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.