Exercises 0.2 #1

a) g(f(x1)) = g(f(x2)) ⇒ f(x1) = f(x2) ⇒ x1 = x2

b) Suppose f is not injective. There exists x1, x2 such that f( x1) = f( x2) and x1 ≠ x2. Hence g(f( x1)) = g(f( x2)) and x1 ≠ x2. Contradiction.

c) z = g(y) for some y, y = f(x) for some x ⇒ z = g(f(x))

d) Suppose g is not surjective. There exists z ≠ g(y) for every y ⇒ z ≠ g(f(x)) for every x. Contradiction.

e) X = ℕ, f: n → 2n+1, is injective and not surjective. f(n) = { 0 if n=0, n-1 otherwise } is surjective and not injective.

f) By (b) and (d), f is injective and g is surjective. Suppose f is surjective and g is not injective. There exists y1, y2 such that g( y1) = g( y2) and y1 ≠ y2. So g(f( x1)) = g(f( x2)) and [y1 = f(x1)] ≠ [y2 = f(x2)] for some x1, x2. By definition of function x1 ≠ x2, contradicting gf being bijective. Thus f is surjective ⇒ g is injective.

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