# Preimage of Subset under Composite Mapping

## Theorem

Let $S_1, S_2, S_3$ be sets.

Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.

Denote with $g \circ f: S_1 \to S_3$ the composition of $g$ and $f$.

Let $S_3' \subseteq S_3$ be a subset of $S_3$.

Then:

$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$

where $g^{-1} \sqbrk {S_3'}$ denotes the preimage of $S_3'$ under $g$.

## Proof 1

A mapping is a specific kind of relation.

Hence, Inverse of Composite Relation applies, and it follows that:

$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$

$\blacksquare$

## Proof 2

Let $x \in S_1$.

Then:

 $\ds x$ $\in$ $\ds \paren {g \circ f}^{-1} \sqbrk {S_3'}$ $\ds \leadstoandfrom \ \$ $\ds \map {\paren {g \circ f} } x$ $\in$ $\ds S_3'$ Definition of Preimage of Subset under Mapping $\ds \leadstoandfrom \ \$ $\ds \map g {\map f x}$ $\in$ $\ds S_3'$ Definition of Composition of Mappings $\ds \leadstoandfrom \ \$ $\ds \map f x$ $\in$ $\ds g^{-1} \sqbrk {S_3'}$ Definition of Preimage of Subset under Mapping $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds f^{-1} \sqbrk {g^{-1} \sqbrk {S_3'} }$ Definition of Preimage of Subset under Mapping $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$ Definition of Composition of Mappings

Hence the result.

$\blacksquare$