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 $f$.

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