User:Jmichayls/Sandbox

Theorem
Let $$E$$ be a semilattice. Then the Munn semigroup $$T_E$$ is an inverse semigroup whose semilattice of idempotents is isomorphic to $$E$$.

Theorem
Let $$S$$ be an inverse semigroup with semilattice of idempotents $$E$$. Then there is a homomorphism $$\phi:S\to T_{E}$$ whose kernel is the maximum idempotent-separating congruence on $$S$$.

Example
Let $$E$$ be the semilattice defined by the Hasse diagram in Figure 1. Then the principal ideals of $$E$$ are:
 * $$1E=\{1\},\ 2E=\{1,2\},\ 3E=\{1,3\}, \ 4E=\{1,4\},\ 5E=\{1,5\}, \ 6E=\{1,3,6\}, \ 7E=\{1,4,5,7\},$$
 * $$8E=\{1,3,4,5,6,7,8\}$$.

Clearly, $$1E\cong 2E\cong 3E \cong 4E$$ and none of the remaining ideals are isomorphic to any of the other ideals. So,

T_{1,1}= \left\{ \begin{pmatrix} 1\\ 1 \end{pmatrix}\right\}, T_{i,j}=\left\{ \begin{pmatrix} i&8\\ j&8 \end{pmatrix}\right\}\ (i,j\in\{1,2,3,4\}),\ T_{6,6}=\left\{ \begin{pmatrix} 1&3&6\\ 1&3&6 \end{pmatrix}\right\},$$

T_{7,7}=\left\{ \begin{pmatrix} 1&4&5&7\\ 1&4&5&7 \end{pmatrix}, \begin{pmatrix} 1&4&5&7\\ 1&5&4&7 \end{pmatrix} \right\}, T_{8,8}=\left\{ \begin{pmatrix} 1&3&4&5&6&7&8\\ 1&3&4&5&6&7&8 \end{pmatrix}, \begin{pmatrix} 1&3&4&5&6&7&8\\ 1&3&5&4&6&7&8 \end{pmatrix} \right\} $$
 * $$\text{ and }

T_{k,l}=\emptyset \text{ otherwise.} $$ The Munn semigroup $$T_E=\bigcup_{i,j=1}^{8} T_{i,j}$$. The partial order of the $$\mathcal{D}$$-classes of $$T_E$$ is shown in Figure 2.