Suprema Preserving Mapping on Ideals is Increasing

Theorem
Let $\left({S, \preceq}\right)$, $\left({T, \precsim}\right)$ be ordered sets.

Let $f: S \to T$ be a mapping.

For every ideal $I$ in $\left({S, \preceq}\right)$, let $f$ preserve the supremum on $I$.

Then $f$ is increasing.

Proof
Let $x, y \in S$ such that:
 * $x \preceq y$

By Supremum of Singleton:
 * $\left\{ {x}\right\}$ and $\left\{ {y}\right\}$ admit suprema in $\left({S, \preceq}\right)$

By Supremum of Lower Closure of Set:
 * $\left\{ {x}\right\}^\preceq$ and $\left\{ {y}\right\}^\preceq$ admit suprema in $\left({S, \preceq}\right)$

where $\left\{ {x}\right\}^\preceq$ denotes the lower closure of $\left\{ {x}\right\}$

By Lower Closure of Singleton:
 * $x^\preceq$ and $y^\preceq$ admit suprema in $\left({S, \preceq}\right)$

By Lower Closure of Element is Ideal:
 * $x^\preceq$ and $y^\preceq$ are ideals in $\left({S, \preceq}\right)$

By assumption and definition of mapping preserves the supremum on subset:
 * $f^\to \left({x^\preceq}\right)$ and $f^\to \left({y^\preceq}\right)$ admit suprema in $\left({T, \precsim}\right)$

and
 * $\sup \left({f^\to \left({x^\preceq}\right)}\right) = f \left({\sup \left({x^\preceq}\right)}\right)$ and $\sup \left({f^\to \left({y^\preceq}\right)}\right) = f \left({\sup \left({y^\preceq}\right)}\right)$

By Supremum of Lower Closure of Element:
 * $\sup \left({x^\preceq}\right) = x$ and $\sup \left({y^\preceq}\right) = y$

By Lower Closure is Increasing:
 * $x^\preceq \subseteq y^\preceq$

By Image of Subset under Relation is Subset of Image/Corollary 2:
 * $f^\to \left({x^\preceq}\right) \subseteq f^\to \left({y^\preceq}\right)$

Thus by Supremum of Subset:
 * $f\left({x}\right) \precsim f\left({y}\right)$

Thus by definition:
 * $f$ is increasing.