Preimage of Subset under Composite Mapping

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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:

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

Hence the result.

$\blacksquare$


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