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:

$\forall B \in \powerset T: \map {\paren {f^\to \circ f^\gets} } B \subseteq B$

Let $B \subseteq T$.

Then:

$\ds y$

$\in$

$\ds \paren {f \circ f^{-1} } \sqbrk B$

$\ds \leadsto \ \ $

$\ds y$

$\in$

$\ds f \sqbrk {f^{-1} \sqbrk B}$

$\ds \leadsto \ \ $

$\ds \exists x \in f^{-1} \sqbrk B: \, $

$\ds \map f x$

$=$

$\ds y$

$\ds \leadsto \ \ $

$\ds y$

$\in$

$\ds B$

So by definition of subset:

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

$\blacksquare$

1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.8$