Order Embedding into Image is Isomorphism

Theorem
Let $$\left({S; \preceq_1}\right)$$ and $$\left({T; \preceq_2}\right)$$ be posets.

Let $$S'$$ be the image of a mapping $$\phi: \left({S; \preceq_1}\right) \to \left({T; \preceq_2}\right)$$.

Then $$\phi$$ is an order monomorphism from $$\left({S; \preceq_1}\right)$$ into $$\left({T; \preceq_2}\right)$$ iff $$\phi$$ is an order isomorphism from $$\left({S; \preceq_1}\right)$$ into $$\left({S'; \preceq_2 \restriction_{S'}}\right)$$.

Proof

 * Let $$\phi$$ be an order monomorphism from $$\left({S; \preceq_1}\right)$$ into $$\left({T; \preceq_2}\right)$$.

Then $$\phi$$ is an injection into $$\left({T; \preceq_2}\right)$$ by definition.

From Surjection iff Image equals Range, any mapping from a set to the image of that mapping is a surjection.

Thus the surjective restriction of $$\phi$$ onto $$S'$$ is an order monomorphism which is also a surjection.

Hence the result from Order Isomorphism is Surjective Order Monomorphism.