Epimorphism Preserves Semigroups

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.

If $\left({S, \circ}\right)$ is a semigroup, then so is $\left({T, *}\right)$.

Proof

If $\left({S, \circ}\right)$ is a semigroup, then by definition it is closed.

From Morphism Property Preserves Closure, $\left({T, *}\right)$ is therefore also closed.

If $\left({S, \circ}\right)$ is a semigroup, then by definition $\circ$ is associative.

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

So $\left({T, *}\right)$ is closed, and $*$ is associative, and therefore by definition, $\left({T, *}\right)$ is a semigroup.

$\blacksquare$

Also see

• Isomorphism Preserves Semigroups

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

• Epimorphism Preserves Groups

• Epimorphism Preserves Distributivity

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 12$: Theorem $12.2$: Corollary