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 $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.

Suppose $S$ is the empty set.

It follows from the definition of an epimorphism that: $\phi$ is a surjective homomorphism

By Empty Mapping to Empty Set is Bijective, the empty map is bijective By definition of bijection, the empty map is an epimorphism.

Therefore, suppose $\phi$ is the empty map, which is indeed an epimorphism.

By definition of a homomorphism, $\phi$ can be defined as:


 * $\forall \O \in S: \map \phi {\O \circ \O} = \map \phi \O * \map \phi \O$

By Image of Empty Set is Empty Set, $T$ is also the empty set.

It follows from the definition of the homomorphism that the binary operations $\circ$ and $*$ are also the empty map.

Hence, it is vacuously true that $\circ$ is commutative on $S$, when $S$ is empty, as required.

Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.

Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.

Suppose $S$ is non-empty.

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