Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence/Corollary

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Corollary to Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence

Let $\struct {A, \oplus}$ be an algebraic structure.

Let $\RR$ and $\TT$ be congruence relations on $\struct {A, \oplus}$ such that $R \subseteq T$.

Let $\SS$ be the relation on the quotient structure $\struct {A / \RR, \oplus_\RR}$ which satisfies:

$\forall X, Y \in A / \RR: X \mathrel \SS Y \iff \exists x \in X, y \in Y: x \mathrel \TT y$


Then:

$\SS$ is a congruence relation on $\struct {A / \RR, \oplus_\RR}$

and:

there exists a unique isomorphism $\phi$ from $\paren {A / \RR} / \SS$ to $A / \TT$ which satisfies:
$\phi \circ q_\SS \circ q_\RR = q_\TT$
where $q_\SS$, $q_\RR$ and $q_\TT$ denote the quotient epimorphisms as appropriate.


Proof

Recall that by definition $\RR$ and $\TT$ are a fortiori equivalence relations.


First it is demonstrated that $\SS$ is an equivalence relation.

Checking in turn each of the criteria for equivalence:


Reflexivity

Let $X \in A / \RR$.

Then as $\TT$ is an equivalence relation, therefore a fortiori reflexive:

$\forall x \in X: x \mathrel \TT x$

Thus by definition of $\SS$:

$\forall X \in A / \RR: X \mathrel \SS X$

Thus $\SS$ is seen to be reflexive.

$\Box$


Symmetry

\(\ds \forall X, Y \in A / \RR: \, \) \(\ds X\) \(\SS\) \(\ds Y\)
\(\ds \leadsto \ \ \) \(\ds \exists x \in X, y \in Y: \, \) \(\ds x\) \(\TT\) \(\ds y\) Definition of $\SS$
\(\ds \leadsto \ \ \) \(\ds \exists x \in X, y \in Y: \, \) \(\ds y\) \(\TT\) \(\ds x\) as $\TT$ is an equivalence relation, therefore a fortiori symmetric
\(\ds \leadsto \ \ \) \(\ds Y\) \(\SS\) \(\ds X\) Definition of $\TT$

Thus $\SS$ is seen to be symmetric.

$\Box$


Transitivity

Let $X, Y, Z \in A / \RR$ such that

\(\ds X\) \(\SS\) \(\ds Y\)
\(\, \ds \land \, \) \(\ds Y\) \(\SS\) \(\ds Z\)

Then we have:

\(\ds X\) \(\SS\) \(\ds Y\)
\(\, \ds \land \, \) \(\ds Y\) \(\SS\) \(\ds Z\)
\(\ds \leadsto \ \ \) \(\ds \exists x \in X, y \in Y, z \in Z: \, \) \(\ds x\) \(\TT\) \(\ds y\) Definition of $\SS$
\(\, \ds \land \, \) \(\ds y\) \(\TT\) \(\ds z\)
\(\ds \leadsto \ \ \) \(\ds \exists x \in X, z \in Z: \, \) \(\ds x\) \(\TT\) \(\ds z\) as $\TT$ is an equivalence relation, therefore a fortiori transitive
\(\ds \leadsto \ \ \) \(\ds X\) \(\SS\) \(\ds Z\) Definition of $\SS$

Thus $\SS$ is seen to be transitive.

$\Box$


Hence $\TT$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\Box$


It remains to be demonstrated that $\SS$ is a congruence relation.

Let $X_1, Y_1, X_2, Y_2 \in A / \RR$ such that:

\(\ds X_1\) \(\SS\) \(\ds Y_1\)
\(\, \ds \land \, \) \(\ds X_2\) \(\SS\) \(\ds Y_2\)

Then we have:

\(\ds \exists x_1, x_2 \in X, y_1, y_2 \in Y: \, \) \(\ds x_1\) \(\TT\) \(\ds y_1\) Definition of $\SS$
\(\, \ds \land \, \) \(\ds x_2\) \(\TT\) \(\ds y_2\)
\(\ds \leadsto \ \ \) \(\ds \exists x_1 \oplus x_2 \in X_1 \oplus_\RR X_2, \exists y_1 \oplus y_2 \in Y_1 \oplus_\RR Y_2: \, \) \(\ds x_1 \oplus x_2\) \(\TT\) \(\ds y_1 \oplus y_2\) Definition of Congruence Relation
\(\ds \leadsto \ \ \) \(\ds X_1 \oplus_\RR X_2\) \(\SS\) \(\ds Y_1 \oplus_\RR Y_2\) Definition of $\SS$

Hence by definition of congruence relation:

$\SS$ is a congruence relation on $\struct {A / \RR, \oplus_\RR}$.

$\Box$


Thus we have that:

$\RR$ is a congruence relation on $\struct {A, \oplus}$

and:

$\SS$ is a congruence relation on the quotient structure $\struct {A / \RR, \oplus_\RR}$ defined by $\RR$.


By definition of quotient structure:

$\forall x, y \in A: x \mathrel \TT y \iff \eqclass x \RR \mathrel \SS \eqclass y \RR$

where $\eqclass x \RR$ is the equivalence class under $\RR$ of $x$.


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Hence the criteria of Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence are fulfilled.


Hence from Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence:

there exists a unique isomorphism $\phi$ from $\paren {A / \RR} / \SS$ to $A / \TT$ which satisfies:
$\phi \circ q_\SS \circ q_\RR = q_\TT$
where $q_\SS$, $q_\RR$ and $q_\TT$ denote the quotient epimorphisms as appropriate.

$\blacksquare$


Sources

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