Unique Isomorphism from Quotient Mapping to Epimorphism Domain

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Theorem

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

Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ 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 $\struct {S / \RR_\phi, \oplus_{\RR_\phi} }$ be the quotient structure defined by $\RR_\phi$.


Then there is one and only one isomorphism:

$\psi: \struct {S / \RR_\phi, \oplus_{\RR_\phi} } \to \struct {T, *}$

which satisfies:

$\psi \circ q_{\RR_\phi} = \phi$

where $\circ$ denotes composition of mappings.


Proof

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


Also:

\(\ds \forall x, y \in S: \, \) \(\ds \map \psi {\eqclass x {\RR_\phi} \oplus_{\RR_\phi} \eqclass y {\RR_\phi} }\) \(=\) \(\ds \map \psi {\eqclass {x \oplus y} {\RR_\phi} }\) Definition of Quotient Structure
\(\ds \) \(=\) \(\ds \map \phi {x \oplus y}\) 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.

$\blacksquare$


Sources

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