# Cross-Relation is Congruence Relation

## 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:

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

The cross-relation $\boxtimes$ is a congruence relation on $\struct {S_1 \times S_2, \oplus}$.

## Proof

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

We now need to show that:

 $\ds \tuple {x_1, y_1}$ $\boxtimes$ $\ds \tuple {x_2, y_2}$ $\, \ds \land \,$ $\ds \tuple {u_1, v_1}$ $\boxtimes$ $\ds \tuple {u_2, v_2}$ $\ds \leadsto \ \$ $\ds \paren {\tuple {x_1, y_1} \oplus \tuple {u_1, v_1} }$ $\boxtimes$ $\ds \paren {\tuple {x_2, y_2} \oplus \tuple {u_2, v_2} }$

First we note that:

 $\text {(1)}: \quad$ $\ds \tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2}$ $\iff$ $\ds x_1 \circ y_2 = y_1 \circ x_2$ $\text {(2)}: \quad$ $\ds \tuple {u_1, v_1} \boxtimes \tuple {u_2, v_2}$ $\iff$ $\ds u_1 \circ v_2 = v_1 \circ u_2$

Then:

 $\ds \paren {x_1 \circ u_1} \circ \paren {y_2 \circ v_2}$ $=$ $\ds x_1 \circ \paren {u_1 \circ y_2 } \circ v_2$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds x_1 \circ \paren { y_2 \circ u_1 } \circ v_2$ Commutativity of $\circ$ $\ds$ $=$ $\ds \paren {x_1 \circ y_2} \circ \paren {u_1 \circ v_2}$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds \paren {y_1 \circ x_2} \circ \paren {v_1 \circ u_2}$ from $(1)$ and $(2)$ $\ds$ $=$ $\ds \paren {x_2 \circ y_1} \circ \paren {u_2 \circ v_1}$ Commutativity of $\circ$ $\ds$ $=$ $\ds x_2 \circ \paren {y_1 \circ u_2} \circ v_1$ Semigroup Axiom $\text S 1$: Associativity $\ds$ $=$ $\ds x_2 \circ \paren {u_2 \circ y_1} \circ v_1$ Commutativity of $\circ$ $\ds \leadsto \ \$ $\ds \paren {x_1 \circ u_1} \circ \paren {y_2 \circ v_2}$ $=$ $\ds \paren {x_2 \circ u_2} \circ \paren {y_1 \circ v_1}$ Semigroup Axiom $\text S 1$: Associativity $\ds \leadsto \ \$ $\ds \tuple {x_1 \circ u_1, y_1 \circ v_1}$ $\boxtimes$ $\ds \tuple {x_2 \circ u_2, y_2 \circ v_2}$ Definition of $\boxtimes$ $\ds \leadsto \ \$ $\ds \paren {\tuple {x_1, y_1} \oplus \tuple {u_1, v_1} }$ $\boxtimes$ $\ds \paren {\tuple {x_2, y_2} \oplus \tuple {u_2, v_2} }$ Definition of $\oplus$

So $\boxtimes$ is a congruence relation on $\struct {S \times C, \oplus}$.

$\blacksquare$