Let f be a function f: A → B, and let A1, A2 ⊆ A. Then f(A1 ∩ A2) = f(A1) ∩ f(A2).

Question

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?

Answer 1

For $A=\{1,2,3,4,5\}$ and $B=\{0,1\}$, define $f$ as: $$f(1)=0$$ $$f(2)=0$$ $$f(3)=1$$ $$f(4)=1$$ $$f(5)=0$$ So lets say $A_1=\{1,2,3\}$ and $A_2=\{3,4,5\}$, then: $$A_1 \cap A_2=\{3\}$$ So we get that; $f(A_1 \cap A_2)=\{1\}$, but $f(A_1) \cap f(A_2)=\{0,1\}$, so:$$f(A_1 \cap A_2)=\{1\})\neq f(A_1) \cap f(A_2)$$ As pointed out in the comments, this follows from non injectivity of $f$.