Bridge to Abstract Mathematics: Mathematical Proof and Structures

ANSWERS AND SOLUTIONS TO SELECTED EXERCISES 375

(-n/2, d2). The latter is one to one over that interval and has range R.

Hence tan-' is one to one, but since its image equals (-42, n/2) c

[- 42,n/2], the mapping f: R -, [- n/2,n/2] is not onto.

(c) Proof => Suppose g 0 f: X -, Z is a surjection. Then g is surjective,

by Theorem 2(d). (: Conversely, suppose g is surjective. Then f and g

are both surjective so that g 0 f is surjective, by Theorem 2(a). 0

(b) (ii) Define f: R -, R by f(x) = 2x - 3 and g: R -, R by g(x) = sin x.

Note that f is clearly bijective. But the mapping g 0 f: R -, R, (g 0 f)(x) =

sin (2x - 3) has range [- 1, 11, a proper subset of the range R, and so is

not surjective, and consequently is not bijective.

(a) (iii) Given f: R -, R, f(x) = 3x - 7, we have f - I([- 7,2]) = [O,3].

(b) (iv) f - '((e, 0)) = {A, C, D, E, G, H). (c) (iii) x E f - '([ - 1, 11) -

-11(x+1)251~05(x+1)251~-l~~+1~1CS.-25x50.

Hence f -'(I- 1, 1)) = [-2,0].

(a) (i) Proof Let x E f - '(N, u N,). Then f (x) E N, u N,, that is, either

f(x)~ N1 or f(x)~ N,. If f(x)~. N1, then XE f-'(NJ c f-'(N1) u f-'(N,),

so x E f - '(NJ u f - '(N,), as desired. Similarly, iff (x) E N,, then

x E f -'(N,) c f-'(N,) u f -'(N,), so x E f -'(N1) u f -'(N2), again, as

desired. Conversely, suppose x E f -'(Nl) u f - '(N,). Then either

x E f -'(N1) or x E fA1(N2). If x E f -'(N1), then f(x) E N1 c Nl u N2 so

that x E f -'(N1 u N,). An identical argument leads to the same conclusion

[i.e.,x~f-'(N~ u N2)]incasex~ f-'(N,).

(a) Proof => Assume f (f - '(Y)) = Y and let y E Y. To prove y E rng f, we

must prove y = f (x) for some x E A. Now y E Y = f (f - '(Y)) so

y E f (f - '(Y)); that is, y = f (x) for some x E f - l(Y). But f - '(Y) c A and so

y = f(x) for some x E A, as desired. Conversely, assume y E mg f. Since we

know f( f - '(Y)) c Y in general [by using Theorem 5(a)], we need only

prove Y G f (f - '(9). So let y E Y. To show y E f (f - '(Y)), we must show

y = f (x) for some x E f -'(Y). Now y E Y and Y E mg f means

y E mg f, SO that y = f (x) for some x E A. Now f (x) = y E Y so f (x) E Y so

that this x, in fact, is contained in f - '(Y). Hence y = f(x) for some

x E f -'(Y), as required. As to the consistency between this result and

Theorem 5(d), note that f (f - '(Y)) = Y for all subsets Y of B CS. Y c mg f

for all subsets Y of B - B c rng f - B = mg f CS. f is onto.

(c) Proof Assume f is one to one. If (b) has been proved, we need only

prove f (M - M,) G f (M,) - f(M,) in order to establish equality. Hence

let y E f (M - M,). Then y = f (x) for some x E MI - M,. To prove

y E f (M - f (M,), we must show that y E f (MI), but y 4 f (M,). Now since

y = f (x), where x E M1 - M2 E MI, we know that y E f (M Proceeding

indirectly, suppose y E f (M,). Then y = f (m) for some m E M,. Hence we

have y = f(x) = f(m), where m E M2 and x E MI - M2 so that x 4 M,. Thus

x # m. But f(x) = f(m) and x # m contradicts the one-to-one property

off.

Article 8.3

1. (c) Proof The mapping f: (0, 1) -, (7,13) given by f(x) = 6x + 7 is clearly a
   one-to-one mapping of (0, 1) onto (7, 13).