# Construction of Inverse Completion/Invertible Elements in Quotient Structure

## Theorem

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

Let $\left({C, \circ {\restriction_C}}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.

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

Let $\boxtimes$ be the cross-relation on $S \times C$, defined as:

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

This cross-relation is a congruence relation on $S \times C$.

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 $\oplus$.

Every cancellable element of $S'$ is invertible in $T'$.

## Proof

From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.

Call this identity $e_{T'}$.

Let the mapping $\psi: S \to T'$ be defined as:

$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
$C' = \psi \sqbrk C$

So:

 $\displaystyle x'$ $\in$ $\displaystyle C'$ $\displaystyle \leadsto \ \$ $\displaystyle \exists x \in C: x'$ $=$ $\displaystyle \map \psi x$ as $\psi$ is a surjection $\displaystyle \leadsto \ \$ $\displaystyle \forall a \in C: x'$ $=$ $\displaystyle \eqclass {\tuple {x \circ a, a} } \boxtimes$ Definition of $\psi$

The inverse of $x'$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$, as follows:

 $\displaystyle a \circ x$ $\in$ $\displaystyle C$ Cancellable Elements of Semigroup form Subsemigroup $\displaystyle a \circ a \circ x$ $\in$ $\displaystyle C$ Cancellable Elements of Semigroup form Subsemigroup $\displaystyle \leadsto \ \$ $\displaystyle \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ x} } \boxtimes$ $=$ $\displaystyle \eqclass {\tuple {x \circ a \circ a, a \circ a \circ x} } \boxtimes$ by definition of $\oplus'$ $\displaystyle$ $=$ $\displaystyle \eqclass {\tuple {a \circ a \circ x, a \circ a \circ x} } \boxtimes$ Commutativity of $\circ$ $\displaystyle$ $=$ $\displaystyle e_{T'}$ Identity of Quotient Structure

thus showing that the inverse of $\eqclass {\tuple {x \circ a, a} } \boxtimes$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$.

$\blacksquare$