under what conditions is f(A ∪B)=f(A) ∪f(B) and f(A∩B)=f(A)∩f(B)?

Question

Does the function need to be bijective? I know for f(A∩B)=f(A)∩f(B) the function has to be injective, but what about the first equation?

Answer 1

We always have $f(A \cup B) = f(A) \cup f(B)$ (even for infinite unions). You can prove it easily by showing both directions if you want.

If $x \in f(A \cup B)$, then $x = f(y)$ for some $y \in A \cup B$. We must have $y \in A$ or $y \in B$. Thus $x \in f(A) \cup f(B)$. It's essentially the same thing for the other direction.