Image of Set Difference under Mapping/Corollary 1

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

Let $S_1 \subseteq S_2 \subseteq S$.

Then:
 * $\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$

where $\complement$ (in this context) denotes relative complement.

Proof
From Image of Set Difference under Relation: Corollary 1 we have:
 * $\relcomp {\mathcal R \sqbrk {S_2} } {\mathcal R \sqbrk {S_1} } \subseteq \mathcal R \sqbrk {\relcomp {S_2} {S_1} }$

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:
 * $\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$