# Intersection of Image with Subset of Codomain

## Theorem

Let $S$ and $T$ be sets.

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

Let $A \subseteq S$ and $B \subseteq T$.

Then:

$f \sqbrk {A \cap f^{-1} \sqbrk B} = f \sqbrk A \cap B$

## Proof

 $\displaystyle f \sqbrk {A \cap f^{-1} \sqbrk B}$ $=$ $\displaystyle \set {\map f x: x \in A \cap f^{-1} \sqbrk B}$ Definition of Image of Subset under Mapping $\displaystyle$ $=$ $\displaystyle \set {\map f x: x \in A \land x \in f^{-1} \sqbrk B}$ Definition of Set Intersection $\displaystyle$ $=$ $\displaystyle \set {\map f x: x \in A \land \map f x \in B}$ Definition of Preimage of Subset under Mapping $\displaystyle$ $=$ $\displaystyle \set {\map f x: x \in A} \cap \set {\map f x: \map f x \in B}$ Definition of Set Intersection $\displaystyle$ $=$ $\displaystyle f \sqbrk A \cap B$ Definition of Image of Subset under Mapping

$\blacksquare$