leq b\vee z$ for all $z\in Z$, then there exists $c\in S$ such that $a\leq b\vee c$ 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 $\langle\vee,0\rangle$-semilattice. We prove that for every conditionally co-Brouwerian lattice $S$ and every partial lattice $P$, every $\langle\vee,0\rangle$-homomorphism $\varphi: \mathop{\rm Con}\nolimits_{\rm c} P\to S$ can be lifted to a lattice homomorphism $f: P\to L$ for some relatively complemented lattice $L$. Here, $\mathop{\rm Con}\nolimits_{\rm c} P$ denotes the $\langle\vee,0\rangle$-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 these 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 $\mathop{\rm Con}_{\rm c}L\cong S$.

Authors

  • Friedrich WehrungCNRS, UMR 6139
    Département de Mathématiques, BP 5186
    Université de Caen, Campus 2
    14032 Caen Cedex, France
    e-mail

Search for IMPAN publications

Query phrase too short. Type at least 4 characters.

Rewrite code from the image

Reload image

Reload image