# Epimorphism Preserves Commutativity

## Theorem

Let $\left({S, \circ}\right)$ and $\left({T, *}\right)$ be algebraic structures.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

Let $\circ$ be a commutative operation.

Then $*$ is also a commutative operation.

## Proof

Let $\left({S, \circ}\right)$ be an algebraic structure in which $\circ$ is commutative.

Let $\phi: \left({S, \circ}\right) \to \left({T, *}\right)$ be an epimorphism.

As an epimorphism is surjective, it follows that:

$\forall u, v \in T: \exists x, y \in S: \phi \left({x}\right) = u, \phi \left({y}\right) = v$

So:

 $\displaystyle u * v$ $=$ $\displaystyle \phi \left({x}\right) * \phi \left({y}\right)$ $\phi$ is a surjection $\displaystyle$ $=$ $\displaystyle \phi \left({x \circ y}\right)$ Morphism Property $\displaystyle$ $=$ $\displaystyle \phi \left({y \circ x}\right)$ Commutativity of $\circ$ $\displaystyle$ $=$ $\displaystyle \phi \left({y}\right) * \phi {\left({x}\right)}$ Morphism Property $\displaystyle$ $=$ $\displaystyle v * u$ 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.