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Look at the example and illustrations on pages 214 and 215.

DEFINITION A function f : A  B is a one-to-one correspondence (called a bijection ) iff f is one-to-one and onto B . We write f : A  B to indicate that f is a bijection. 1-1. onto. Look at the example and illustrations on pages 214 and 215. 1-1. 1-1. 1-1.

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Look at the example and illustrations on pages 214 and 215.

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  1. DEFINITION A function f : A B is a one-to-one correspondence (called a bijection) iff f is one-to-one and onto B. We write f : A B to indicate that f is a bijection. 1-1 onto Look at the example and illustrations on pages 214 and 215. 1-1 1-1 1-1 Theorem 4.4.1 If f : A  B and g : B  C, then g◦f : AC. That is, the composite of one-to-one correspondences is a one-to-one correspondence. onto onto onto Proof: This is a combination of Theorems 4.3.1 and 4.3.3.

  2. Theorem 4.4.2 Let F : A  B (i.e., F is a function from set A to set B.) (a) F–1 is a function from Rng(F) to A iff F is one-to-one. (b) If F–1 is a function, then F–1 is one-to-one. Proof of (a): Suppose F : A  B. Suppose F–1 is a function from Rng(F) to A. To show F is one-to-one, suppose F(x) = F(y) = z. (x, z) F and (y, z) F change of notation (z, x) F–1 and (z, y) F–1 _______________________ definition of F–1 x = y supposition that F–1 is a function _______________________ F is one-to-one F(x) = F(y)  x = y Suppose F is one-to-one.

  3. Proof of (a): Suppose F : A  B. Suppose F–1 is a function from Rng(F) to A. To show F is one-to-one, suppose F(x) = F(y) = z. (x, z) F and (y, z) F change of notation (z, x) F–1 and (z, y) F–1 _______________________ definition of F–1 x = y supposition that F–1 is a function _______________________ F is one-to-one F(x) = F(y)  x = y Suppose F is one-to-one. Dom(F–1) = Rng(F) Theorem _______________________ 3.1.2(a) To show F–1 is a function, suppose (x, y) F–1 and (x, z) F–1 . (y, x) F and (z, x) F _______________________ definition of F–1 F(y) = F(z) follows from previous line y = z supposition that F is one-to-one _______________________ F–1 is a function (x, y) F–1 and (x, z) F–1 y = z We have now shown that F–1 is a function from Rng(F) to A

  4. Theorem 4.4.2 Let F : A  B (i.e., F is a function from set A to set B.) (a) F–1 is a function from Rng(F) to A iff F is one-to-one. (b) If F–1 is a function, then F–1 is one-to-one. Proof of (b): Suppose F : A  B. Suppose F–1 is a function from Rng(F) to A. Theorem _______________________ 3.1.3(a) F = (F–1)–1 applying part (a) to _______________ F–1 F–1 is one-to-one 1-1 1-1 Corollary 4.4.3 If F : A  B, then F–1: B  A. That is, the inverse of a one-to-one correspondence is a one-to-one correspondence. onto onto Theorem 4.4.4 Let F : A  B and G : B  A. Then (a) G = F–1 iff G◦F = IA and F ◦ G = IB . (b) If F is one-to-one and onto B, then G = F–1 iff G◦F = IA or F ◦ G = IB .

  5. Theorem 4.4.4 Let F : A  B and G : B  A. Then (a) G = F–1 iff G◦F = IA and F ◦ G = IB . Proof of (a): Suppose F : A  B and G : BA. Note: The textbook proof erroneously references Theorem 4.2.3. Suppose G = F–1 B = Dom(G) = Dom(F–1) = Rng(F) Theorem _______________ 3.1.2(a) G ◦ F = IA and F ◦ G = IB Theorem _______________ 4.2.4 Suppose G ◦ F = IA and F ◦ G = IB . G ◦ F = IA is one-to-one every identity function is one-to-one F is one-to-one Theorem _______________ 4.3.4 F ◦ G = IB is onto B every identity function is onto F is onto B Theorem _______________ 4.3.2 F–1 is a function on B F is one-to-one & Theorem ______________ 4.4.2(a) F–1 = F–1 ◦ IB = F–1 ◦ (F ◦ G) = (F–1 ◦ F ) ◦ G = IA ◦ G = G properties of the identity function We have now proven part (a).

  6. Theorem 4.4.4 Let F : A  B and G : B  A. Then (a) G = F–1 iff G◦F = IA and F ◦ G = IB . (b) If F is one-to-one and onto B, then G = F–1 iff G◦F = IA or F ◦ G = IB . 1-1 Proof of (b): Suppose F : A  B. onto Suppose G = F–1. We can say G◦F = IA or F ◦ G = IB from part (a) Now suppose that G ◦ F = IA or F ◦ G = IB . We first show GF–1. Case 1: G ◦ F = IA Let (b, a)  G (i.e, b B and a A) properties of the identity function (a, a)  IA (a, a)  G ◦ F supposition that _____________ G ◦ F = IA definition of G ◦ F (a, c)  F /\ (c, a)  G for some c B _____________________ F is one-to-one and onto B (a, b)  F /\ (b, a)  G, that is, c =b

  7. 1-1 Proof of (b): Suppose F : A  B. onto Suppose G = F–1. We can say G◦F = IA or F ◦ G = IB from part (a) Now suppose that G ◦ F = IA or F ◦ G = IB . We first show GF–1. Case 1: G ◦ F = IA Let (b, a)  G (i.e, b B and a A) properties of the identity function (a, a)  IA (a, a)  G ◦ F supposition that _____________ G ◦ F = IA definition of G ◦ F (a, c)  F /\ (c, a)  G for some c B _____________________ F is one-to-one and onto B (a, b)  F /\ (b, a)  G, that is, c =b (b, a) F–1 _____________________ definition of F–1 GF–1 (b, a)  G  (b, a) F–1

  8. 1-1 Proof of (b): Suppose F : A  B. onto Suppose G = F–1. We can say G◦F = IA or F ◦ G = IB from part (a) Now suppose that G ◦ F = IA or F ◦ G = IB . We first show GF–1. Case 2: F ◦ G = IB Let (b, a)  G (i.e, b B and a A) properties of the identity function (b, b)  IB (b, b)  F ◦ G supposition that _____________ F ◦ G = I\B definition of F ◦ G (b, d)  G /\ (d, b)  F for some d A _____________________ G is a function (b, a)  G /\ (b, d)  G, that is, d =a (b, a) F–1 (a, b)  F & definition of F–1 _____________________ GF–1 (b, a)  G  (b, a) F–1 In either case, (b, a)  F–1 which shows that GF–1.

  9. We now show F–1G. Case 1: G ◦ F = IA Let (b, a)  F–1 (i.e, b B and a A) properties of the identity function (a, a)  IA (a, a)  G ◦ F supposition that _____________ G ◦ F = IA definition of G ◦ F (a, c)  F /\ (c, a)  G for some c B _____________________ (a, c)  F /\ (a, b)  F, that is, c =b a function (b, a)  F–1 and F is ________ (b, a) G _____________________ (c, a)  G and c =b F–1G (b, a)  F–1  (b, a) G

  10. We now show F–1G. Case 2: F ◦ G = IB Let (b, a)  F–1 (i.e, b B and a A) properties of the identity function (b, b)  IB (b, b)  F ◦ G supposition that _____________ F ◦ G = I\B definition of F ◦ G (b, d)  G /\ (d, b)  F for some d A _____________________ (a, b)  F /\ (d, b)  F, that is, d =a (b, a)  F–1 and F is ________ 1-1 (b, a) F–1 (a, b)  F & definition of F–1 _____________________ F–1G (b, a)  G  (b, a) F–1 Look at the examples on page 216. In either case, (b, a)  G which shows that F–1G. We conclude G = F–1, since GF–1 and F–1G.

  11. Exercises 4.4 (pages 218-219) 1  (a)

  12. 1  (b)

  13. 2  (b)

  14. 2  (c)

  15. 2  (d)

  16. 3  (d)

  17. Section 5.1 3 

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