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.

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.

_____________________

_____________________

_____________________

_____________________

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 = 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

_____________________

_____________________

_____________________

_____________________

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) = 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)) =

h(g(f(x))) = h((g ◦ f)(x)) = (h ◦ (g ◦ f))(x)previously defined notation

(h ◦ g) ◦ f = h ◦ (g ◦ f) Theorem 4.1.1_____________

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_____________

substitution using Dom(IA) = ADom(f ◦ IA) = A

Dom(f ◦ IA) = Dom(f) ______________ f : A Bsupposition that

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

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

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 functionDom(f –1 ◦ f) = Dom(f) Theorem 4.2.1_____________

substitution using Dom(f) = ADom(f –1 ◦ 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

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 functionDom(f ◦ f –1) = Dom(f –1) Theorem 4.2.1______________

TheoremDom(f ◦ f –1) = Rng(f) ______________ 3.1.2(a)supposition thatDom(f ◦ f –1) = C ______________ Rng(f) = C

Dom(f –1 ◦ f) = Dom(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

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.

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 Ag(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 B __________________________ definition of A B

Dom(h g) A B x Dom(h g) x A B

We first show that Dom(h g) = A BLet 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

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)

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 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 functionx 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.

1 (b)

(d)

Exercises 4.2 (pages 202-205)

(f)

(g)

1 - continued

(h)

(j)

2 (b)

(d)

2 - continued

(f)

(g)

(h)

(j)

3 (b)

(c)

8

14 (b)

(c)

(d)

(e)