Subset of Codomain is Superset of Image of Preimage

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

$B \subseteq T \implies \left({f \circ f^{-1} }\right) \left[{B}\right] = B \cap f \left[{S}\right]$

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:

$B \subseteq T \implies \paren {f \circ f^{-1} } \sqbrk B \subseteq B$

$\blacksquare$