Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence

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

Let $\RR$ be a congruence relation on $\struct {A, \oplus}$.

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

Let $\TT$ be the relation on $A$ defined as:
 * $\forall x, y \in A: x \mathrel \TT y \iff \eqclass x \RR \mathrel \SS \eqclass y \RR$

Then:
 * $\TT$ is a congruence relation on $\struct {A, \oplus}$.

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 $\SS$ are equivalence relations.

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

Checking in turn each of the criteria for equivalence:

Reflexivity
We have:
 * $\forall a \in A: a \in \eqclass a \RR$

Hence as $\SS$ is an equivalence relation, therefore reflexive:
 * $\forall a \in A: \eqclass a \RR \mathrel \SS \eqclass a \RR$

That is:
 * $\forall a \in A: a \mathrel \TT a$

Thus $\TT$ is seen to be reflexive.

Symmetry
Thus $\TT$ is seen to be symmetric.

Transitivity
Let $a, b, c \in A$ such that

Then we have:

Thus $\TT$ is seen to be transitive.

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

Hence by definition it is an equivalence relation.

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

Let $x_1, y_1, x_2, y_2 \in A$ such that:

Then we have:

Hence by definition of congruence relation:
 * $\TT$ is a congruence relation on $\struct {A, \oplus}$.

Now we have:

From Composite of Epimorphisms is Epimorphism, $q_\SS \circ q_\RR$ is an epimorphism.

We also have that $q_\TT$ is an epimorphism.

From the Quotient Theorem for Epimorphisms, there exists a unique isomorphism $\phi: \paren {A / \RR} / \SS \to A / \TT$ which satisfies $\phi \circ \paren {q_\SS \circ q_\RR} = q_\TT$.

This theorem can be illustrated by means of the following commutative diagram:


 * $\begin{xy}\xymatrix@L+2mu@+1em{

A \ar[r]^*{q_\TT} \ar[d]^*{q_\RR} & A / \TT \\ A / \RR \ar[r]_*{q_\SS} & \paren {A / \RR} / \SS \ar@{-->}[u]_*{\phi}^*{\cong} }\end{xy}$