Image of Set Difference under Mapping/Corollary 2

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

Let $X$ be a subset of $S$.

Then:
 * $\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$

where:
 * $\Img f$ denotes the image of $f$
 * $\complement_{\Img f}$ denotes the complement relative to $\Img f$.

This can be expressed in the language and notation of direct image mappings as:
 * $\forall X \in \powerset S: \relcomp {\Img f} {\map {f^\to} X} \subseteq \map {f^\to} {\relcomp S X}$

That is:
 * $\forall X \in \powerset S: \map {\paren {\complement_{\Img f} \circ f^\to} } X \subseteq \map {\paren {f^\to \circ \complement_S} } X$

where $\circ$ denotes composition of mappings.

Proof
From Image of Set Difference under Relation: Corollary 2 we have:
 * $\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk X} \subseteq \mathcal R \sqbrk {\relcomp S X}$

where $\mathcal R \subseteq S \times T$ is a relation on $S \times T$.

As $f$, being a mapping, is also a relation, it follows directly that:
 * $\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$