Epimorphism Preserves Commutativity

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

If $\circ$ is commutative, then so is $*$.

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:

Comment
Note that this result is applied to epimorphisms. For a general homomorphism which is not surjective, we can say nothing definite about the behaviour of the elements of its codomain which are not part of its image.