Construction of Inverse Completion/Invertible Elements in Quotient Structure

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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, $\left({T', \oplus'}\right)$ has an identity, and it is $\left[\!\left[{\left({c, c}\right)}\right]\!\right]_\boxtimes$ for any $c \in C$. Call this identity $e_{T'}$.


From Image of Cancellable Elements in Quotient Mapping, $C' = \psi \left({C}\right)$.

So:

\(\displaystyle x'\) \(\in\) \(\displaystyle C'\) $\quad$ $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \exists x \in C: x'\) \(=\) \(\displaystyle \psi \left({x}\right)\) $\quad$ as $\psi$ is a surjection $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \forall a \in C: x'\) \(=\) \(\displaystyle \left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes\) $\quad$ Definition of $\psi$ $\quad$


The inverse of $x'$ is $\left[\!\left[{\left({a, a \circ x}\right)}\right]\!\right]_\boxtimes$, as follows:

\(\displaystyle a \circ x\) \(\in\) \(\displaystyle C\) $\quad$ Cancellable Elements of Semigroup form Subsemigroup $\quad$
\(\displaystyle a \circ a \circ x\) \(\in\) \(\displaystyle C\) $\quad$ Cancellable Elements of Semigroup form Subsemigroup $\quad$
\(\displaystyle \implies \ \ \) \(\displaystyle \left[\!\left[{ \left({x \circ a, a}\right)} \right]\!\right]_\boxtimes \oplus' \left[\!\left[{ \left({a, a \circ x}\right)} \right]\!\right]_\boxtimes\) \(=\) \(\displaystyle \left[\!\left[{ \left({x \circ a \circ a, a \circ a \circ x}\right)} \right]\!\right]_\boxtimes\) $\quad$ by definition of $\oplus'$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{\left({a \circ a \circ x, a \circ a \circ x}\right)}\right]\!\right]_\boxtimes\) $\quad$ Commutativity of $\circ$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e_{T'}\) $\quad$ Identity of Quotient Structure $\quad$


... thus showing that the inverse of $\left[\!\left[{\left({x \circ a, a}\right)}\right]\!\right]_\boxtimes$ is $\left[\!\left[{\left({a, a \circ x}\right)}\right]\!\right]_\boxtimes$.

$\blacksquare$


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