Preimage of Subset under Composite Mapping/Proof 1
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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
A mapping is a specific kind of relation.
Hence, Inverse of Composite Relation applies, and it follows that:
- $\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
$\blacksquare$