# 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

$\relcomp {\Img \RR} {\RR \sqbrk X} \subseteq \RR \sqbrk {\relcomp S X}$

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:

$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$

$\blacksquare$