Unique Isomorphism from Quotient Mapping to Epimorphism Domain

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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

Let $\mathcal R_\phi$ be the equivalence induced by $\phi$.

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

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

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


Then there is one and only one isomorphism:

$\psi: \struct {S / \mathcal R_\phi}, {\circ_{\mathcal R_\phi} } \to \struct {T, *}$

which satisfies:

$\psi \bullet q_{\mathcal R_\phi} = \phi$

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


Proof

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


Also:

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


Therefore $\psi$ is an isomorphism.

$\blacksquare$


Sources