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 $\oplus$.

$\blacksquare$


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