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:

Also see

 * Epimorphism Preserves Associativity
 * Epimorphism Preserves Identity
 * Epimorphism Preserves Inverses


 * Epimorphism Preserves Semigroups
 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity