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