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:
\(\ds x\) | \(\in\) | \(\ds \paren {g \circ f}^{-1} \sqbrk {S_3'}\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \map {\paren {g \circ f} } x\) | \(\in\) | \(\ds S_3'\) | 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.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 8$: Composition of Functions and Diagrams: Exercise $4$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.4 \ \text{(a)}$