# Composition of Direct Image Mappings of Mappings

## Theorem

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

Let $f: A \to B$ and $g: B \to C$ be mappings.

Let:

$f^\to: \powerset A \to \powerset B$

and

$g^\to: \powerset B \to \powerset C$

be the direct image mappings of $f$ and $g$.

Then:

$\paren {g \circ f}^\to = g^\to \circ f^\to$

## Proof 1

Let $S \subseteq A$ such that $S \ne \O$.

Then:

 $\ds \map {\paren {g^\to \circ f^\to} } S$ $=$ $\ds \map {g^\to} {\map {f^\to} S}$ $\ds$ $=$ $\ds \set {\map g x: x \in \map {f^\to} S}$ $\ds$ $=$ $\ds \set {\map g x: \exists y \in S: x = \map f y}$ $\ds$ $=$ $\ds \set {\map g {\map f y}: y \in S}$ $\ds$ $=$ $\ds \set {\map {\paren {g \circ f} } y: y \in S}$ $\ds$ $=$ $\ds \map {\paren {g \circ f}^\to} S$

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

 $\ds \map {\paren {g^\to \circ f^\to} } \O$ $=$ $\ds \O$ $\ds$ $=$ $\ds \map {\paren {g \circ f}^\to} \O$

and the result is complete.

$\blacksquare$

## Proof 2

We have that a mapping is a relation.

Hence Composition of Direct Image Mappings of Relations applies.

$\blacksquare$