Construction of Inverse Completion/Generator for Quotient Structure

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Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $C \subseteq S$ be the set of cancellable elements of $S$.


Let $\left({S \times C, \oplus}\right)$ be the external direct product of $\left({S, \circ}\right)$ and $\left({C, \circ {\restriction_C}}\right)$, where:

$\circ {\restriction_C}$ is the restriction of $\circ$ to $C \times C$

and :

$\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.


Let $\boxtimes$ be the congruence relation $\boxtimes$ defined on $\left({S \times C, \oplus}\right)$ by:

$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

Let the quotient structure defined by $\boxtimes$ be:

$\left({T', \oplus'}\right) := \left({\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.


$T' = S' \cup \left({C'}\right)^{-1}$ is a generator for the semigroup $T'$.


Proof

Let $\left({x, y}\right) \in S \times C$. Then:

\(\displaystyle \psi \left({x}\right) \oplus' \left({\psi \left({y}\right)}\right)^{-1}\) \(=\) \(\displaystyle \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes \oplus' \left[\!\left[{\left({a, a \circ y}\right)}\right]\!\right]_\boxtimes\) Invertible Elements in Quotient Structure above
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left({x \circ a \circ a, a \circ a \circ y}\right)}\right]\!\right]_\boxtimes\) Definition of $\oplus'$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left({x \circ a \circ a, y \circ a \circ a}\right)}\right]\!\right]_\boxtimes\) Commutativity of $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left({x, y}\right)}\right]\!\right]_\boxtimes\) Cancellability of $a \in C$

$\blacksquare$


Sources