Order Monomorphism is Injection

Theorem

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

Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an order monomorphism, i.e.:

$\forall x, y \in S: x \preceq_1 y \iff f \left({x}\right) \preceq_2 f \left({y}\right)$

Then $\phi$ is an injection.

Proof

Suppose $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ is a mapping such that:

$\forall x, y \in S: x \preceq_1 y \iff \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

We have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \phi \left({x}\right)\)	\(=\)	\(\displaystyle \phi \left({y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \phi \left({x}\right) \preceq_2 \phi \left({y}\right)\)	\(\land\)	\(\displaystyle \phi \left({y}\right) \preceq_2 \phi \left({x}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by antisymmetry of $\preceq_2$
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \preceq_1 y\)	\(\land\)	\(\displaystyle y \preceq_1 x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\preceq_2$ is increasing
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by antisymmetry of $\preceq_1$

So $\phi$ is an injection.

$\blacksquare$