Extension of Directed Suprema Preserving Mapping to Complete Lattice Preserves Directed Suprema

Theorem
Let $L_1 = \struct {S_1, \preceq_1}$ and $L_2 = \struct {S_2, \preceq_2}$ be ordered sets.

Let $L_3 = \struct {S_3, \preceq_3}$ be a complete lattice.

Suppose that
 * $L_2$ is directed suprema inheriting ordered subset of $L_3$.

Let $f:S_1 \to S_2$ be a mapping such that
 * $f$ preserves directed suprema.

Then $f:S_1 \to S_3$ preserves directed suprema.

Proof
By definition of ordered subset:
 * $S_2 \subseteq S_3$

Then define $g = f: S_1 \to S_3$

Let $X$ be a directed subset of $S_1$ such that
 * $X$ admits a supremum in $L_1$.

Thus by definition of complete lattice:
 * $g \sqbrk X$ admits a supremum in $L_3$.

By definition of mapping preserves directed suprema:
 * $f \sqbrk X$ admits a supremum in $L_2$ and $\map {\sup_{L_2} } {f \sqbrk X} = \map f {\sup_{L_1} X}$

By Directed Suprema Preserving Mapping is Increasing:
 * $f$ is an increasing mapping.

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

Thus by definition of directed suprema inheriting:
 * $\map {\inf_{L_3} } {g \sqbrk X} = \map g {\inf_{L_1} X}$