Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image

Theorem
Let $\struct {A, \odot}$ and $\struct {B, \otimes}$ be algebraic structures.

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

Let $f: \struct {A, \odot} \to \struct {B, \otimes}$ be an epimorphism.

Let $\struct {A / \RR, \odot_\RR}$ denote the quotient structure defined by $\RR$.

Let $q_\RR: A \to A / \RR$ denote the quotient mapping induced by $\RR$:
 * $\forall x \in A: \map {q_\RR} x = \eqclass x \RR$

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

Then:
 * there exists an epimorphism $g$ from $\struct {A / \RR, \odot_\RR}$ to $\struct {B, \otimes}$ which satisfies $g \circ q_\RR = f$


 * $\RR \subseteq \RR_f$
 * $\RR \subseteq \RR_f$

where $\RR_f$ denotes the equivalence relation induced by $f$.

Necessary Condition
Let $\RR \subseteq \RR_f$.

Recall the definition of $\RR_f$:
 * $\forall x, y \in A: x \mathrel {\RR_f} y \iff \map f x = \map f y$

Let us define $g: \struct {A / \RR, \odot_\RR} \to \struct {B, \otimes}$ as:
 * $\forall \eqclass x \RR \in A / \RR: \map g {\eqclass x \RR} = \map f x$

We show that $g$ is well-defined.

Let $x \mathrel \RR y$.

Then as $\RR \subseteq \RR_f$, it follows that:
 * $x \mathrel {\RR_f} y$

By definition of the equivalence class of $x$:
 * $\eqclass x \RR = \eqclass y \RR$

Thus:

Thus we have that:
 * $\forall x, y \in A: x \mathrel \RR y \implies \map g {\eqclass x \RR} = \map g {\eqclass y \RR}$

and so $g$ is well-defined.

Then:

and so we have that:
 * $g \circ q_\RR = f$

Let $x, y \in A$.

We have:

Thus we see that $g$ is a homomorphism.

We have that $g \circ q_\RR = f$ is an epimorphism.

Hence $g \circ q_\RR$ is a surjection.

So from Surjection if Composite is Surjection we have that $g$ is likewise a surjection.

Thus we have that $g$ is a surjective homomorphism.

Hence by definition $g$ is an epimorphism.

We note that if it is not the case that $\RR \subseteq \RR_f$, then there exist $\tuple {x, y} \in \RR$ such that $\map f x \ne \map f y$.

In that case we would see that while $x \mathrel y$ we would have that:


 * $\map g {\eqclass x \RR} \ne \map g {\eqclass y \RR}$

meaning that $g$ is not a well-defined mapping.

So, in summary, given that $\RR \subseteq \RR_f$, we have demonstrated the existence of an epimorphism $g$ which satisfies $g \circ q_\RR = f$.

Sufficient Condition
Let there exist an epimorphism $g$ from $\struct {A / \RR, \odot_\RR}$ to $\struct {B, \otimes}$ which satisfies $g \circ q_\RR = f$.

Let $x, y \in A$ be arbitrary such that $x \mathrel \RR y$.

Then we have: