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 $\left({S \times C, \oplus}\right)$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): $\S 20$
