# 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

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$