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
- Epimorphism Preserves Associativity
- Epimorphism Preserves Commutativity
- Epimorphism Preserves Identity
- Epimorphism Preserves Inverses
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 $12.1$