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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.

Presentation on theme: "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."— Presentation transcript:

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 215. 1-1 onto 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. 1-1 Proof: This is a combination of Theorems 4.3.1 and 4.3.3. onto

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.

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

Proof of (b): Suppose F : A  B. 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. 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 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. 1-1 onto 1-1 onto Theorem 4.4.4 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 (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 4.2.3. Theorem 4.4.4 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).

Theorem 4.4.4 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 –1. 1-1 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

Proof of (b): Suppose F : A  B. Suppose G = F –1. 1-1 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

Proof of (b): Suppose F : A  B. Suppose G = F –1. 1-1 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.

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

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.

1  (a) Exercises 4.4 (pages 218-219)

1  (b)

2  (b)

2  (c)

2  (d)

3  (d)

Section 5.1 3 

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