Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Supremum of Image is Mapping at Supremum of Directed Subset

Theorem
Let $\struct {S, \vee_1, \wedge_1, \preceq_1}$ and $\struct {T, \vee_2, \wedge_2, \preceq_2}$ be up-complete lattices.

Let $f: S \to T$ be a mapping such that
 * for all directed set $\struct {D, \precsim}$ and Moore-Smith sequence $N: D \to S$ in $S: \map f {\liminf N} \preceq_2 \map \liminf {f \circ N}$

Let $D$ be a directed subset of $S$.

Then $\map \sup {f \sqbrk D} = \map f {\sup D}$

where $f \sqbrk D$ denotes the image of $D$ under $f$.

Proof
By Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping is Increasing:
 * $f$ is an increasing mapping.

By Image of Directed Subset under Increasing Mapping is Directed:
 * $f \sqbrk D$ is directed.

By definition of up-complete:
 * $D$ and $f \sqbrk D$ admit suprema.

By Subset and Image Admit Suprema and Mapping is Increasing implies Supremum of Image Precedes Mapping at Supremum:
 * $\map \sup {f \sqbrk D} \preceq_2 \map f {\sup D}$

By Limit Inferior of Inclusion Moore-Smith Sequence is Supremum of Directed Subset:
 * $\sup D = \liminf i_D$

By assumption:
 * $\map f {\sup D} \preceq_2 \map \liminf {f \circ i_D}$

By Composition of Mapping and Inclusion is Restriction of Mapping:
 * $f \circ i_D = f {\restriction} D$

By Limit Inferior of Restriction Moore-Smith Sequence is Supremum of Image of Directed Subset:
 * $\map \sup {f \sqbrk D} = \map \liminf {f \circ i_D}$

Then
 * $\map f {\sup D} \preceq_2 \map \sup {f \sqbrk D}$

Thus by definition of antisymmetry:
 * $\map \sup {f \sqbrk D} = \map f {\sup D}$