Construction of Inverse Completion/Properties of Quotient Structure

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 $\mathcal R$ be the relation $\mathcal R$ defined on $S \times C$ by:
 * $\left({x_1, y_1}\right) \mathcal R \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

is a congruence relation on $\left({S \times C, \oplus}\right)$.

Let the quotient structure defined by $\mathcal R$ be $\left({T', \oplus'}\right)$

where:


 * $T'$ denotes the quotient set $\displaystyle \frac {S \times C} {\mathcal R}$


 * $\oplus'$ denotes the operation $\oplus_{\mathcal R}$ induced on $\displaystyle \frac {S \times C} {\mathcal R}$ by $\oplus$.

Identity of Quotient Structure
We have that:


 * $\forall c \in C: \left[\!\left[{\left({c, c}\right)}\right]\!\right]_\mathcal R$

is the identity of $T'$.

We denote the identity of $T'$ as $e_{T'}$, as usual.

Invertible Elements in Quotient Structure
Every cancellable element of $S'$ is invertible in $T'$.

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

Quotient Structure is Inverse Completion
$T'$ is an inverse completion of its subsemigroup $S'$.

Proof of Invertible Elements in Quotient Structure
From Identity of Quotient Structure, $\left({T', \oplus'}\right)$ has an identity, and it is $\left[\!\left[{\left({c, c}\right)}\right]\!\right]_\mathcal R$ for any $c \in C$. Call this identity $e_{T'}$.


 * First we note that, from Image of Cancellable Elements in Quotient Mapping, $C' = \psi \left({C}\right)$. So:


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

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

Proof of Generator for Quotient Structure
Let $\left({x, y}\right) \in S \times C$. Then:

Proof that Quotient Structure is Inverse Completion

 * Every cancellable element of $S'$ is invertible in $T'$, from Invertible Elements in Quotient Structure.


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