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Exploring Function Composition Theorems

Dive into the theorems on function composition, inverses, and restrictions. Learn the proofs behind these fundamental concepts in mathematics.

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Exploring Function Composition Theorems

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  1. DEFINITIONS For a function F : A B, theinverse 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  Care 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.

  2. 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. We first show that Dom(G ◦ F) = A Dom(G ◦ F)  Dom(F) = A Exercise 3.1-9(a) ___________ We now want to show that A  Dom(G ◦ F) Let a A = Dom(F) _____________________ definition of Dom(F) (a, b)  F for some b B b B = Dom(G) _____________________ (b, c)  G for some c C _____________________ definition of G ◦ F (a, c)  G ◦ F definition of Dom(G ◦ F) _____________________ a  Dom(G ◦F) a A  a Dom(G ◦F) A  Dom(G ◦F) Dom(G ◦ F) A and A  Dom(G ◦ F) Dom(G ◦ F) = A

  3. To prove (ii), let (x, y)  G ◦ F and (x, z)  G ◦ F; we must show y = z. (x, y)  G ◦ F _____________________ (x, u)  F /\ (u, y)  G for some u  B (x, z)  G ◦ F _____________________ (x, v)  F /\ (v, z)  G for some v  B u = v _____________________ F is a function y = z _____________________ u = v and G is a function G◦F is a function (x, y)  G ◦ F /\ (x, z)  G ◦ F  y = z

  4. 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  Dare 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) = A Theorem 4.2.1 _____________ Dom(h ◦ (g ◦ f)) = Dom(g ◦ f) = Dom(f) = A Theorem 4.2.1 Now, suppose xA ((h ◦ g) ◦ f)(x) = (h ◦ g)(f(x)) = previously defined notation h(g(f(x))) = h((g◦f)(x)) = (h◦ (g ◦ f))(x) (h ◦ g) ◦ f = h ◦ (g ◦ f) _____________ Theorem 4.1.1

  5. Theorem 4.2.3 Let f : A  B. Then f◦IA = f and IB◦ f = f. Proof Dom(f ◦ IA) = Dom(IA) _____________ Theorem 4.2.1 Dom(f ◦ IA) = A substitution using Dom(IA) = A Dom(f ◦ IA) = Dom(f) ______________ supposition that f : A  B Now, suppose xA. (f ◦ IA)(x) = f(IA(x)) = f(x) previously defined notation f ◦ IA = f the two conditions of Theorem _____ are satisfied 4.1.1

  6. Theorem 4.2.3 Let f : A  B. Then f◦IA = f and IB◦ f = f. Dom(IB ◦ f) = Dom(f) _____________ Theorem 4.2.1 Now, suppose xA. (IB ◦ f)(x) = IB(f(x)) = f(x) previously defined notation IB ◦ f = f the two conditions of Theorem _____ are satisfied 4.1.1

  7. Theorem 4.2.4 Let f : A  B with Rng(f) = C. If f–1 is a function, then f –1 ◦ f = IA and f ◦ f –1 = IC. Proof: Suppose f : A  B with Rng(f) = C, and f –1 is a function _____________ Dom(f–1 ◦ f) = Dom(f) Theorem 4.2.1 Dom(f–1 ◦ f) = A substitution using Dom(f) = A Dom(f–1 ◦ f) = Dom(IA) ____________________________ Dom(IA) = 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) = IA(x) IA(x) = x f –1 ◦ f = IA the two conditions of Theorem _____ are satisfied 4.1.1

  8. Theorem 4.2.4 Let f : A  B with Rng(f) = C. If f–1 is a function, then f –1 ◦ f = IA and f ◦ f –1 = IC. Proof: Suppose f : A  B with Rng(f) = C, and f –1 is a function ______________ Dom(f ◦ f–1) = Dom(f–1) Theorem 4.2.1 ______________ Dom(f ◦ f–1) = Rng(f) Theorem 3.1.2(a) ______________ Dom(f ◦ f–1) = C supposition that Rng(f) = C Dom(f–1 ◦ f) = Dom(IC) ____________________________ Dom(IC) = 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) = IC(x) IC(x) = x f ◦ f –1 = IC the two conditions of Theorem _____ are satisfied 4.1.1

  9. DEFINITIONS Let f : A B, and let D  A. Therestriction 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.

  10. 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  gis 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  B __________________________ definition of A  B xDom(h  g)  x AB Dom(h  g) AB

  11. 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  B __________________________ definition of A  B xDom(h  g)  x AB Dom(h  g) AB Let xAB x  A \/ x B __________________________ definition of AB x  A = Dom(h)  y such that (x, y)  h _____________________ definition of Dom(h) x  B = Dom(g) y such that (x, y)  g _____________________ definition of Dom(g) (x, y)  h  g _____________________ (x, y)  h or (x, y)  g x Dom(h  g) definition of Dom(h  g) _____________________ xAB  xDom(h  g) AB Dom(h  g) Dom(h  g) AB/\ AB Dom(h  g) Dom(h  g) =AB

  12. 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. __________________________ definition of Dom(h  g) x Dom(h  g) __________________________ AB = Dom(h  g) was just proven x AB x  A \/ x B __________________________ definition of AB (x  A /\ x B) \/ (x B /\ x A) __________________________ supposition that AB =  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 havey = z, which is what we wanted to show.

  13. Exercises 4.2 (pages 202-205) 1  (b)  (d)

  14.  (f)  (g)

  15. 1 - continued  (h)  (j)

  16. 2  (b)  (d)

  17. 2 - continued  (f)  (g)

  18.  (h)  (j)

  19. 3  (b)  (c)

  20. 8 

  21. 14  (b)  (c)

  22.  (d)  (e)

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