semilattice decomposition of a semigroup

A semigroup $S$ has a semilattice decomposition if we can write $S={\bigcup }_{\gamma \in \mathrm{\Gamma }}{S}_{\gamma }$ as a disjoint union of subsemigroups, indexed by elements of a semilattice $\mathrm{\Gamma }$, with the additional condition that $x\in S_{\alpha }$ and $y\in S_{\beta }$ implies $xy\in S_{\alpha \beta }$.

Semilattice decompositions arise from homomorphims of semigroups onto semilattices. If $\varphi :S\to \mathrm{\Gamma }$ is a surjective homomorphism, then it is easy to see that we get a semilattice decomposition by putting $S_{\gamma }={\varphi }^{-1}(\gamma )$ for each $\gamma \in \mathrm{\Gamma }$. Conversely, every semilattice decomposition defines a map from $S$ to the indexing set $\mathrm{\Gamma }$ which is easily seen to be a homomorphism.

A third way to look at semilattice decompositions is to consider the congruence $\rho$ defined by the homomorphism $\varphi :S\to \mathrm{\Gamma }$. Because $\mathrm{\Gamma }$ is a semilattice, $\varphi (x^2)=\varphi (x)$ for all $x$, and so $\rho$ satisfies the constraint that $x\rho x^2$ for all $x\in S$. Also, $\varphi (xy)=\varphi (yx)$ so that $xy\rho yx$ for all $x,y\in S$.A congruence $\rho$ which satisfies these two conditions is called a semilattice congruence.

Conversely, a semilattice congruence $\rho$ on $S$ gives rise to a homomorphism from $S$ to a semilattice $S/\rho$. The $\rho$-classes are the components of the decomposition.