Epimorphism Preserves Commutativity

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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 $\circ$ be a commutative operation.


Then $*$ is also a commutative operation.


Proof

Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.

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

As an epimorphism is surjective, it follows that:

$\forall u, v \in T: \exists x, y \in S: \map \phi x = u, \map \phi y = v$

So:

\(\ds u * v\) \(=\) \(\ds \map \phi x * \map \phi y\) Definition of Surjection
\(\ds \) \(=\) \(\ds \map \phi {x \circ y}\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds \map \phi {y \circ x}\) Definition of Commutative Operation
\(\ds \) \(=\) \(\ds \map \phi y * \map \phi x\) Definition of Morphism Property
\(\ds \) \(=\) \(\ds v * u\) by definition as above

$\blacksquare$


Warning

Note that this result is applied to epimorphisms.

For a general homomorphism which is not surjective, nothing definite can be said about the behaviour of the elements of its codomain which are not part of its image.


Also see


Sources

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