Page 263 - HOW TO PROVE IT: A Structured Approach, Second Edition
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                                                  Inverses of Functions                249
                            Commentary . The assumption that there is a g : B → A such that g ◦ f = i A
                            is an existential statement, so we immediately imagine that a particular function
                            g has been chosen. The proof that f is one-to-one follows the usual pattern for
                            such proofs, based on Theorem 5.2.3.


                              We have come full circle. In Theorem 5.3.1 we found that if f is a one-to-
                            one, onto function from A to B, then f  −1  is a function from B to A. From
                            this conclusion it follows, as we showed in Theorem 5.3.2, that the composi-
                            tion of f with its inverse must be the identity function. And in Theorem 5.3.3
                            we found that when the composition of two functions is the identity func-
                            tion, we are led back to the properties one-to-one and onto! Thus, combining
                            Theorems 5.3.1–5.3.3, we get the following theorem.

                            Theorem 5.3.4. Suppose f : A → B. Then the following statements are
                            equivalent.

                            1. f is one-to-one and onto.
                            2. f  −1  : B → A.
                            3. There is a function g : B → A such that g ◦ f = i A and f ◦ g = i B .

                            Proof. 1 → 2. This is precisely what Theorem 5.3.1 says.
                              2 → 3. Suppose f  −1  : B → A. Let g = f  −1  and apply Theorem 5.3.2.
                              3 → 1. Apply Theorem 5.3.3.

                            Commentary . As we saw in Section 3.6, the easiest way to prove that several
                            statementsareequivalentistoproveacircleofimplications.Inthiscasewehave
                            proven the circle 1 → 2 → 3 → 1. Note that the proofs of these implications
                            are quite sketchy. You should make sure you know how to fill in all the details.

                              For example, let f and g be functions from R to R defined by the following
                            formulas:
                                                   x + 7
                                             f (x) =    ,    g(x) = 5x − 7 .
                                                     5
                            Then for any real number x,

                                                     x + 7      x + 7
                              (g ◦ f )(x) = g( f (x)) = g   = 5 ·     − 7 = x + 7 − 7 = x.
                                                       5          5
                            Thus, g ◦ f = i R . A similar computation shows that f ◦ g = i R . Thus, it fol-
                            lows from Theorem 5.3.4 that f must be one-to-one and onto, and f  −1  must
                            also be a function from R to R. What is f  −1 ? Of course, a logical guess
                            would be that f  −1  = g, but this doesn’t actually follow from the theorems
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