Question:
Let f be a function f: A → B, and let A1, A2 ⊆ A. Then f(A1 ∩ A2) = f(A1) ∩ f(A2).
Answer: $$ f(A_1 \cap A_2) \subseteq f(A_1) \cap f(A_2) :$$ Let $ y $ be an arbitrary element in $ f(A_1 \cap A_2) $. This means there exists an $ x \in A_1 \cap A_2 $ such that $ f(x) = y $. Since $ x $ is in both $ A_1 $ and $ A_2 $, it follows that $ y $ is in both $ f(A_1) $ and $ f(A_2) $. Thus, $ y \in f(A_1) \cap f(A_2) $.
y ∈ f(A1 ∩ A2) $⇒$ y = f(x) for some x ∈ A1 ∩ A2
$⇒$ y = f(x) with x ∈ A1 and x ∈ A2
$⇒$ y ∈ f(A1) and y ∈ f(A2)
$⇒$ y ∈ f(A1) ∩ f(A2)
Would that be correct approach?