Page 264 - HOW TO PROVE IT: A Structured Approach, Second Edition

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250 Functions
we’ve proven. You could check it directly by solving for f −1 (x), using the
fact that f −1 (x) must be the unique solution for y in the equation f (y) = x.
However, there is no need to check. The next theorem shows that f −1 must be
equal to g.

Theorem 5.3.5. Suppose f : A → B, g : B → A, g ◦ f = i A , and f ◦ g = i B .
Then g = f −1 .
Proof. By Theorem 5.3.4, f −1 : B → A. Therefore, by Theorem 5.3.2,
f −1 ◦ f = i A . Thus,

g = i A ◦ g (exercise 9 of Section 4.3)
= ( f −1 ◦ f ) ◦ g
= f −1 ◦ ( f ◦ g) (Theorem 4.2.5)
= f −1 ◦ i B
= f −1 (exercise 9 of Section 4.3).

Commentary . This proof gets the desired conclusion quickly by clever use of
previous theorems and exercises. For a more direct but somewhat longer proof,
see exercise 10.

Example 5.3.6. In each part, determine whether or not f is one-to-one and
onto. If it is, find f −1 .

1. Let A = R \{0} and B = R \{2}, and define f : A → B by the formula
1
f (x) = + 2.
x
(Note that for all x ∈ A, 1/x is defined and nonzero, so f (x) = 2 and
therefore f (x) ∈ B.)
2. Let A = R and B ={x ∈ R | x ≥ 0}, and define f : A → B by the formula
2
f (x) = x .

Solutions

1. You can check directly that f is one-to-one and onto, but we won’t bother
to check. Instead, we’ll simply try to find a function g : B → A such that
g ◦ f = i A and f ◦ g = i B . We know by Theorems 5.3.4 and 5.3.5 that if
we find such a g, then we can conclude that f is one-to-one and onto and
g = f −1 .