Subset of Codomain is Superset of Image of Preimage
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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 1
From Image of Preimage under Mapping:
- $B \subseteq T \implies \left({f \circ f^{-1} }\right) \left[{B}\right] = B \cap f \left[{S}\right]$
The result follows from Intersection is Subset.
$\blacksquare$
Proof 2
Let $y \in B$.
Then:
- $\exists x \in S: y = f \left({x}\right)$
Therefore by definition of preimage of subset:
- $\exists x \in f^{-1} \left[{B}\right]$
It follows by definition of image of subset that:
- $y \in f \left[{f^{-1} \left[{B}\right]}\right]$
Thus by definition of composition $f$ with $f^{-1}$:
- $y \in \left({f \circ f^{-1}}\right) \left[{B}\right]$
The result follows by definition of subset.
$\blacksquare$
Proof 3
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
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $1 \ \text{(b})$
- 2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions