# Composition of Direct Image Mappings of Relations

## Theorem

Let $A, B, C$ be non-empty sets.

Let $\RR_1 \subseteq A \times B, \RR_2 \subseteq B \times C$ be relations.

Let:

${\RR_1}^\to: \powerset A \to \powerset B$

and

${\RR_2}^\to: \powerset B \to \powerset C$

be the direct image mappings of $\RR_1$ and $\RR_2$.

Then:

$\paren {\RR_2 \circ \RR_1}^\to = {\RR_2}^\to \circ {\RR_1}^\to$

## Proof

Let $S \subseteq A, S \ne \O$.

Then:

 $\ds \map {\paren { {\RR_2}^\to \circ {\RR_1}^\to} } S$ $=$ $\ds \map { {\RR_2}^\to} {\map { {\RR_1}^\to } S}$ $\ds$ $=$ $\ds \set {\map {\RR_2} x: x \in \map { {\RR_1}^\to } S}$ $\ds$ $=$ $\ds \set {\map {\RR_2} x: \exists y \in S: \tuple {x, y} \in \RR_1}$ $\ds$ $=$ $\ds \set {\map {\RR_2} {\map {\RR_1} y}: y \in S}$ $\ds$ $=$ $\ds \set {\map {\RR_2 \circ \RR_1} y: y \in S}$ $\ds$ $=$ $\ds \map {\paren {\RR_2 \circ \RR_1}^\to} S$

Now we treat the case where $S = \O$:

 $\ds \map {\paren { {\RR_2}^\to \circ {\RR_1}^\to} } \O$ $=$ $\ds \O$ $\ds$ $=$ $\ds \map {\paren {\RR_2 \circ \RR_1}^\to} \O$

$\blacksquare$