Mapping at Element is Supremum implies Mapping is Increasing

Theorem
Let $\struct {S, \vee, \wedge, \preceq}$ be a lattice.

Let $\struct {T, \vee_2, \wedge_2, \precsim}$ be a complete lattice.

Let $f: S \to T$ be a mapping such that:
 * $\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$

Then $f$ is an increasing mapping.

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

By Preceding implies Way Below Closure is Subset of Way Below Closure:
 * $x^\ll \subseteq y^\ll$

By definitions of image of set and way below closure:
 * $f \sqbrk {x^\ll} = \set {\map f w: w \in S \land w \ll x}$

and
 * $f \sqbrk {y^\ll} = \set {\map f w: w \in S \land w \ll y}$

where $f \sqbrk {x^\ll}$ denotes the image of $x^\ll$ under $f$.

By Image of Subset under Relation is Subset of Image/Corollary 2:
 * $f \sqbrk {x^\ll} \subseteq f \sqbrk {y^\ll}$

Thus