Let  f:X→Y  be onto. Also, let A, B be subsets  of X and C,D be subsets of Y. Then

(a)  y∈f−1(C∪D)⇔f(y) ∈C∪D⇔f(y)∈D  or  f(y)∈C ⇔y∈f−1 (C)  or  y∈f−1(D) ⇔y∈f−1(C)∪f−1(D)                                            (b)  Similarly f−1(C∩D)=f−1(C)∩f−1(D) c) y∈f(A∪B)⇔∃ x∈A∪B such that f(x)=y⇔f(x)=y  for some x∈A  or  x∈B   ⇔y∈f(A)  or  y∈f(B) ⇔y∈f(A)∪f(B) d) f(A∩B)  is only a subset of f(A)∩f(B)  but may not be equal to f(A)∩f(B) Take X={1,2,3,4},   y={a,b,c}  and  f={(1,a),(2,b),(3,b),(4,c)} Then f is onto. If A={1,2}  and  B={3,4} , thenf(A)={a,b}  and  f(B)={b,c} f(A)∩f(B)={b}≠f(A∩B),  since A∩B=θ