Unformatted text preview: is the inclusion, then there is a surjection g : X → im f with f = j ◦ g . Solution. True. (vi) If f : X → Y is a function for which there is a function g : Y → X with f ◦ g = 1 Y , then f is a bijection. Solution. False. (vii) The formula f ( a b ) = ( a + b )( a − b ) is a wellde f ned function Q → Z . Solution. False. (viii) If f : N → N is given by f ( n ) = n + 1 and g : N → N is given by g ( n ) = n 2 , then the composite g ◦ f is n 7→ n 2 ( n + 1 ) . Solution. False. (ix) Complex conjugation z = a + ib 7→ z = a − ib is a bijection C → C . Solution. True....
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 Fall '11
 KeithCornell
 Gregorian calendar, Solution., Bijection, October Revolution

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