Intersection of Image with Subset of Codomain

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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

$\blacksquare$


Sources