# Epimorphism Preserves Commutativity

## 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.