Subset of Codomain is Superset of Image of Preimage/Proof 3
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Theorem
Let $f: S \to T$ be a mapping.
Then:
- $B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
where:
- $f \sqbrk B$ denotes the image of $B$ under $f$
- $f^{-1}$ denotes the inverse of $f$
- $f \circ f^{-1}$ denotes composition of $f$ and $f^{-1}$.
This can be expressed in the language and notation of direct image mappings and inverse image mappings as:
- $\forall B \in \powerset T: \map {\paren {f^\to \circ f^\gets} } B \subseteq B$
Proof
Let $B \subseteq T$.
Then:
\(\ds y\) | \(\in\) | \(\ds \paren {f \circ f^{-1} } \sqbrk B\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds f \sqbrk {f^{-1} \sqbrk B}\) | Definition of Composition of Mappings | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \exists x \in f^{-1} \sqbrk B: \, \) | \(\ds \map f x\) | \(=\) | \(\ds y\) | Definition of Image of Subset under Mapping | |||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds B\) | Definition of Preimage of Subset under Mapping |
So by definition of subset:
- $B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$
$\blacksquare$
Sources
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$