# Cross-Relation is Congruence Relation

## 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:

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

The cross-relation $\boxtimes$ is a congruence relation on $\left({S_1 \times S_2, \oplus}\right)$.

## Proof

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

We now need to show that:

 $\displaystyle \left({x_1, y_1}\right)$ $\boxtimes$ $\displaystyle \left({x_2, y_2}\right)$ $\, \displaystyle \land \,$ $\displaystyle \left({u_1, v_1}\right)$ $\boxtimes$ $\displaystyle \left({u_2, v_2}\right)$ $\displaystyle \implies \ \$ $\displaystyle \left({\left({x_1, y_1}\right) \oplus \left({u_1, v_1}\right)}\right)$ $\boxtimes$ $\displaystyle \left({\left({x_2, y_2}\right) \oplus \left({u_2, v_2}\right)}\right)$

So:

 $\displaystyle \left({x_1, y_1}\right)$ $\boxtimes$ $\displaystyle \left({x_2, y_2}\right)$ $\, \displaystyle \land \,$ $\displaystyle \left({u_1, v_1}\right)$ $\boxtimes$ $\displaystyle \left({u_2, v_2}\right)$ By assumption $\displaystyle \left({x_1 \circ u_1}\right) \circ \left({y_2 \circ v_2}\right)$ $=$ $\displaystyle \left({x_1 \circ y_2}\right) \circ \left({u_1 \circ v_2}\right)$ Commutativity and associativity of $\circ$ $\displaystyle$ $=$ $\displaystyle \left({x_2 \circ y_1}\right) \circ \left({u_2 \circ v_1}\right)$ By assumption $\displaystyle$ $=$ $\displaystyle \left({x_2 \circ u_2}\right) \circ \left({y_1 \circ v_1}\right)$ Commutativity and associativity of $\circ$ $\displaystyle \implies \ \$ $\displaystyle \left({x_1 \circ u_1, y_1 \circ v_1}\right)$ $\boxtimes$ $\displaystyle \left({x_2 \circ u_2, y_2 \circ v_2}\right)$ Definition of $\boxtimes$ $\displaystyle \implies \ \$ $\displaystyle \left({\left({x_1, y_1}\right) \oplus \left({u_1, v_1}\right)}\right)$ $\boxtimes$ $\displaystyle \left({\left({x_2, y_2}\right) \oplus \left({u_2, v_2}\right)}\right)$ Definition of $\oplus$

So $\boxtimes$ is a congruence relation on $\left({S \times C, \oplus}\right)$.

$\blacksquare$