Cross-Relation is Congruence Relation

From ProofWiki
Jump to: navigation, search

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:

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


So:

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


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

$\blacksquare$


Sources