Maximizing a congruence with respect to its partition of idempotents (Q1113292)

For a congruence \(\sigma\) on a semigroup \({\mathcal S}\) a congruence \(\mu(\sigma)\) on \({\mathcal S}\), containing \(\sigma\), is defined such that the semigroup \({\mathcal S}/\sigma\) is fundamental if and only if \(\sigma=\mu(\sigma)\). The congruence \(\mu(\sigma)\) is shown to possess maximality properties and for idempotent-surjective semigroups, \(\mu(\sigma)\) is the maximum congruence with respect to the partition of the idempotents determined by \(\sigma\). Thus \(\mu\) is the maximum idempotent-separating congruence on any idempotent-surjective semigroup. It is shown that \(\mu(\mu(\sigma))=\mu(\sigma)\). If \(\rho\) is another congruence on \({\mathcal S}\), possibly with the same partition of the idempotents as \(\sigma\), then it is of interest to know when \(\rho\subseteq\sigma\) (or \(\rho\subseteq\mu(\sigma))\) implies \(\mu(\rho)\subseteq\mu(\sigma)\) or even \(\mu(\rho)=\mu(\sigma)\). These implications are not true in general but if \(\sigma\subseteq\rho\subseteq\mu(\sigma)\) then \(\mu(\rho)\subseteq\mu(\sigma)\). If \({\mathcal S}\) is an idempotent-surjective semigroup and \(\rho\) and \(\sigma\) have the same partition of the idempotents then \(\mu(\rho)=\mu(\sigma)\).