Elements of Semigroup with Equal Images under Homomorphisms form Subsemigroup

Theorem

Let $\struct {A, \circ}$ and $\struct {B, *}$ be semigroups.

Let $f: A \to B$ and $g: A \to B$ be semigroup homomorphisms.

Then the set:

$S = \set {x \in A: \map f x = \map g x}$

is a subsemigroup of $A$.

Proof

Let $x, y \in A$. Then:

\(\ds \map f {x \circ y}\)

\(=\)

\(\ds \map f x * \map f y\)

Morphism Property

\(\ds \)

\(=\)

\(\ds \map g x * \map g y\)

Definition of $A$

\(\ds \)

\(=\)

\(\ds \map g {x \circ y}\)

Morphism Property

Thus $x \circ y \in A$.

So, by the Subsemigroup Closure Test:

$S$ is a subsemigroup of $A$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.13$