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# compositionkey - f is onto Let T ∈ P B and let S = x ∈...

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McCombs Math 381 Function Composition 1. Suppose g : A B and f : B C where A = { 1 , 2 , 3 , 4 } , B = { a , b , c } , C = { 2 , 7 , 10 } , and f and g are defined by g = { (1 , b ) , (2 , a ) , (3 , a ) , (4 , b ) } and f = { ( a , 10) , ( b , 7) , ( c , 2) } . Find f g . Ans: { (1 , 7) , (2 , 10) , (3 , 10) , (4 , 7) } . In the questions below suppose g : A B and f : B C where A = B = C = { 1 , 2 , 3 , 4 } , g = { (1 , 4) , (2 , 1) , (3 , 1) , (4 , 2) } and f = { (1 , 3) , (2 , 2) , (3 , 4) , (4 , 2) } . 2. Find f g . Ans: { (1 , 2) , (2 , 3) , (3 , 3) , (4 , 2) } . 3. Find g f . Ans: { (1 , 1) , (2 , 1) , (3 , 2) , (4 , 1) } . 4. g g . Ans: { (1 , 2) , (2 , 4) , (3 , 4) , (4 , 1) } . 5. For any function f : A B , define a new function g : P ( A ) P ( B ) as follows: for every S A , g ( S ) = { f ( x ) | x S } . Prove that f is onto if and only if g is onto. Ans: Suppose
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Unformatted text preview: f is onto. Let T ∈ P ( B ) and let S = { x ∈ A | f ( x ) ∈ T } . Then g ( S ) = T , and g is onto. If f is not onto B , let y ∈ B − f ( A ). Then there is no subset S of A such that g ( S ) = { y } . 6. Suppose g : A → B and f : B → C , where f g is 1-1 and g is 1-1. Must f be 1-1? Ans: No. 7. Suppose g : A → B and f : B → C , where f g is 1-1 and f is 1-1. Must g be 1-1? Ans: Yes....
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