one pigeon. Let f : X rightarrow Y and g : Y rightarrow Z be functions. lf both

Question

one pigeon. Let f : X rightarrow Y and g : Y rightarrow Z be functions. lf both f and g are injective, prove that g f is injective. If both f and g are surjective, prove that g f is surjective.

Explanation / Answer

i) Assume f and g are both injective. Let x.y be in X. Then gof(x) = gof(y) => g(f(x)) = g(f(y)) => f(x) = f(y) since g is injective => x = y since f is injective Therefore gof is injective. ii) Assume f and g are both surjective. Let z be in Z. Since g is surjective, there exists a y in Y such that g(y) = z. But since f is surjective and y is in Y, there exists an x in X such that f(x) = y. Thus g(f(x)) = z and hence gof is surjective. iii) Assume f and g are bijective. That means f and g are both injective and surjective. By part (i), we showed gof is injective if both f and g are injective, and by part (ii) we showed gof is surjective if both f and g are surjective. Therefore gof is also bijective. iv) Assume gof is a bijection. That is, gof is both injective and surjective. So if there exists a z in Z, then there is an x in X such that gof(x) = z. But gof(x) = g(f(x)) = z. Thus g is bijective since f(x) maps to z. Now assume f(x) = f(y). Taking g of both sides: g(f(x)) = g(f(y)) => gof(x) = gof(y) => x = y since gof is injective. Therefore f is injective.

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