Subset of Codomain is Superset of Image of Preimage/Proof 1

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

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.

$\blacksquare$