Direct Image of Intersection with Inverse Image

Theorem
Let $S$ and $T$ be sets.

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:
 * $\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$

Proof
Let $A \in \powerset S, B \in \powerset T$ be arbitrary.

Then:

Let $x \in \map {f^\to} A \cap B$.

Thus we have:
 * $\map {f^\to} {A \cap \map {f^\gets} B} \subseteq \map {f^\to} A \cap B$

and:
 * $\map {f^\to} A \cap B \subseteq \map {f^\to} {A \cap \map {f^\gets} B}$

and so:
 * $\forall A \in \powerset S, B \in \powerset T: \map {f^\to} {A \cap \map {f^\gets} B} = \map {f^\to} A \cap B$

by definition of set equality.