Quotient Theorem for Epimorphisms

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Theorem

Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.

Let $\RR_\phi$ be the equivalence induced by $\phi$.

Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.

Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induced by $\RR_\phi$.

Let $\struct {S / \RR_\phi}, {\circ_{\RR_\phi} }$ be the quotient structure defined by $\RR_\phi$.


Then:

The induced equivalence $\RR_\phi$ is a congruence relation for $\circ$
There is one and only one isomorphism $\psi: \struct {S / \RR_\phi}, {\circ_{\RR_\phi} } \to \struct {T, *}$ which satisfies $\psi \bullet q_{\RR_\phi} = \phi$.

where, in order not to cause notational confusion, $\bullet$ is used as the symbol to denote composition of mappings.


Proof

Proof of Congruence Relation

Let $x, x', y, y' \in S$ such that:

$x \mathrel {\RR_\phi} x' \land y \mathrel {\RR_\phi} y'$


By definition of induced equivalence:

\(\ds x \mathrel {\RR_\phi} x'\) \(\leadsto\) \(\ds \map \phi x = \map \phi {x'}\)
\(\ds y \mathrel {\RR_\phi} y'\) \(\leadsto\) \(\ds \map \phi y = \map \phi {y'}\)


Then:

\(\ds \map \phi {x \circ y}\) \(=\) \(\ds \map \phi x * \map \phi y\) Definition of Epimorphism (Abstract Algebra)
\(\ds \) \(=\) \(\ds \map \phi {x'} * \map \phi {y'}\) equality shown above
\(\ds \) \(=\) \(\ds \map \phi {x' \circ y'}\) Definition of Epimorphism (Abstract Algebra)


Thus $\paren {x \circ y} \mathrel {\RR_\phi} \paren {x' \circ y'}$ by definition of induced equivalence.

So $\RR_\phi$ is a congruence relation for $\circ$.

$\Box$


Proof of Unique Isomorphism

From the Quotient Theorem for Surjections, there is a unique bijection from $S / \RR_\phi$ onto $T$ satisfying $\psi \bullet q_{\RR_\phi} = \phi$.


Also:

\(\ds \forall x, y \in S: \, \) \(\ds \map \psi {\eqclass x {\RR_\phi} \circ_{\RR_\phi} \eqclass y {\RR_\phi} }\) \(=\) \(\ds \map \psi {\eqclass {x \circ y} {\RR_\phi} }\) Definition of Quotient Structure
\(\ds \) \(=\) \(\ds \map \phi {x \circ y}\) Definition of Epimorphism (Abstract Algebra)
\(\ds \) \(=\) \(\ds \map \phi x * \map \phi y\) Definition of Epimorphism (Abstract Algebra)
\(\ds \) \(=\) \(\ds \map \psi {\eqclass x {\RR_\phi} } * \map \psi {\eqclass y {\RR_\phi} }\) Definition of Quotient Mapping


Therefore $\psi$ is an isomorphism.

$\blacksquare$


Also known as

Some authors call this the Factor Theorem for Epimorphisms.


Sources