Order Monomorphism iff Strictly Increasing

Theorem

Let $\left({S, \preceq_1}\right)$ be a totally ordered set and let $\left({T, \preceq_2}\right)$ be a poset.

A mapping $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is an order monomorphism iff $\phi$ is strictly increasing.

That is, iff:

$\forall x, y \in S: x \ \prec_1 \ y \iff \phi \left({x}\right) \ \prec_2 \ \phi \left({y}\right)$

Corollary

A mapping $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is an order isomorphism iff:

• $\phi$ is a surjection
• $\forall x, y \in S: x \ \prec_1 \ y \iff \phi \left({x}\right) \ \prec_2 \ \phi \left({y}\right)$

Proof

Sufficient Condition

Let $\phi$ be an order monomorphism:

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle x \ \prec_1 \ y\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle x \ \prec_1 \ y \land x \ne y\)	\(\displaystyle \)	Strictly Precedes is a Strict Ordering
	\(\displaystyle \)	\(\displaystyle \)	\(\implies\)	\(\displaystyle \phi \left({x}\right) \ \prec_2 \ \phi \left({y}\right)\)	\(\displaystyle \)	Definition of Order Monomorphism, and $\phi$ is injective

So by definition, $\phi$ is strictly increasing.

$\Box$

Necessary Condition

Now let $\phi$ be strictly increasing.

Then $\phi$ is a strictly monotone mapping by definition.

Then $\phi$ is injective, by Strictly Monotone Mapping is Injective.

By Strictly Increasing is Increasing:

$x \ \preceq_1 \ y \implies \phi \left({x}\right) \ \preceq_2 \ \phi \left({y}\right)$

Now suppose $\phi \left({x}\right) \ \preceq_2 \ \phi \left({y}\right)$.

Then, if $y \ \prec_1 \ x$, by the fact that $\phi$ is strictly increasing, we would have $\phi \left({y}\right) \ \prec_2 \ \phi \left({x}\right)$, which contradicts $\phi \left({x}\right) \ \preceq_2 \ \phi \left({y}\right)$.

Thus:

$\phi \left({x}\right) \ \preceq_2 \ \phi \left({y}\right) \iff x \ \preceq_1 \ y$

and $\phi$, being injective, has been proved to be an order monomorphism from its definition.

$\blacksquare$

Proof of Corollary

Follows directly from Order Isomorphism is Surjective Order Monomorphism.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 14$: Theorem $14.9: \ 2^\circ$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 7$: Theorem $7.2$