8.1 FUNCTIONS AND MAPPINGS 261both functions, together with the condition "there exists y.. ." in Defi-
nition 5, it clearly has far fewer ordered pairs. For example, we may ask
whether (6, 17) is an element off 0 k. Since (6,s) E k, we would need
(5, 17) E f in order to have (6, 17) E f 0 k. Since (5, 17) 4 f, the answer to
our question is "no." We may quickly list the ordered pairs in f 0 k, in
fact, by "splicing together" ordered pairs in k and f. Since (4,3) E k
and (3, 10) E f, then (4, 10) E f 0 k. Since (6,s) E k and (5,20) E f, then
(6,20) E f 0 k. Now (lo, 20) E k, but 20 4 dorn f, so that this ordered pair
from k does not lead to an ordered pair in f 0 k. In fact, f 0 k =
((4, lo), (6,20)} so that, for instance, (f 0 k)(4) = 10. Similarly, g 0 k =
((4,4), (10, lo)), while k O g = {(2,5), (3, 3), (7,20), (17,20), (20,20)}.You should calculate k 0 f, f 0 g, and g 0 f in the preceding example.
Also, noting-that f and k are one-to-one functions, so that f - ' and k- are
functions, you should calculate f 0 f - l, f - 0 f, k 0 k- ', and k- ' 0 k. Gen-
eral conclusions from these calculations are contained in Exercise 9. Fi-
nally, you should calculate and compare f 0 (k 0 g) and (f 0 k) 0 g.
The definition of composition, together with the results from Example 3,
suggest a number of observations about function composition. For in-
stance, using standard functional notation, we see that z = (g 0 f)(x) if and
only if y = f(x) and z = g(y) for some y. If such a y exists, we then have
z = g(y) = g( f (x)). The equation z = g( f (x)) is the formula commonly used
to calculate the rule of correspondence for g 0 f, given rules for g and for f.
Thus if f(x) = 3x - 4 and g(x) = ex, then (g 0 f)(x) = g(f(x)) = g(3x - 4) =
e3x - 4 for each x E R.
Second, we note that dorn (g 0 f) is clearly contained as a subset in
dorn f. That is, in order to be in dorn (g 0 f), an object x must first be
in dorn f. Furthermore, for each such x, the object f(x) (i.e., y) must be in
dorn g. Indeed, x E dorn (g 0 f) if and only if both these conditions are sat-
isfied, so that dorn (g 0 f) = (x 1 x E dorn f and f (x) E dorn g) [see Exercise
8(a)l.
Third, the results of the calculations of g 0 k and k 0 g indicate that com-
position of functions is a noncommutative operation; that is, it is not the
case that f 0 g = g 0 f for all functions f and g; indeed, this relationship
seldom holds (see Exercise 3).
Fourth, our results in Example 3 suggest the following fact.
THEOREM 2
If f and g are functions, then the composite g 0 f is a function.
Proof To prove that g 0 f is a function, assume that the ordered pairs
(x, z,) and (x, z,) are in g 0 f. Our claim is that z, = z,. By Definition
5, there exists objects y, and y, such that (x, y,) E f, (y,, z,) E g, (x, y,) E
f, and (y,, z2) E g. Since f is a function, we conclude y, = y,. Let us
denote this common value by y. Since g is a function and since both
(y, z,) and (y, 2,) are elements of g, we conclude z, = z,, as desired.