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Appendix 1: Solutions to Selected Exercises 349
quadrants of the plane, the positive and negative x-axes, the positive and
negative y-axes, and the origin.
23. (a) Hint: Let T ={(X, Y) ∈ A/S × A/S |∃x ∈ X∃y ∈ Y(xRy)}.
(b) Suppose x, y, x , y ∈ A, xSx , and ySy . Then [x] S = [x ] S and

[y] S = [y ] S ,so xRy iff [x] S T [y] S iff [x ] S T [y ] S iff x Ry .

Chapter 5
Section 5.1
1. (a) Yes.
(b) No.
(c) Yes.
3. (a) f (a) = b, f (b) = b, f (c) = a.
(b) f (2) = 0.
(c) f (π) = 3 and f (−π) =−4.
5. L ◦ H : N → N, and for every n ∈ N, (L ◦ H)(n) = n. Thus, L ◦ H =
i N .
H ◦ L : C → C, and for every c ∈ C, (H ◦ L)(c) = the capital of the
country in which c is located.
7. (a) Suppose that c ∈ C. We must prove that there is a unique b ∈ B such
that (c, b) ∈ f C.
Existence: Let b = f (c) ∈ B. Then (c, b) ∈ f and (c, b) ∈ C × B,
and therefore (c, b) ∈ f ∩ (C × B) = f C.
Uniqueness: Suppose that (c, b 1 ) ∈ f C and (c, b 2 ) ∈ f C. Then
(c, b 1 ) ∈ f and (c, b 2 ) ∈ f , so since f is a function, b 1 = b 2 .
This proves that f C is a function from C to B. Finally, to derive
the formula for ( f C)(c), suppose that c ∈ C, and let b = f (c). We
showed in the existence half of the proof that (c, b) ∈ f C. It follows
that
f (c) = b = ( f C)(c).

(b) (→) Suppose g = f C. Then g = f ∩ (C × B), so clearly g ⊆ f .
(←) Suppose g ⊆ f . Suppose c ∈ C, and let b = g(c). Then (c, b) ∈
g,so(c, b) ∈ f , and therefore f (c) = b. But then by part (a),
( f C)(c) = f (c) = b = g(c). Since c was arbitrary, it follows by
Theorem 5.1.4 that g = f C.
(c) h Z = h ∩ (Z × R) ={(x, y) ∈ R × R | y = 2x + 3}∩ (Z × R) =
{(x, y) ∈ Z × R | y = 2x + 3}= g.