Page 168 - Discrete Mathematics and Its Applications

2.3 Functions 147

( f g)(a)

g(a) f(g(a))

a g(a) f(g(a))
g f

A B C

f g
FIGURE 7 The Composition of the Functions f and g.

EXAMPLE 22 Let g be the function from the set {a, b, c} to itself such that g(a) = b, g(b) = c, and g(c) = a.
Let f be the function from the set {a, b, c} to the set {1, 2, 3} such that f(a) = 3, f(b) = 2, and
f(c) = 1. What is the composition of f and g, and what is the composition of g and f ?

Solution: The composition f ◦ g is defined by (f ◦ g)(a) = f(g(a)) = f(b) = 2,
(f ◦ g) (b) = f(g(b)) = f(c) = 1, and (f ◦ g)(c) = f(g(c)) = f(a) = 3.
Note that g ◦ f is not defined, because the range of f is not a subset of the domain of g. ▲

EXAMPLE 23 Let f and g be the functions from the set of integers to the set of integers defined by
f(x) = 2x + 3 and g(x) = 3x + 2. What is the composition of f and g? What is the com-
position of g and f ?

Solution: Both the compositions f ◦ g and g ◦ f are defined. Moreover,

(f ◦ g)(x) = f(g(x)) = f(3x + 2) = 2(3x + 2) + 3 = 6x + 7

and
(g ◦ f )(x) = g(f (x)) = g(2x + 3) = 3(2x + 3) + 2 = 6x + 11. ▲

Remark: Note that even though f ◦ g and g ◦ f are defined for the functions f and g in
Example 23, f ◦ g and g ◦ f are not equal. In other words, the commutative law does not hold
for the composition of functions.

When the composition of a function and its inverse is formed, in either order, an identity
function is obtained. To see this, suppose that f is a one-to-one correspondence from the set A
to the set B. Then the inverse function f −1 exists and is a one-to-one correspondence from B
to A. The inverse function reverses the correspondence of the original function, so f −1 (b) = a
when f(a) = b, and f(a) = b when f −1 (b) = a. Hence,

(f −1 ◦ f )(a) = f −1 (f (a)) = f −1 (b) = a,

and

(f ◦ f −1 )(b) = f(f −1 (b)) = f(a) = b.
Consequently f −1 ◦ f = ι A and f ◦ f −1 = ι B , where ι A and ι B are the identity functions on
the sets A and B, respectively. That is, (f −1 −1 = f .
)