Page 245 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Functions 231
from A to C. In fact, it turns out that g ◦ f is a function from A to C, as the next
theorem shows.

Theorem 5.1.5. Suppose f : A → B and g : B → C. Then g ◦ f : A → C,
and for any a ∈ A, the value of g ◦ f at a is given by the formula (g ◦ f )(a) =
g( f (a)).

Scratch work
Before proving this theorem, it might be helpful to discuss the scratch work
for the proof. According to the definition of function, to show that g ◦ f :
A → C we must prove that ∀a ∈ A∃!c ∈ C((a, c) ∈ g ◦ f ), so we will start
out by letting a be an arbitrary element of A and then try to prove that ∃!c ∈
C((a, c) ∈ g ◦ f ). As we saw in Section 3.6, we can prove this statement by
proving existence and uniqueness separately. To prove existence, we should
try to find a c ∈ C such that (a, c) ∈ g ◦ f . For uniqueness, we should assume
that (a, c 1 ) ∈ g ◦ f and (a, c 2 ) ∈ g ◦ f , and then try to prove that c 1 = c 2 .

Proof. Let a be an arbitrary element of A. We must show that there is a unique
c ∈ C such that (a, c) ∈ g ◦ f .
Existence: Let b = f (a) ∈ B. Let c = g(b) ∈ C. Then (a, b) ∈ f and
(b, c) ∈ g, so by the definition of composition of relations, (a, c) ∈ g ◦ f .
Thus, ∃c ∈ C((a, c) ∈ g ◦ f ).
Uniqueness: Suppose (a, c 1 ) ∈ g ◦ f and (a, c 2 ) ∈ g ◦ f . Then by the def-
inition of composition, we can choose b 1 ∈ B such that (a, b 1 ) ∈ f and
(b 1 , c 1 ) ∈ g, and we can also choose b 2 ∈ B such that (a, b 2 ) ∈ f and
(b 2 , c 2 ) ∈ g. Since f is a function, there can be only one b ∈ B such that
(a, b) ∈ f . Thus, since (a, b 1 ) and (a, b 2 ) are both elements of f, it follows
that b 1 = b 2 . But now applying the same reasoning to g, since (b 1 , c 1 ) ∈ g and
(b 1 , c 2 ) = (b 2 , c 2 ) ∈ g, it follows that c 1 = c 2 , as required.
This completes the proof that g ◦ f is a function from A to C. Finally, to
derivetheformulafor(g ◦ f )(a),notethatweshowedintheexistencehalfofthe
proof that for any a ∈ A,ifwelet b = f (a) and c = g(b), then (a, c) ∈ g ◦ f .
Thus,

(g ◦ f )(a) = c = g(b) = g( f (a)).

When we first introduced the idea of the composition of two relations
in Chapter 4, we pointed out that the notation was somewhat peculiar and
promised to explain the reason for the notation in this chapter. We can now
provide this explanation. The reason for the notation we’ve used for composi-
tion of relations is that it leads to the convenient formula (g ◦ f )(x) = g( f (x))