Look at the illustration on page 205 and the examples on pages 206 and 207.

1. DEFINITION A function f : A B isonto B (called a surjection) iff Rng(f) = B. We write f : A B to indicate that f is a surjection. 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. Look at the illustration on page 205 and the examples on pages 206 and 207. Theorem 4.3.1 If f : A  B and g : B  C, then g◦f : A  C. That is, the composite of surjective functions is a surjection. onto onto onto onto onto 4.2.1 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. __________________________  b  B such that g(b) = c by supposition, g is onto C by supposition, f is onto B __________________________  a  A such that f(a) = b (g◦f)(a) = g(f(a)) = g(b) = c two previous lines onto g◦f : A  C  a  A such that (g◦f)(a) = c

2. onto Theorem 4.3.2 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. __________________________  a  A such that (g ◦ f)(a) = c by supposition, g ◦ f is onto C __________________________ f (a) = b for some b  B f : A  B (g◦f)(a) = g(f(a)) = g(b) and (g ◦ f)(a) = c two previous lines g(b) = c substitution onto g : B  C  b  B such that g(b) = c

3. DEFINITION A function f : A B isone-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 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. Look at the examples on pages 208 and 209. 1-1 1-1 1-1 Theorem 4.3.3 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 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). g(f(x)) = g(f(y)) change of notation by supposition g is 1-1 __________________________ f(x) = f(y) __________________________ x = y by supposition f is 1-1 1-1 g◦f : AC (g ◦ f)(x) = (g ◦ f)(y)  x = y

4. 1-1 Theorem 4.3.4 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. 1-1 1-1 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. by supposition g is a function __________________________ g(f(x)) = g(f(y)) (g◦f)(x) = (g◦f)(y) change of notation by supposition g◦f is 1-1 __________________________ x = y 1-1 f : AB f (x) = f (y)  x = y

5. Theorem 4.3.5 (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. 1-1 onto onto onto 1-1 1-1 1-1 Proof of (a): 1-1 Suppose f : A  B and C  A. Let f |C(x) = f |C(y) for x, y C. Then f(x) = f(y). Since f is _____________, we have x = y. one-to-one Proof of (b): onto onto Suppose h:AC, g:BD, and AB = . h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

6. Proof of (b): onto onto Suppose h:AC, g:BD, and AB = . h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 Case 1: y C  x A such that h(x) = y by supposition h is onto C _______________________ (hg)(x) = h(x) = y AB =  and Theorem 4.2.5 _______ Case 2: y D  x B such that g(x) = y by supposition h is onto C _______________________ (hg)(x) = g(x) = y AB =  and Theorem 4.2.5 _______ 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).

7. Theorem 4.3.5 (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. 1-1 onto onto onto 1-1 1-1 1-1 Proof of (c): 1-1 1-1 Suppose h:AC, g:BD, AB = , and CD = . h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.

8. 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g)(x) = (hg)(y) __________________________ x = y by supposition h is 1-1 __________________________ Case (ii): x, y B (hg)(x) = g(x) and (hg)(y) = g(y) Theorem ____________ 4.2.5 by supposition (hg)(x) = (hg)(y) g(x) = g(y) __________________________ x = y _________________________ by supposition g is 1-1 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) h(x) = g(y) __________________________ h(x)  C and g(y)  D by supposition h:ACandg:BD __________________________ by supposition CD =  This is a contradiction __________________________ 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).

9. Exercises 4.3 (pages 210-213) 1  (b)  (d)

10.  (e)  (f)

11. 1 - continued  (g)  (h)

12.  (i)  (j)

13. 1 - continued  (l)

14. 2  (b)  (d)

15. 2 - continued  (e)  (f)

16.  (g)  (h)

17. 2 - continued  (i)  (j)

18.  (l)

19. 9  (a)  (b)

20.  (d)

24. 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 onto onto 1-1 1-1 1-1