Page 260 - HOW TO PROVE IT: A Structured Approach, Second Edition

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246 Functions
Theorem 5.3.1. Suppose f : A → B. If f is one-to-one and onto, then f −1 :
B → A.
Proof. Suppose f is one-to-one and onto, and let b be an arbitrary element
of B. To show that f −1 is a function from B to A, we must prove that
∃!a ∈ A((b, a) ∈ f −1 ), so we prove existence and uniqueness separately.
Existence: Since f is onto, there is some a ∈ A such that f (a) = b. Thus,
(a, b) ∈ f ,so(b, a) ∈ f −1 .
Uniqueness: Suppose (b, a 1 ) ∈ f −1 and (b, a 2 ) ∈ f −1 for some a 1 , a 2 ∈ A.
Then (a 1 , b) ∈ f and (a 2 , b) ∈ f ,so f (a 1 ) = b = f (a 2 ). Since f is one-to-one,
it follows that a 1 = a 2 .

Commentary . The form of the proof is guided by the logical form of the state-
ment that f −1 : B → A. Because this means ∀b ∈ B∃!a ∈ A((b, a) ∈ f −1 ),
we let b be an arbitrary element of B and then prove existence and uniqueness
for the required a ∈ A separately. Note that the assumption that f is onto is the
key to the existence half of the proof, and the assumption that f is one-to-one
is the key to the uniqueness half.

Suppose f is any function from a set A to a set B. Theorem 5.3.1 says that
a sufficient condition for f −1 to be a function from B to A is that f be one-to-
one and onto. Is it also a necessary condition? In other words, is the converse
of Theorem 5.3.1 true? (If you don’t remember what the words sufficient,
necessary, and converse mean, you should review Section 1.5!) We will show
in Theorem 5.3.4 that the answer to this question is yes. In other words, if f −1
is a function from B to A, then f must be one-to-one and onto.
If f −1 : B → A then, by the definition of function, for every b ∈ B there is
exactly one a ∈ A such that (b, a) ∈ f −1 , and

f −1 (b) = the unique a ∈ A such that (b, a) ∈ f −1
= the unique a ∈ A such that (a, b) ∈ f
= the unique a ∈ A such that f (a) = b.

This gives another useful way to think about f −1 .If f −1 is a function from B
to A, then it is the function that assigns, to each b ∈ B, the unique a ∈ A such
that f (a) = b. The assumption in Theorem 5.3.1 that f is one-to-one and onto
guarantees that there is exactly one such a.
As an example, consider again the function f that assigns, to each person in
the audience at a sold-out concert, the seat in which that person is sitting. As
we saw at the end of the last section, f is a one-to-one, onto function from the
set A of all members of the audience to the set S of all seats in the concert hall.