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 $$\frac {S \times C} {\mathcal R}$$;


 * $$\oplus'$$ denotes the operation $\oplus_{\mathcal R}$ induced on $\frac {S \times C} \mathcal R$ 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 Identity of Quotient Structure
$$ $$ $$

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.