Image of Set Difference under Mapping/Corollary 1

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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:

$\relcomp {\RR \sqbrk {S_2} } {\RR \sqbrk {S_1} } \subseteq \RR \sqbrk {\relcomp {S_2} {S_1} }$

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

As $f$, being a mapping, is also a relation, it follows directly that:

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

$\blacksquare$