DEFINITION A function f : A  B is onto B (called a surjection) iff Rng(f) = B. We write f : A  B to indicate that f is a surjection. Look at the illustration on page 205 and the examples on pages 206 and 207. Theorem If f : A  B and g : B  C, then g ◦ f : A  C. That is, the composite of surjective functions is a surjection. Proof: Suppose f : A  B and g : B  C. By Theorem ______ we have g ◦ f : A  C. Let c  C. We need to show that  a  A such that (g ◦ f)(a) = c. onto To show that f is onto B, we can show that for any b  B, there must be some a  A such that f(a) = b.  b  B such that g(b) = c by supposition, g is onto C __________________________  a  A such that f(a) = b by supposition, f is onto B __________________________ (g ◦ f)(a) = g(f(a)) = g(b) = ctwo previous lines g ◦ f : A  C  a  A such that (g ◦ f)(a) = c onto

Theorem If f : A  B, g : B  C, and g ◦ f : A  C, then g is onto C. That is, when the composite of two functions maps onto a set C, then the second function applied must map onto the set C. onto Proof: Suppose f : A  B, g : B  C, and g ◦ f : A  C. Let c  C. To show g is onto C, we must find b  B such that g(b) = c. onto  a  A such that (g ◦ f)(a) = c by supposition, g ◦ f is onto C __________________________ f (a) = b for some b  Bf : A  B __________________________ (g ◦ f)(a) = g(f(a)) = g(b) and (g ◦ f)(a) = c two previous lines g(b) = csubstitution g : B  C  b  B such that g(b) = c onto

DEFINITION A function f : A  B is one-to-one (called an injection) iff whenever f(x) = f(y), then x = y. We write f : A  B to indicate that f is an injection. 1-1 Look at the examples on pages 208 and 209. To show that f is one-to-one, we show that for any x, y  A for which f(x) = f(y), we must have x = y. Theorem If f : A  B and g : B  C, then g ◦ f : A  C. That is, the composite of injective functions is an injection. 1-1 Proof: Suppose f : A  B and g : B  C. To show g ◦ f is one-to-one, suppose (g ◦ f)(x) = (g ◦ f)(y). 1-1 g(f(x)) = g(f(y))change of notation f(x) = f(y) by supposition g is 1-1 __________________________ x = yby supposition f is 1-1 __________________________ g ◦ f : A  C 1-1 (g ◦ f)(x) = (g ◦ f)(y)  x = y

Theorem If f : A  B, g : B  C, and g ◦ f : A  C, then f : A  B. That is, if the composite of two functions is one-to-one, then the first function applied must be one-to-one. Proof: Suppose f : A  B, g : B  C, and g ◦ f : A  C. To show f is one-to-one, suppose f(x) = f(y) for some x, y  A. 1-1 g(f(x)) = g(f(y)) by supposition g is a function __________________________ (g ◦ f)(x) = (g ◦ f)(y)change of notation x = y by supposition g ◦ f is 1-1 __________________________ f : A  B 1-1 f (x) = f (y)  x = y

Theorem (a) Suppose f : A  B and C  A. Then f | C is one-to-one. That is, a restriction of a one-to-one function is one-to-one. (b) If h : A  C, g : B  D, and A  B = , then h  g : A  B  C  D. (c) If h : A  C, g : B  D, A  B = , and C  D = , then h  g : A  B  C  D. Suppose f : A  B and C  A. Let f | C (x) = f | C (y) for x, y  C. onto 1-1 Then f(x) = f(y).Since f is _____________, we have x = y. Proof of (a): 1-1 one-to-one Suppose h : A  C, g : B  D, and A  B = . onto Proof of (b): h  g : A  B  C  Dh  g : A  B  C  D From Theorem ____________4.2.5 Let y  C  D We want to show  x  A  B such that (h  g)(x) = y. y  C \/ y  D ________________________________ definition of C  D

Suppose h : A  C, g : B  D, and A  B = . onto Proof of (b): h  g : A  B  C  Dh  g : A  B  C  D From Theorem ____________4.2.5 Let y  C  D We want to show  x  A  B such that (h  g)(x) = y. y  C \/ y  D _______________________ definition of C  D  x  A such that h(x) = y _______________________ by supposition h is onto C Case 1: y  C (h  g)(x) = h(x) = y _______ A  B =  and Theorem  x  B such that g(x) = y _______________________ by supposition h is onto C Case 2: y  D (h  g)(x) = g(x) = y _______ A  B =  and Theorem In either case, we have (h  g)(x) = y for some x  A  B. We have shown that in each case (h  g)(x) = y for some x  A  B, and thus proven part (b).

Theorem (a) Suppose f : A  B and C  A. Then f | C is one-to-one. That is, a restriction of a one-to-one function is one-to-one. (b) If h : A  C, g : B  D, and A  B = , then h  g : A  B  C  D. (c) If h : A  C, g : B  D, A  B = , and C  D = , then h  g : A  B  C  D. onto 1-1 Proof of (c): Suppose h : A  C, g : B  D, A  B = , and C  D = . 1-1 h  g : A  B  C  Dh  g : A  B  C  D Theorem ____________4.2.5 Suppose (h  g)(x) = (h  g)(y) where x, y  A  B. We want to show that One of the following cases must be true: (i) x, y  A, (ii) x, y  B,(iii) x  A and y  B,(iv) x  B and y  A.

Case (i): x, y  A (h  g)(x) = h(x) and (h  g)(y) = h(y) Theorem ____________4.2.5 h(x) = h(y) by supposition h is 1-1 __________________________ x = y Case (ii): x, y  B (h  g)(x) = g(x) and (h  g)(y) = g(y) Theorem ____________4.2.5 g(x) = g(y) by supposition g is 1-1_________________________ x = y Case (iii): x  A and y  B (h  g)(x) = h(x) and (h  g)(y) = g(y) Theorem ____________4.2.5 by supposition (h  g)(x) = (h  g)(y) __________________________ by supposition (h  g)(x) = (h  g)(y) __________________________ h(x) = g(y) by supposition (h  g)(x) = (h  g)(y) __________________________ h(x)  C and g(y)  D by supposition h : A  C and g : B  D __________________________ This is a contradiction__________________________ by supposition C  D =  This case is not possible; similarly, Case (iv) is not possible. We have shown that in each possible case x = y, and thus proven part (c).

1  (b)  (d) Exercises 4.3 (pages )

 (e)  (f)

1 - continued  (g)  (h)

 (i)  (j)

1 - continued  (l)

2  (b)  (d)

2 - continued  (e)  (f)

 (g)  (h)

2 - continued  (i)  (j)

 (l)

9  (a)  (b)

 (d)

Theorem 4.15 (a) Suppose f : A  B and C  A. Then f | C is one-to-one. (b) If h : A  C, g : B  D, and A  B = , then h  g : A  B  C  D. (c) If h : A  C, g : B  D, A  B = , and C  D = , then h  g : A  B  C  D. 1-1 onto 1-1