Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes

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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:

$\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$


Then:

$\forall x, y \in S, a, b \in C:$

$(1): \quad \tuple {x \circ a, a} \boxtimes \tuple {y \circ b, b} \iff x = y$
$(2): \quad \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$

where $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$.


Proof

From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation.

Hence the equivalence class of $\tuple {x, y}$ under $\boxtimes$ is defined for all $\tuple {x, y} \in S \times C$.


From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.

Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.

The result follows from Elements of Cross-Relation Equivalence Class.

$\blacksquare$


Sources