Condition for Existence of Epimorphism from Quotient Structure to Epimorphic Image

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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 $g \circ q_\RR = f$

if and only if:

$\RR \subseteq \RR_f$

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


Proof

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:

\(\ds \map g {\eqclass x \RR}\) \(=\) \(\ds \map f x\) Definition of $g$
\(\ds \) \(=\) \(\ds \map f y\) as $x \mathrel {\RR_f} y$
\(\ds \) \(=\) \(\ds \map g {\eqclass y \RR}\) Definition of $g$

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:

\(\ds \forall x \in A: \, \) \(\ds \map {g \circ q_R} x\) \(=\) \(\ds \map g {\map {q_R} x}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map g {\eqclass x \RR}\) Definition of Quotient Mapping
\(\ds \) \(=\) \(\ds \map f x\) Definition of $g$

and so we have that:

$g \circ q_\RR = f$


Let $x, y \in A$.

We have:

\(\ds \map g {\eqclass x \RR \odot_\RR \eqclass y \RR}\) \(=\) \(\ds \map g {\eqclass {x \odot y} \RR}\) Quotient Structure is Well-Defined
\(\ds \) \(=\) \(\ds \map f {x \odot y}\) Definition of $g$
\(\ds \) \(=\) \(\ds \map f x \otimes \map f y\) $f$ is an epimorphism
\(\ds \) \(=\) \(\ds \map g {\eqclass x \RR} \otimes \map g {\eqclass y \RR}\) Definition of $g$: a priori $g$ is well-defined

Thus we see that $g$ is a homomorphism.


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

Hence a fortiori $g \circ q_\RR$ 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 $g \circ q_\RR = f$.

$\Box$


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:

\(\ds x\) \(\RR\) \(\ds y\)
\(\ds \leadsto \ \ \) \(\ds \eqclass x \RR\) \(=\) \(\ds \eqclass y \RR\) Definition of Equivalence Class
\(\ds \leadsto \ \ \) \(\ds \map g {\eqclass x \RR}\) \(=\) \(\ds \map g {\eqclass y \RR}\) as $g$ is functional
\(\ds \leadsto \ \ \) \(\ds \map g {\map {q_\RR} x}\) \(=\) \(\ds \map g {\map {q_\RR} y}\) Definition of Quotient Mapping
\(\ds \leadsto \ \ \) \(\ds \map {g \circ q_\RR} x\) \(=\) \(\ds \map {g \circ q_\RR} y\) Definition of Composition of Mappings
\(\ds \leadsto \ \ \) \(\ds \map f x\) \(=\) \(\ds \map f y\) by hypothesis: $g \circ q_\RR = f$
\(\ds \leadsto \ \ \) \(\ds x\) \(\RR_f\) \(\ds y\) Definition of Equivalence Relation Induced by Mapping
\(\ds \leadsto \ \ \) \(\ds \RR\) \(\subseteq\) \(\ds \RR_f\) Definition of Subset

$\blacksquare$


Sources

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