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:

Also see

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


 * Epimorphism Preserves Semigroups
 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity