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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. Look at the example and illustrations on pages 214 and onto Theorem 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. 1-1 Proof: This is a combination of Theorems and onto

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Theorem 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.

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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. 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 Dom(F –1 ) = Rng(F)Theorem _______________________3.1.2(a) We have now shown that F –1 is a function from Rng(F) to A

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Proof of (b): Suppose F : A B. Theorem 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. Suppose F –1 is a function from Rng(F) to A. F = (F –1 ) –1 Theorem _______________________3.1.3(a) F –1 is one-to-one applying part (a) to _______________F –1 Corollary 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. 1-1 onto 1-1 onto Theorem Let F : A B and G : B A. Then (a) G = F –1 iff G ◦ F = I A and F ◦ G = I B. (b) If F is one-to-one and onto B, then G = F –1 iff G ◦ F = I A or F ◦ G = I B.

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Proof of (a): Suppose F : A B and G : B A. Suppose G = F –1 Theorem _______________3.1.2(a)B = Dom(G) = Dom(F –1 ) = Rng(F) Theorem _______________4.2.4G ◦ F = I A and F ◦ G = I B Suppose G ◦ F = I A and F ◦ G = I B. every identity function is one-to-oneG ◦ F = I A is one-to-one Theorem _______________4.3.4F is one-to-one every identity function is ontoF ◦ G = I B is onto B Theorem _______________4.3.2F is onto B F is one-to-one & Theorem ______________4.4.2(a)F –1 is a function on B F –1 = F –1 ◦ I B = F –1 ◦ (F ◦ G) = (F –1 ◦ F ) ◦ G = I A ◦ G = G properties of the identity function Note: The textbook proof erroneously references Theorem Theorem Let F : A B and G : B A. Then (a) G = F –1 iff G ◦ F = I A and F ◦ G = I B. We have now proven part (a).

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

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

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Proof of (b): Suppose F : A B. Suppose G = F – onto We can say G ◦ F = I A or F ◦ G = I B from part (a) Now suppose that G ◦ F = I A or F ◦ G = I B. We first show G F –1. Let (b, a) G (i.e, b B and a A) Case 2: F ◦ G = I B properties of the identity function (b, b) I B (b, b) F ◦ G supposition that _____________F ◦ G = I \B (b, d) G /\ (d, b) F for some d A _____________________ definition of F ◦ G (b, a) G /\ (b, d) G, that is, d = a G is a function (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.

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

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

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1 (a) Exercises 4.4 (pages )

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1 (b)

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2 (b)

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2 (c)

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2 (d)

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3 (d)

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Section

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