1. 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.

2. 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. _____________________

3. 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 _____________________

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  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 _____________

5. 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

6. 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

7. 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

8. 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

9. 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.

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  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

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  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)

12. 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.

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

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 8 

21. 14  (b)  (c)

22.  (d)  (e)