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Function Composition

Proposition: The composition of two bijective functions is bijective:

forall A, B, C, f: A ->bijective B, g: B ->bijective C:

f o g: A ->bijective C.

Proof: Take arbitrary f: A ->^bijective B, g: B ->^bijective C.

• We show f o g is injective. Take arbitrary x₀ in A and x₁ in A with (f o g)(x₀) = (f o g)(x₁). We have to show x₀ = x₁.

  We know, by definition of o , that g(f(x₀)) = g(f(x₁)) and thus, because g is injective, f(x₀) = f(x₁). Since f is injective, we then have x₀ = x₁.

• We show f o g is surjective. Take arbitrary z in C; we have to find some x such that (f o g)(x) = z.

  Since g is surjective, we have some y in B such that g(y) = z. Since f is surjective, we have some x in A such that f(x) = y. Thus (f o g)(x) = g(f(x)) = g(y) = z.