Diagonal Relation is Universally Congruent
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Theorem
The diagonal relation $\Delta_S$ on a set $S$ is universally congruent on $S$.
Proof
We have that the diagonal relation is an equivalence relation.
Let $\struct {S, \circ}$ be any algebraic structure.
\(\ds \) | \(\) | \(\ds x_1 \mathrel {\Delta_S} x_2 \land y_1 \mathrel {\Delta_S} y_2\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x_1 = x_2 \land y_1 = y_2\) | Definition of Diagonal Relation | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds x_1 \circ y_1 = x_2 \circ y_2\) | as a consequence of equality | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \paren {x_1 \circ y_1} \mathrel {\Delta_S} \paren {x_2 \circ y_2}\) | Definition of Diagonal Relation |
$\Delta_S$ can therefore be described as universally congruent.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Example $11.3$