Final answer:
a) If gof is injective and f is injective, then g is injective. b) If gof is injective and g is injective, then f is injective. c) If gof is surjective and f is surjective, then g is surjective. d) If gof is surjective and g is surjective, then f is surjective.
Step-by-step explanation:
a) Given that gof is injective and f is injective, we want to prove that g is injective. To do this, we assume that g is not injective and reach a contradiction. Since f is injective, for every element a and b in the domain A of f, if a ≠ b, then f(a) ≠ f(b). Now, if gof is injective, for every element a and b in the domain A of gof, if a ≠ b, then gof(a) ≠ gof(b). Since gof(a) = gof(b) = g(f(a)) = g(f(b)), if we assume that g is not injective, then there exist elements a and b in A such that f(a) ≠ f(b) but g(f(a)) = g(f(b)), which contradicts the fact that gof is injective. Therefore, if gof is injective and f is injective, then g is injective.
b) Given that gof is injective and g is injective, we want to prove that f is injective. To do this, we assume that f is not injective and reach a contradiction. Since gof is injective, for every element a and b in the domain A of gof, if a ≠ b, then gof(a) ≠ gof(b). Now, if g is injective, for every element a and b in the domain B of g, if a ≠ b, then g(a) ≠ g(b). Since gof(a) = gof(b) = g(f(a)) = g(f(b)), if we assume that f is not injective, then there exist elements a and b in B such that f(a) ≠ f(b) but g(f(a)) = g(f(b)), which contradicts the fact that gof is injective. Therefore, if gof is injective and g is injective, then f is injective.
c) Given that gof is surjective and f is surjective, we want to prove that g is surjective. To do this, we assume that g is not surjective and reach a contradiction. Since g is not surjective, there exists an element c in the codomain C of g such that there is no element a in the domain A of gof such that g(a) = c. Now, since f is surjective, for every element c in C, there exists an element b in the domain B of f such that f(b) = c. Let c be an element in C such that there is no element a in A satisfying g(a) = c. Since gof(a) = g(f(a)), when f(a) = b, gof(a) = g(f(a)) = g(b) = c. Therefore, we have found an element a in the domain A of gof such that gof(a) = c, which contradicts the fact that gof is surjective. Therefore, if gof is surjective and f is surjective, then g is surjective.
d) Given that gof is surjective and g is surjective, we want to prove that f is surjective. To do this, we assume that f is not surjective and reach a contradiction. Since f is not surjective, there exists an element b in the codomain B of f such that there is no element a in the domain A of gof satisfying f(a) = b. Now, since g is surjective, for every element b in B, there exists an element c in the codomain C of g such that g(c) = b. Let b be an element in B such that there is no element a in A satisfying f(a) = b. Since gof(a) = g(f(a)), when g(f(a)) = c, gof(a) = g(f(a)) = g(b) = c. Therefore, we have found an element a in the domain A of gof such that f(a) = b, which contradicts the fact that f is not surjective. Therefore, if gof is surjective and g is surjective, then f is surjective.