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DEFINITION A function f : A B is onto B (called a surjection) onto iff Rng(f) = B. We write f : A B to indicate that f is a surjection. 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 we Proof: Suppose f : A B and g : B C. By Theorem ______ have g ◦ f : A C. Let c C. We need to show that a A such that (g ◦ f)(a) = c. __________________________ by supposition, g is onto C b B such that g(b) = 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:AC a A such that (g ◦ f)(a) = c 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 f (a) = b for some b B __________________________ by supposition, g ◦ f is onto C __________________________ f:AB (g ◦ f)(a) = g(f(a)) = g(b) and (g ◦ f)(a) = c g(b) = c two previous lines onto g:BC substitution b B such that g(b) = c DEFINITION A function f : A B is one-to-one (called an 1-1 injection) iff whenever f(x) = f(y), then x = y. We write f : A B to indicate that f is an injection. 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) __________________________ by supposition f is 1-1 x=y 1-1 g◦f:AC (g ◦ f)(x) = (g ◦ f)(y) x = y 1-1 Theorem 4.3.4 If f : A B, g : B C, and g ◦ f : A C, then 1-1 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 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:AB f (x) = f (y) x = y Theorem 4.3.5 1-1 (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. onto onto 1-1 1-1 onto (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 1-1 hg:ABCD. 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 _____________, one-to-one we have x = y. Proof of (b): onto onto Suppose h : A C , g : B D , and A B = . From Theorem ____________ 4.2.5 hg:ABCD 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 Proof of (b): onto onto Suppose h : A C , g : B D , and A B = . From Theorem ____________ 4.2.5 hg:ABCD 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 (h g)(x) = h(x) = y Case 2: y D by supposition h is onto C _______________________ 4.2.5 A B = and Theorem _______ x B such that g(x) = y (h g)(x) = g(x) = y by supposition h is onto C _______________________ 4.2.5 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 4.3.5 1-1 (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. onto onto 1-1 1-1 onto (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 1-1 hg:ABCD. Proof of (c): 1-1 1-1 Suppose h : A C , g : B D , A B = , and C D = . Theorem ____________ 4.2.5 hg:ABCD 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 Theorem ____________ 4.2.5 (h g)(x) = h(x) and (h g)(y) = h(y) 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 Theorem ____________ 4.2.5 (h g)(x) = g(x) and (h g)(y) = g(y) 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 Theorem ____________ 4.2.5 (h g)(x) = h(x) and (h g)(y) = g(y) by supposition (h g)(x) = (h g)(y) h(x) = g(y) __________________________ h(x) C and g(y) D by supposition h : A C and g : 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). Exercises 4.3 (pages 210-213) 1 (b) (d) (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 1-1 (a) Suppose f : A B and C A. Then f |C is one-to-one. onto onto 1-1 1-1 onto (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 1-1 hg:ABCD.