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Unformatted text preview: Solutions to Homework Assignment 3 Math 74, Fall 2006 October 1, 2006 1. (a) Proposition 1. Let f : X Y . Then f is surjective if and only if f ( f 1 [ B ] ) = B for every subset B Y . Proof. First, assume that f is surjective. Choose an arbitrary subset B Y . We will show that f ( f 1 [ B ]) = B . From Problem 7(b) on Homework Assignment 2, we have that f ( f 1 [ B ]) B . Hence it suffices to show that B f ( f 1 [ B ]). To that end, choose an arbitrary point y B . Since f is surjective, there exists x X such that y = f ( x ). Clearly x { z  f ( z ) B } = f 1 [ B ] . Thus y = f ( x ) f ( f 1 [ B ]). This demonstrates that B f ( f 1 [ B ]). Conversely, assume that f ( f 1 [ B ] ) = B for every subset B Y . Then f ( f 1 [ Y ] ) = Y. Since X = f 1 [ Y ], we see that Y = f ( X ), so f is surjective. (b) Proposition 2. Let f : X Y . Then f is injective if and only if f 1 [ f ( A )] = A for every subset A X . 1 Proof. Suppose that f is injective. We will show that f 1 [ f ( A )] = A for every subset A X . Choose an arbitrary subset A X . By Problem 7(a) on Homework Assignment 2, we have that A f 1 [ f ( A )]. So if suffices to show that f 1 [ f ( A )] A . To that end, choose a point x f 1 [ f ( A )]. Then f ( x ) f ( A ). Thus there exists x A such that f ( x ) = f ( x ). Since f is injective, x = x , and so x A . Thus we have demonstrated that f 1 [ f ( A )] A . Now suppose that f 1 [ f ( A )] = A for every subset A X . We will show that f is injective. Choose points x 1 ,x 2 X for which f ( x 1 ) = f ( x 2 ). Then { x 1 } = f 1 [ f ( { x 1 } )] = f 1 [ { f ( x 1 ) } ] . Since x 2 f 1 [ { f ( x 1 ) } ], it follows that x 1 = x 2 . 2. (a) This statement is false. Consider the map f : { 1 } { 1 , 2 } given by 1 7 2. Then f ( f 1 [ f ( )]) = = f ( ), and f ( f 1 [ f ( { 1 } )] ) = { 2 } = f ( { 1 } ) ....
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 Winter '06
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