a1sol - MATH 242 Assignment 1 Solutions 1. TRUE. We have y...

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Unformatted text preview: MATH 242 Assignment 1 Solutions 1. TRUE. We have y f ( A f- 1 ( B )) x A f- 1 ( B ) such that f ( x ) = y x A such that f ( x ) B and f ( x ) = y ( x A such that f ( x ) = y ) and y B y f ( A ) and y B y f ( A ) B 2. FALSE. You can start out by arguing as in question 1. y f ( A f- 1 ( B )) x A f- 1 ( B ) such that f ( x ) = y x A such that f ( x ) B or f ( x ) = y ( x A such that f ( x ) = y ) or ( x f- 1 ( B ) such that f ( x ) = y ) y f ( A ) or y f ( f- 1 ( B )) The argument shows that f ( A f- 1 ( B )) = f ( A ) f ( f- 1 ( B )) . It is clear that f ( f- 1 ( B )) B , but f- 1 ( B ) can be empty when B isnt. For a specific example take X = Y = { 1 , 2 } , f (1) = 1 , f (2) = 1 , A = , B = { 2 } . Then f- 1 ( B ) and A f- 1 ( B ) are both empty, and hence, so is f ( A f- 1 ( B )) , but f ( A ) B = B = { 2 } . So f ( A f- 1 ( B )) 6 = f ( A ) B in this case. 3. TRUE. Probably best to show containment in both directions. We will show that ( g f )( T ) g ( f ( T )) . Let z ( g f )( T ) , then there exists x T such that z = ( g f )( x ) = g ( f ( x )) . But f ( x ) f ( T ) and z = g ( f ( x )) g ( f ( T )) . To show that ( g f )( T ) g ( f ( T )) , let z g ( f ( T )) . Then there exists y f ( T ) such that z = g ( y ) . Further, there exists x T such that y = f ( x ) . Hence z = g ( f ( x )) = ( g f )( x ) and x T so that z ( g f )( T ) ....
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a1sol - MATH 242 Assignment 1 Solutions 1. TRUE. We have y...

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