Inverse Image of Direct Image of Inverse Image equals Inverse Image Mapping

Theorem
Let $f: S \to T$ be a mapping.

Let:
 * $f^\to: \powerset S \to \powerset T$ denote the direct image mapping of $f$
 * $f^\gets: \powerset T \to \powerset S$ denote the inverse image mapping of $f$

where $\powerset S$ denotes the power set of $S$.

Then:
 * $f^\gets \circ f^\to \circ f^\gets = f^\gets$

where $\circ$ denotes composition of mappings.

Proof
Thus we have:


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

and the result follows.

Also see

 * Direct Image of Inverse Image of Direct Image equals Direct Image Mapping