Cross-Relation is Equivalence Relation

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Theorem

Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements.

Let $\struct {C, \circ {\restriction_C} } \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.


Let $\struct {S_1, \circ {\restriction_1} } \subseteq \struct {S, \circ}$ be a subsemigroup of $S$.

Let $\struct {S_2, \circ {\restriction_2} } \subseteq \struct {C, \circ {\restriction_C} }$ be a subsemigroup of $C$.


Let $\left({S_1 \times S_2, \oplus}\right)$ be the (external) direct product of $\struct {S_1, \circ {\restriction_1} }$ and $\struct {S_2, \circ {\restriction_2} }$, where $\oplus$ is the operation on $S_1 \times S_2$ induced by $\circ {\restriction_1}$ on $S_1$ and $\circ {\restriction_2}$ on $S_2$.


Let $\boxtimes$ be the cross-relation on $S_1 \times S_2$, defined as:

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


Then $\boxtimes$ is an equivalence relation on $\struct {S_1 \times S_2, \oplus}$.


Proof

Reflexivity

$\forall \tuple {x_1, y_1} \in S_1 \times S_2: x_1 \circ y_1 = x_1 \circ y_1 \implies \tuple {x_1, y_1} \boxtimes \tuple {x_1, y_1}$

So $\boxtimes$ is a reflexive relation.

$\Box$


Symmetry

\(\ds \tuple {x_1, y_1}\) \(\boxtimes\) \(\ds \tuple {x_2, y_2}\)
\(\ds \leadsto \ \ \) \(\ds x_1 \circ y_2\) \(=\) \(\ds x_2 \circ y_1\)
\(\ds \leadsto \ \ \) \(\ds x_2 \circ y_1\) \(=\) \(\ds x_1 \circ y_2\) $\circ$ is commutative
\(\ds \leadsto \ \ \) \(\ds \tuple {x_2, y_2}\) \(\boxtimes\) \(\ds \tuple {x_1, y_1}\)

So $\boxtimes$ is a symmetric relation.

$\Box$


Transitivity

\(\ds \tuple {x_1, y_1}\) \(\boxtimes\) \(\ds \tuple {x_2, y_2}\)
\(\, \ds \land \, \) \(\ds \tuple {x_2, y_2}\) \(\boxtimes\) \(\ds \tuple {x_3, y_3}\)
\(\ds \leadsto \ \ \) \(\ds x_1 \circ y_2\) \(=\) \(\ds x_2 \circ y_1\)
\(\, \ds \land \, \) \(\ds x_2 \circ y_3\) \(=\) \(\ds x_3 \circ y_2\)
\(\ds \leadsto \ \ \) \(\ds x_1 \circ y_3 \circ y_2\) \(=\) \(\ds x_1 \circ y_2 \circ y_3\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds x_2 \circ y_1 \circ y_3\) substituting $x_2 \circ y_1$ for $x_1 \circ y_2$
\(\ds \) \(=\) \(\ds x_2 \circ y_3 \circ y_1\) $\circ$ is commutative
\(\ds \) \(=\) \(\ds x_3 \circ y_2 \circ y_1\) substituting $x_3 \circ y_2$ for $x_2 \circ y_3$
\(\ds \) \(=\) \(\ds x_3 \circ y_1 \circ y_2\) $\circ$ is commutative
\(\ds \leadsto \ \ \) \(\ds x_1 \circ y_3\) \(=\) \(\ds x_3 \circ y_1\) as $y_2 \in S_2$, therefore $y_2 \in C$, and so is cancellable for $\circ$
\(\ds \leadsto \ \ \) \(\ds \tuple {x_1, y_1}\) \(\boxtimes\) \(\ds \tuple {x_3, y_3}\)

So $\boxtimes$ is a transitive relation.

$\Box$


All the criteria are therefore seen to hold for $\boxtimes$ to be an equivalence relation.

$\blacksquare$


Also see

Examples


Sources

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