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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 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.
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.