1. (a) i.
(w, z) ∈ (X ∩ Y) × Z ⟺ w ∈ X ∩ Y
⟺ w ∈ X and w ∈ Y

⟺ (w, z) ∈ X × Z and (w, z) ∈ Y × Z

⟺ (w, z) ∈ (X × Z) ∩ (Y × Z)

Hence, (X ∩ Y) × Z = (X × Z) ∩ (Y × Z). ii.

x ∈ Z \ (Y ∪ Z) ⟺ x ∈ Z, x ∉ (Y ∪ Z)
⟺ x ∈ X, y ∉ Y and x ∉ Z
⟺ x ∈ X, x ∉ Y and x ∈ X, x ∉ Z
⟺ x ∈ X \ Y and x ∈ X \ Z
⟺ x ∈ (X \ Y) ∩ (X \ Z)

Hence, X \ (Y ∪ Z) = (X \ Y) ∩ (X \ Z). iii. x ∈ X \ (Y ∩ Z) ⟺ x ∈ X, x ∉ (Y ∩ Z). Since x ∉ (Y ∩ Z),

we know that x ∉ Y, which holds iff x ∈ X \ Y, or x ∉ Z, which is true iff x ∈ X \ Z. So x is either in X \ Y or x is in X \ Z. Hence, x ∈ (X \ Y) ∪ (X \ Z), which implies that X \ (Y ∩ Z) = (X \ Y) ∪ (X \ Z).

(b) i.
α ∈ f^{−1}(Y ∩ Z) ⟺ f(α) ∈ Y ∩ Z
⟺ f(α) ∈ Y and f(α) ∈ Z
⟺ α ∈ f^{−1}(Y) and α ∈ f^{−1}(z)
⟺ α ∈ f^{−1}(Y) ∩ f^{−1}(Z)

Hence, f^{−1}(Y ∩ Z) = f^{−1}(Y) ∩ f^{−1}(Z). ii.

b ∈ f^{−1}(Y \ Z) ⟺ f(b) ∈ Y \ Z
⟺ f(b) ∈ Y, f(b) ∉ Z
⟺ b ∈ f^{−1}(Y), b ∉ f^{−1}(Z)
⟺ b ∈ f^{−1}(Y) \ f^{−1}(Z)

And so f^{−1}(Y \ Z) = f^{−1}(Y) \ f^{−1}(Z).

(c) i. The statement is true. Take x in Y ∪ Z and observe that f(x) = f(Y ∪ Z) means that, if x ∈ Y, then f(x) ∈ f(Y), or if x ∈ Z, then f(x) ∈ f(Z). Thus, f(x) ∈ f(Y ∪ Z) ⟺ f(x) ∈ f(Y) ∪ f(Z), which means that f(Y ∪ Z) = f(Y) ∪ f(Z). ii. The statement is false. Counterexample via Hilary Havens: Take

f(x) = x^{2} and let Y = {−1}, Z = {1}. Then Y ∩ Z = ∅ and f(Y ∩ Z) = ∅, but f(Y) ∩ f(Z) = 1, which is a contradiction. Hence f(Y ∩ Z) ≠ f(Y) ∩ f(Z).

(d) i. We’ll first show that given a function f ∈ (A^{B})^{C}, we can find a corresponding h ∈ A^{B×C}. So take f ∈ (A^{B})^{C}, c_{i} ∈ C, b_{j} ∈ B, a_{k} ∈ A. By definition, f : C ⟶ A^{B}, and so f : c_{i} ⟶ g_{i}, g_{i} ∈ G the set of all maps g that send B to A. Thus, g_{i}(b_{j}) = a_{k}, and given (c_{i}, b_{j}), we can get an a_{k} ∈ A, which means that the desired function h is one such that h(c_{i}, b_{j}) = a_{k}, for h ∈ A^{B×C}. Now, going in the other direction, given an h ∈ A^{B×C}, we again take c_{i} ∈ C, b_{j} ∈ B, a_{k} ∈ A such that h(c_{i}, b_{j}) = a_{k}. Given the set of all maps g from B to A, we know there exists one such g_{i} with g_{i}(b_{j}) = a_{k}. We then create a map f such that f(c_{i}) = g_{i}, and we have f ∈ (A^{B})^{C}. Therefore we have a bijection between (A^{B})^{C} and A^{B×C}.