# Subset of Codomain is Superset of Image of Preimage/Proof 3

## 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:

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

$\blacksquare$