Equivalence Relation induced by Congruence Relation on Quotient Structure is Congruence

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

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

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

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 the result by definition of congruence relation.