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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

