Order Embedding into Image is Isomorphism

Theorem
Let $$\left({S; \le_1}\right)$$ and $$\left({T; \le_2}\right)$$ be posets.

Let $$S'$$ be the image of a mapping $$\phi: \left({S; \le_1}\right) \to \left({T; \le_2}\right)$$.

Then $$\phi$$ is a monomorphism from $$\left({S; \le_1}\right)$$ into $$\left({T; \le_2}\right)$$ iff $$\phi$$ is an isomorphism from $$\left({S; \le_1}\right)$$ into $$\left({S'; \le_2|_{S'}}\right)$$.

Proof

 * Let $$\phi$$ be a monomorphism from $$\left({S; \le_1}\right)$$ into $$\left({T; \le_2}\right)$$.

Then $$\phi$$ is an injection into $$\left({T; \le_2}\right)$$ by definition.

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

Thus $$\phi: \left({S; \le_1}\right) \to \left({S'; \le_2|_{S'}}\right)$$ is a monomorphism which is both an injection and a surjection, and thus a bijection.

By definition, a monomorphism is order-preserving.

A bijection which is order-preserving is an isomorphism.


 * Let $$\phi: \left({S; \le_1}\right) \to \left({S'; \le_2|_{S'}}\right)$$ be an isomorphism.

By definition, an isomorphism is injective, and changing the range of the mapping does not change that.

Thus $$\phi: \left({S; \le_1}\right) \to \left({S'; \le_2|_{S'}}\right)$$ is a monomorphism.