# Composition of Inverse Image Mappings of Mappings

## Theorem

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

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

Let:

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

and

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

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

Then:

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

## Proof

Let $T \subseteq C$.

We have:

 $\ds \map {\paren {f \circ g}^\gets} T$ $=$ $\ds \begin {cases} \set {x \in A: \map g {\map f x} \in T} & : \Img {g \circ f} \cap T \ne \O \\ \O & : \Img {g \circ f} \cap T = \O \end {cases}$ and $\ds \map {f^\gets \circ g^\gets} T$ $=$ $\ds \begin {cases} \set {x \in A: \map f x \in \map {g^\gets} T} & : \Img f \cap \map {g^\gets} T \ne \O \\ \O & : \Img f \cap \map {g^\gets} T = \O \end {cases}$ $\ds$ $=$ $\ds \begin {cases} \set {x \in A: \map g {\map f x} \in T} & : \Img f \cap \map {g^\gets} T \ne \O \\ \O & : \Img f \cap \map {g^\gets} T = \O \end {cases}$

It remains to be shown that:

$\Img f \cap \map {g^\gets} T = \O \iff \Img {g \circ f} \cap T = \O$

So:

 $\ds \Img f \cap \map {g^\gets} T$ $=$ $\ds \O$ $\ds \leadstoandfrom \ \$ $\ds \Img f$ $\subseteq$ $\ds \relcomp B {\map {g^\gets} T}$ Empty Intersection iff Subset of Complement $\ds$ $=$ $\ds \map {g^\gets} {\relcomp C T}$ Complement of Preimage equals Preimage of Complement $\ds \leadstoandfrom \ \$ $\ds \Img {g \circ f}$ $=$ $\ds \map {g^\gets} {\Img f}$ Intersection with Complement is Empty iff Subset $\ds$ $\subseteq$ $\ds \relcomp C T$ Subset of Preimage under Relation is Preimage of Subset: Corollary $\ds \leadstoandfrom \ \$ $\ds \Img {g \circ f} \cap T$ $=$ $\ds \O$ Empty Intersection iff Subset of Complement

$\blacksquare$