Quotient Mapping of Inverse Completion

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Theorem

Let $\struct {T, \circ'}$ be an inverse completion of a commutative semigroup $\struct {S, \circ}$, where $C$ is the set of cancellable elements of $S$.

Let $f: S \times C: T$ be the mapping defined as:

$\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$

Let $\RR_f$ be the equivalence relation induced by $f$.


Then:

$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$


Proof

By the definition of $\RR_f$:

$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$


Now:

\(\ds x_1 \circ' y_1^{-1}\) \(=\) \(\ds x_2 \circ' y_2^{-1}\)
\(\ds \leadsto \ \ \) \(\ds x_1 \circ' y_1^{-1} \circ' y_1 \circ' y_2\) \(=\) \(\ds x_2 \circ' y_2^{-1} \circ' y_1 \circ' y_2\) Elements $y_1$ and $y_2$ are cancellable
\(\ds \leadsto \ \ \) \(\ds x_1 \circ' y_2\) \(=\) \(\ds x_2 \circ' y_1\) Definition of Inverse Element, and commutativity of $\circ'$
\(\ds \leadsto \ \ \) \(\ds x_1 \circ y_2\) \(=\) \(\ds x_2 \circ y_1\) $\circ'$ extends $\circ$


which leads us to:

$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$

$\blacksquare$