632 Appendix B • Sets and Functions
Next, the ~ direction. Suppose f : A----; B is 1-1 and onto B. Let b E B.
Since f is onto B, ::3a EA 3 f(a) = b. Moreover, since f is 1-1, there is no
more than one such a. Define
g(b) =a.
Then Va E A, (go f)(a) = g(f(a)) = g(b) = a = iA(a). Also, Vb E B,
(! o g)(b) = f(g(b)) = f(a) = b = iB(b). That is, go f = iA and fog = iB.
Therefore, by Definition B.3.11, f is invertible.
Finally, the proof of g = 1-^1 ::::> f = g-^1 is Exercise 16. •
Corollary B .3.13 If f : A----; B is a 1-1 correspondence, so is 1-^1 : B ----; A.
Proof. Apply Theorem B.3.10. •
EXERCISE SET B.3
- Let f(x) = 2x+ 1 and g(x) = x^2 -2. Find(!+ g)(x), (! -g)(x), f(x+2),
f(x) +2, g(x +2), g(x) +2, 3f(x), f(3x), 3g(x), g(3x), (fg)(x), (!jg) (x),
IJl(x), max{/, g }(x), min{/, g }(x), (! o g)(x), and (go f)(x). - Repeat Exercise 1 with f(x) = x and g(x) = 2-.
3x+4 x - For each of the following functions, f, find V (!) and R(f), and tell
whether f is 1-1:
(a) f(x) = 7x+8
(c) f(x) = JX2=l
(e) f(x) = ex
(g) f(x) = sinx
(b) f(x) = v'x+1
(d) f(x) = lnx
x
(f) f(x ) = -
x+l
(h) f(x) = x^3 + 2
- Which of the functions given in Exercise 3, viewed as f : V(f) ----; R(f),
are invertible. Find 1-^1 where possible. - Prove Theorem B.3 .3, (3) and (13).
- Prove Theorem B .3.3, (5). Note: you must define -f.
- Prove Theorem B.3.3, (6).
- Prove Theorem B.3.3, (9).
- Prove Theorem B.3.3, (10).