Subset of Codomain is Superset of Image of Preimage/Proof 2

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 $y \in B$.

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.

$\blacksquare$

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 10$: Inverses and Composites