Epimorphism Preserves Semigroups

Theorem
Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.

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

Let $\struct {S, \circ}$ be a semigroup.

Then $\struct {T, *}$ is also a semigroup.

Proof
As $\struct {S, \circ}$ is a semigroup, then by definition it is closed.

As $\phi$ is an epimorphism, it is by definition surjective.

That is:
 * $T = \phi \sqbrk S$

where $\phi \sqbrk S$ denotes the image of $S$ under $\phi$.

From Morphism Property Preserves Closure it follows that $\struct {T, *}$ is closed.

As $\struct {S, \circ}$ is a semigroup, then by definition $\circ$ is associative.

From Epimorphism Preserves Associativity, $*$ is therefore also associative.

So:
 * $\struct {T, *}$ is closed

and:
 * $*$ is associative.

Therefore, by definition, $\struct {T, *}$ is a semigroup.

Also see

 * Isomorphism Preserves Semigroups


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


 * Epimorphism Preserves Groups


 * Epimorphism Preserves Distributivity