# Ordered Semigroup Isomorphism is Surjective Monomorphism

## Theorem

Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.

Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.

Then $\phi$ is an ordered semigroup isomorphism if and only if:

$(1): \quad \phi$ is an ordered semigroup monomorphism
$(2): \quad \phi$ is a surjection.

## Proof

### Necessary Condition

Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup isomorphism.

Then by definition:

$\phi$ is a semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$
$\phi$ is an order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$.

A semigroup isomorphism is by definition:

A semigroup homomorphism

which is:

A monomorphism and an epimorphism.

From Order Isomorphism is Surjective Order Embedding, an order isomorphism is an order embedding which is also a surjection.

Putting this all together, we see that an ordered semigroup isomorphism is:

A monomorphism
An order embedding
A surjection.

An ordered semigroup monomorphism is by definition:

A monomorphism

which is also

An order embedding

Hence $\phi$ is:

An ordered semigroup monomorphism
A surjection.

$\Box$

### Sufficient Condition

Let $\phi$ be:

An ordered semigroup monomorphism
A surjection.

By definition, that means $\phi$ be:

A monomorphism
An order embedding
A surjection.

From Order Isomorphism is Surjective Order Embedding, an order isomorphism is an order embedding which is also a surjection.

A semigroup isomorphism is by definition:

A semigroup homomorphism

which is:

A semigroup monomorphism and an semigroup epimorphism.

Thus a semigroup monomorphism which is also a surjection is a semigroup isomorphism.

So $\phi$ is:

A semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$
An order isomorphism from the ordered set $\struct {S, \preceq}$ to the ordered set $\struct {T, \preccurlyeq}$.

$\blacksquare$