Image of Union under Mapping

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

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

Let $A$ and $B$ be subsets of $S$.

Then:
 * $f \sqbrk {A \cup B} = f \sqbrk A \cup f \sqbrk B$

Also see

 * Preimage of Union under Mapping
 * Image of Intersection under Mapping
 * Preimage of Intersection under Mapping