Image of Set Difference under Mapping/Corollary 2

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

Let $S_1$ and $S_2$ be subsets of $S$.

Then:
 * $\relcomp {\Img f} {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp S {S_1} }$

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:
 * $\forall S_1 \in \powerset S: \relcomp {\Img f} {\map {f^\to} {S_1} } \subseteq \map {f^\to} {\relcomp S {S_1} }$

Proof
From Image of Set Difference under Relation: Corollary 2 we have:
 * $\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk {S_1} } \subseteq \mathcal R \sqbrk {\relcomp S {S_1} }$

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:
 * $\relcomp {\Img f} {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp S {S_1} }$