Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below

Theorem
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

Let $\left({T, \vee_2, \wedge_2, \precsim}\right)$ be a continuous complete lattice.

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

Let $x \in S, y \in T$.

Then
 * $y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$

Proof
By Mapping at Element is Supremum implies Mapping is Increasing:
 * $f$ is an increasing mapping.

Sufficient Condition
Let $y \ll f\left({x}\right)$

By Way Below Closure is Directed in Bounded Below Join Semilattice:
 * $x^\ll$ is directed

where $x^\ll$ denotes the way below closure of $x$.

By Image of Directed Subset under Increasing Mapping is Directed:
 * $f\left[{x^\ll}\right]$ is directed

where $f\left[{x^\ll}\right]$ denotes the image of $x^\ll$ under $f$.

By assumption:
 * $f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

By definitions of image of set and way below closure:
 * $f\left[{x^\ll}\right] = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

By Way Below iff Second Operand Preceding Supremum of Directed Set There Exists Element of Directed Set First Operand Way Below Element:
 * $\exists z \in f\left[{x^\ll}\right]: y \ll z$

By definition of image of set:
 * $\exists w \in x^\ll: f\left({w}\right) = z$

Thus by definition of way below closure:
 * $w \ll x$

Thus $y \ll f\left({w}\right)$

Necessary Condition
Let
 * $\exists w \in S: w \ll x \land y \ll f\left({w}\right)$

By Way Below implies Preceding:
 * $w \preceq x$

By definition of increasing mapping:
 * $f\left({w}\right) \precsim f\left({x}\right)$

Thus by Preceding and Way Below implies Way Below:
 * $y \ll f\left({x}\right)$