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DEFINITIONS For a function F : A B, the inverse of F is the following relation from B to A: F –1 = {(x, y) : (y, x) F}. For functions F : A B and G : B C the composite of F and G is the following relation from A to C: G ◦ F = {(x, z) A C : (x, y) F and (y, z) G, for some y B}. Look at the examples on page 196. Theorem 4.2.1 Let A, B, and C be sets, and suppose functions F : A B and G : B C are defined. Then G ◦ F is a function from A to C where Dom(G ◦ F) = A. Proof We know that G ◦ F is a relation from A to C. We must show that (i) Dom(G ◦ F) = A, and (ii) if (x, y) G ◦ F and (x, z) G ◦ F, then y = z.
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We first show that Dom(G ◦ F) = A Dom(G ◦ F) Dom(F) = A Exercise We now want to show that A Dom(G ◦ F) Let a A = Dom(F) (a, b) F for some b B definition of Dom(F) (b, c) G for some c C b B = Dom(G) (a, c) G ◦ F definition of G ◦ F a Dom(G ◦ F) definition of Dom(G ◦ F) A Dom(G ◦ F) a A a Dom(G ◦ F) Dom(G ◦ F) = A Dom(G ◦ F) A and A Dom(G ◦ F) ___________3.1-9(a) Proof We know that G ◦ F is a relation from A to C. We must show that (i) Dom(G ◦ F) = A, and (ii) if (x, y) G ◦ F and (x, z) G ◦ F, then y = z. _____________________
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To prove (ii), let (x, y) G ◦ F and (x, z) G ◦ F; we must show y = z. (x, u) F /\ (u, y) G for some u B (x, y) G ◦ F (x, v) F /\ (v, z) G for some v B (x, z) G ◦ F u = vF is a function y = zu = v and G is a function G ◦ F is a function (x, y) G ◦ F /\ (x, z) G ◦ F y = z _____________________
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Note: Theorem 3.1.3(b) tells us that the composition of relations is associative. Theorem 4.2.2 Let A, B, C, and D be sets, and suppose the functions f : A B, g : B C, and h : C D are defined. Then (h ◦ g) ◦ f = h ◦ (g ◦ f), that is, the composition of functions is associative. Proof We must show that Dom((h ◦ g) ◦ f) = Dom(h ◦ (g ◦ f)) and that ((h ◦ g) ◦ f)(x) = (h ◦ (g ◦ f))(x). Dom((h ◦ g) ◦ f) = Dom(f) = ATheorem4.2.1 _____________ Dom(h ◦ (g ◦ f)) = Dom(g ◦ f) = Dom(f) = ATheorem4.2.1 _____________ Now, suppose x A ((h ◦ g) ◦ f)(x) = (h ◦ g)(f(x)) = h(g(f(x))) =h((g ◦ f)(x)) =(h ◦ (g ◦ f))(x) previously defined notation (h ◦ g) ◦ f = h ◦ (g ◦ f) Theorem4.1.1 _____________
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Theorem 4.2.3 Let f : A B. Then f ◦ I A = f and I B ◦ f = f. Proof Dom(f ◦ I A ) = Dom(I A ) Theorem4.2.1 _____________ substitution using Dom(I A ) = ADom(f ◦ I A ) = A Dom(f ◦ I A ) = Dom(f)______________ f : A B supposition that Now, suppose x A. (f ◦ I A )(x) = f(I A (x)) = f(x)previously defined notation f ◦ I A = fthe two conditions of Theorem _____ are satisfied 4.1.1
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Theorem 4.2.3 Let f : A B. Then f ◦ I A = f and I B ◦ f = f. Dom(I B ◦ f) = Dom(f) Theorem4.2.1 _____________ Now, suppose x A. (I B ◦ f)(x) = I B (f(x)) = f(x)previously defined notation I B ◦ f = fthe two conditions of Theorem _____ are satisfied 4.1.1
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Theorem 4.2.4 Let f : A B with Rng(f) = C. If f –1 is a function, then f –1 ◦ f = I A and f ◦ f –1 = I C. Proof: Suppose f : A B with Rng(f) = C, and f –1 is a function Dom(f –1 ◦ f) = Dom(f)Theorem4.2.1 _____________ substitution using Dom(f) = ADom(f –1 ◦ f) = A Dom(f –1 ◦ f) = Dom(I A )____________________________ Dom(I A ) = A Now, suppose x A. (x, f(x)) f definition of Dom(f) = A (f(x), x) f –1 definition of_____________ f –1 (f –1 ◦ f)(x) = f –1 (f(x)) (x, f(x)) f (f –1 ◦ f)(x) = x (f(x), x) f –1 (f –1 ◦ f)(x) = I A (x)I A (x) = x f –1 ◦ f = I A the two conditions of Theorem _____ are satisfied 4.1.1
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Theorem 4.2.4 Let f : A B with Rng(f) = C. If f –1 is a function, then f –1 ◦ f = I A and f ◦ f –1 = I C. Proof: Suppose f : A B with Rng(f) = C, and f –1 is a function Dom(f ◦ f –1 ) = Dom(f –1 )Theorem4.2.1 ______________ TheoremDom(f ◦ f –1 ) = Rng(f) ______________ 3.1.2(a) supposition thatDom(f ◦ f –1 ) = C ______________ Rng(f) = C Dom(f –1 ◦ f) = Dom(I C )____________________________ Dom(I C ) = C Now, suppose x C. (x, f –1 (x)) f –1 definition of Dom(f –1 ) = Rng(f) = C (f –1 (x), x) f definition of_____________ f –1 (f ◦ f –1 )(x) = f(f –1 (x)) (x, f –1 (x)) f –1 (f –1 ◦ f)(x) = x (f –1 (x), x) f (f –1 ◦ f)(x) = I C (x)I C (x) = x f ◦ f –1 = I C the two conditions of Theorem _____ are satisfied 4.1.1
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DEFINITIONS Let f : A B, and let D A. The restriction of f to D is the function f | D = {(x, y) : y = f(x) and x D}. If g and h are functions, and g is a restriction of h, then we say h is an extension of g. Look at the examples on pages 199 and 200.
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Theorem 4.2.5 Let h and g be functions with Dom(h) = A and Dom(g) = B. If A B = , then h g is a function with domain A B. Furthermore, (h g)(x) = h(x)if x A g(x)if x B Proof We know that h g is a relation. We must show that (i) Dom(h g) = A B, and (ii) if (x, y), (x, z) h g, then y = z We first show that Dom(h g) = A B Let x Dom(h g) y such that (x, y) h g __________________________ definition of Dom(h g) (x, y) h or (x, y) g __________________________ definition of h g x Dom(h) = A or x Dom(g) = B __________________________ definitions of Dom(h) & Dom(g) x A Bx A B __________________________ definition of A B Dom(h g) A B x Dom(h g) x A B
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We first show that Dom(h g) = A B Let x Dom(h g) y such that (x, y) h g __________________________ definition of Dom(h g) (x, y) h or (x, y) g __________________________ definition of h g x Dom(h) = A or x Dom(g) = B __________________________ definitions of Dom(h) & Dom(g) x A Bx A B __________________________ definition of A B Dom(h g) A B x Dom(h g) x A B Let x A B x A \/ x B x A = Dom(h) y such that (x, y) h __________________________ definition of A B _____________________ definition of Dom(h) x B = Dom(g) y such that (x, y) g (x, y) h g _____________________ definition of Dom(g) _____________________ (x, y) h or (x, y) g x Dom(h g) _____________________ definition of Dom(h g) A B Dom(h g) x A B x Dom(h g) Dom(h g) = A B Dom(h g) A B /\ A B Dom(h g)
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x A \/ x B (x A /\ x B) \/ (x B /\ x A) __________________________ definition of A B __________________________ supposition that A B = We now show that if (x, y), (x, z) h g, then y = z. Let (x, y), (x, z) h g. We must show that y = z. x Dom(h g) x A Bx A B __________________________ definition of Dom(h g) __________________________ A B = Dom(h g) was just proven x A /\ x B (x, y), (x, z) h __________________________ definition of Dom(h) (x, y), (x, z) h y = z __________________________ g is a function x B /\ x A (x, y), (x, z) g __________________________ definition of Dom(g) (x, y), (x, z) g y = z __________________________ h is a function In either case, we have y = z, which is what we wanted to show.
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1 (b) (d) Exercises 4.2 (pages 202-205)
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(f) (g)
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1 - continued (h) (j)
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2 (b) (d)
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2 - continued (f) (g)
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(h) (j)
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3 (b) (c)
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8 8
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(d) (e)
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