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 range which are not part of its image.