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Inverses of Functions 247
Thus, f −1 must be a function from S to A, and for each s ∈ S,

f −1 (s) = the unique a ∈ A such that f (a) = s
= the unique person a such that the seat in which a is sitting is s
= the person who is sitting in the seat s.

In other words, the function f assigns to each person the seat in which that
person is sitting, and the function f −1 assigns to each seat the person sitting in
that seat.
Because f : A → S and f −1 : S → A, it follows by Theorem 5.1.5 that
f −1 ◦ f : A → A and f ◦ f −1 : S → S. What are these functions? To figure
out what the first function is, let’s let a be an arbitrary element of A and
compute ( f −1 ◦ f )(a).

( f −1 ◦ f )(a) = f −1 ( f (a))
−1
= f (the seat in which a is sitting)
= the person sitting in the seat in which a is sitting

= a.

But recall that for every a ∈ A, i A (a) = a. Thus, we have shown that ∀a ∈
A(( f −1 ◦ f )(a) = i A (a)), so by Theorem 5.1.4, f −1 ◦ f = i A . Similarly, you
should be able to check that f ◦ f −1 = i S .
When mathematicians find an unusual phenomenon like this in an example,
they always wonder whether it’s just a coincidence or if it’s part of a more
general pattern. In other words, can we prove a theorem that says that what
happened in this example will happen in other examples too? In this case, it
turns out that we can.

Theorem 5.3.2. Suppose f is a function from A to B, and suppose that f −1 is
a function from B to A. Then f −1 ◦ f = i A and f ◦ f −1 = i B .
Proof Let a be an arbitrary element of A. Let b = f (a) ∈ B. Then (a, b) ∈ f ,
so (b, a) ∈ f −1 and therefore f −1 (b) = a. Thus,

( f −1 ◦ f )(a) = f −1 ( f (a)) = f −1 (b) = a = i A (a).

Since a was arbitrary, we have shown that ∀a ∈ A(( f −1 ◦ f )(a) = i A (a)), so
f −1 ◦ f = i A . The proof of the second half of the theorem is similar and is left
as an exercise (see exercise 8).