a) To show that f^{-1}(X cap Y)subseteq f^{-1}(X) cap f^{-1}(Y)f−1(X∩Y)⊆f−1(X)∩f−1(Y), assume that a in f^{-1}(X cap Y)a∈f−1(X∩Y). This means that f(a) in X cap Yf(a)∈X∩Y so that f(a) in Xf(a)∈X and f(a) in Yf(a)∈Y. But this implies that a in f^{-1}(X)a∈f−1(X) and a in f^{-1}(Y)a∈f−1(Y), leading to the conclusion that a in f^{-1}(X) cap f^{-1}(Y)a∈f−1(X)∩f−1(Y).
The previous logic is reversible to show that f^{-1}(X) cap f^{-1}(Y) subseteq f^{-1}(X cap Y)f−1(X)∩f−1(Y)⊆f−1(X∩Y). For if a in f^{-1}(X) cap f^{-1}(Y)a∈f−1(X)∩f−1(Y), then a in f^{-1}(X)a∈f−1(X) and a in f^{-1}(Y)a∈f−1(Y), implying that f(a) in Xf(a)∈X and f(a) in Yf(a)∈Y. But this means that f(a) in X cap Yf(a)∈X∩Y so that a in f^{-1}(X cap Y)a∈f−1(X∩Y).
The last two paragraphs lead to the conclusion that f^{-1}(X cap Y)= f^{-1}(X) cap f^{-1}(Y)f−1(X∩Y)=f−1(X)∩f−1(Y).
b) Assume that ff is surjective (onto).
To show that f(f^{-1}(X))subseteq Xf(f−1(X))⊆X, suppose that b in f(f^{-1}(X))b∈f(f−1(X)). This means that there exists a in f^{-1}(X)a∈f−1(X) such that f(a)=bf(a)=b. But a in f^{-1}(X)a∈f−1(X) implies that b=f(a) in Xb=f(a)∈X.
The logic of the previous paragraph is not completely reversible since we did not need the fact that ff is surjective. To reverse the logic, we need that assumption. To show that X subseteq f(f^{-1}(X))X⊆f(f−1(X)), suppose that b in Xb∈X. Since ff is surjective (onto), there exists an a in Aa∈A such that f(a)=bf(a)=b, which also means that a in f^{-1}(X)a∈f−1(X) (since b in Xb∈X). But this then means that b in f(f^{-1}(X))b∈f(f−1(X)).
The last two paragraphs lead to the conclusion that f(f^{-1}(X))=Xf(f−1(X))=X when ff is surjective.
