Mapping from Totally Ordered Set is Order Embedding 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
 * $\forall x, y \in S: x \prec_1 y \iff \phi \left({x}\right) \prec_2 \phi \left({y}\right)$

Sufficient Condition
Let $\phi$ be an order monomorphism:

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

Necessary Condition
Now let $\phi$ be strictly increasing.

Then $\phi$ is strictly monotone 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.

Proof of Corollary
Follows directly from Order Isomorphism is Surjective Order Monomorphism.