Cross-Relation is Congruence Relation

From ProofWiki
Jump to navigation Jump to search

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 $\struct {S \times C, \oplus}$.

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Cross-Relation_is_Congruence_Relation&oldid=628431"