HW11solns

# Mathematical Thinking: Problem-Solving and Proofs (2nd Edition)

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4.33 (a) Let f : A B, g : B C be injections. Suppose ( g f )( x ) = ( g f )( y ). This means g ( f ( x )) = g ( f ( y )). Since g is injective, this implies that f ( x ) = f ( y ). But f is also injective, so x = y . It follows that g f is injective. (b) Let f : A B, g : B C be surjections and let c C . Since g is surjective, there is a b B such that g ( b ) = c . As f is also surjective, there is an a A with f ( a ) = b . Then ( g f )( a ) = g ( f ( a )) = g ( b ) = c . Hence g f is surjective. (c) This follows immediately from (a) and (b). (d) By (c), g f is bijective. Thus it is enough to show that ( f - 1 g - 1 ) ( g f ) = id A . We have ( f - 1 g - 1 ) ( g f ) = f - 1 (( g - 1 ) ( g f )) = f - 1 (( g - 1 g ) f )) = f - 1 (( id B ) f )) = f - 1 f = id A . The result follows. 4.34 (a) True. Use proof by contrapositive. (b) False. One counterexample is given by deﬁning f : R + R by f ( x ) = x and g : R + R by g ( x ) = x 2 . (c) False. One counterexample is given by deﬁning
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Unformatted text preview: f : { } → { , 1 } by f ( x ) = 0 and g : { , 1 } → { } by g ( x ) = 0. (d) True. Use proof by contrapositive. 4.36 (a) If f ◦ g = id B , then for every b ∈ B , f ( g ( b )) = ( f ◦ g )( b ) = id B ( b ) = b which shows that f is surjective. (b) If f ( x ) = f ( y ), then x = id A ( x ) = ( g ◦ f )( x ) = g ( f ( x )) = g ( f ( y )) = ( g ◦ f )( y ) = id A ( y ) = y, so f is injective. 4.47 The maps n 7→ 2 n and n 7→ 2 n + 1 are easily veriﬁed to be bijections from set of natural numbers to the set of even numbers and the set of odd numbers, respectively. 1...
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## This document was uploaded on 03/25/2011.

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