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

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