Equivalence Induced by Epimorphism is Congruence Relation
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Theorem
Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Then the induced equivalence $\RR_\phi$ is a congruence relation for $\oplus$.
Proof
Let $x, x', y, y' \in S$ such that:
- $x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$
By definition of induced equivalence:
\(\ds x \mathrel {\RR_\phi} x'\) | \(\leadsto\) | \(\ds \map \phi x = \map \phi {x'}\) | ||||||||||||
\(\ds y \mathrel {\RR_\phi} y'\) | \(\leadsto\) | \(\ds \map \phi y = \map \phi {y'}\) |
Then:
\(\ds \map \phi {x \oplus y}\) | \(=\) | \(\ds \map \phi x * \map \phi y\) | Definition of Epimorphism (Abstract Algebra) | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x'} * \map \phi {y'}\) | equality shown above | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \phi {x' \oplus y'}\) | Definition of Epimorphism (Abstract Algebra) |
Thus $\paren {x \oplus y} \mathrel {\RR_\phi} \paren {x' \oplus y'}$ by definition of induced equivalence.
So $\RR_\phi$ is a congruence relation for $\oplus$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.5$