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.

$\blacksquare$

Warning

Note that this result is applied to epimorphisms.

For a general homomorphism which is not surjective, nothing definite can be said about the behaviour of the elements of its codomain which are not part of its image.

Also see

• Isomorphism Preserves Semigroups

• Epimorphism Preserves Associativity
• Epimorphism Preserves Commutativity
• Epimorphism Preserves Identity
• Epimorphism Preserves Inverses

• Epimorphism Preserves Groups

• Epimorphism Preserves Distributivity

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Theorem $12.2$: Corollary
• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {II}$: New Structures from Old: $\S 12$: Homomorphisms: Exercise $12.1$