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352 Appendix 1: Solutions to Selected Exercises
10. (a) Suppose that f is one-to-one. Suppose that c 1 ∈ C, c 2 ∈ C, and
( f C)(c 1 ) = ( f C)(c 2 ). By exercise 7(a) of Section 5.1, it follows
that f (c 1 ) = f (c 2 ), so since f is one-to-one, c 1 = c 2 .
(b) Suppose that f C is onto. Suppose b ∈ B. Then since f C is onto,
we can choose some c ∈ C such that ( f C)(c) = b. But then c ∈ A,
and by exercise 7(a) of Section 5.1, f (c) = b.
(c) Let A = B = R and C = R . For (a), use f (x) =|x|, and for (b), use
+
f (x) = x.
14. (a) Suppose R is reflexive and f is onto. Let x ∈ B be arbitrary. Since f
is onto, we can choose some u ∈ A such that f (u) = x. Since R is
reflexive, (u, u) ∈ R. Therefore (x, x) ∈ S.
(b) Suppose R is transitive and f is one-to-one. Suppose that (x, y) ∈ S
and (y, z) ∈ S. Since (x, y) ∈ S, we can choose some u and v in
A such that f (u) = x, f (v) = y, and (u, v) ∈ R. Similarly, since
(y, z) ∈ S we can choose p and q in A such that f (p) = y, f (q) = z,
and (p, q) ∈ R. Since f (v) = y = f (p) and f is one-to-one, v = p.
Therefore (v, q) = (p, q) ∈ R. Since we also have (u, v) ∈ R, by tran-
sitivity of R it follows that (u, q) ∈ R,so(x, z) ∈ S.
17. (a) Let b ∈ B be arbitrary. Since f is onto, we can choose some a ∈ A
such that f (a) = b. Therefore g(b) = (g ◦ f )(a) = (h ◦ f )(a) =
h(b). Since b was arbitrary, this shows that ∀b ∈ B(g(b) = h(b)), so
g = h.
(b) Let c 1 and c 2 be two distinct elements of C. Suppose b ∈ B. Let
g and h be functions from B to C such that ∀x ∈ B(g(x) = c 1 ),
∀x ∈ B \{b}(h(x) = c 1 ), and h(b) = c 2 . (Formally, g = B ×{c 1 } and
h = [(B \{b}) ×{c 1 }] ∪{(b, c 2 )}.) Then g = h, so by assumption
g ◦ f = h ◦ f , and therefore we can choose some a ∈ A such that
g( f (a)) = h( f (a)). But by the way g and h were defined, the only
x ∈ B for which g(x) = h(x)is x = b, so it follows that f (a) = b.
Since b was arbitrary, this shows that f is onto.

Section 5.3
−1
1. R (p) = the person sitting immediately to the right of p.
3. Let g(x) = (3x − 5)/2. Then for any x ∈ R,
2(3x − 5)/2 + 5 3x − 5 + 5 3x
f (g(x)) = = = = x
3 3 3
and
3(2x + 5)/3 − 5 2x + 5 − 5 2x
g( f (x)) = = = = x.
2 2 2