MATH 13 Study Guide - Midterm Guide: Equivalence Class, Bijection, Surjective Function
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Solutions to review problems from the last class: let f : y z and g : x y be bijections. Prove that the composite f g : x z is a bijection. We need to show the composite is injective and bijective. To show injectivity take x1, x2 x with (f g)(x1) = (f g)(x2). By de nition of the composite this assumption implies f (g(x1)) = f (g(x2)), but we have also assumed that f is bijective and therefore in particular injective, so it follows that g(x1) = g(x2). Since g is also assumed injective, we nd x1 = x2, establishing injectivity. To show surjectivity take arbitrary z z. We need to nd x x such that f (g(x)) = z. Regarded as a relation, the function f is the set {(x, f (x)) r2 | x r}. We are told that this relation happens to be an equivalence relation as well as a function.
