Cross-Relation is Equivalence Relation

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Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup with 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 $\left({S_1, \circ {\restriction_1}}\right) \subseteq \left({S, \circ}\right)$ be a subsemigroup of $S$.

Let $\left({S_2, \circ {\restriction_2}}\right) \subseteq \left({C, \circ {\restriction_C}}\right)$ be a subsemigroup of $C$.


Let $\left({S_1 \times S_2, \oplus}\right)$ be the (external) direct product of $\left({S_1, \circ {\restriction_1}}\right)$ and $\left({S_2, \circ {\restriction_2}}\right)$, 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:

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


Then $\boxtimes$ 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

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

So $\boxtimes$ is a symmetric relation.

$\Box$


Transitivity

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

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