Adv Alegbra HW Solutions 41

Adv Alegbra HW Solutions 41 - Solution Let g Y → X and h...

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41 (ii) If f : R R is a bijection whose graph consists of certain points ( a , b ) [of course, b = f ( a ) ], prove that the graph of f 1 is { ( b , a ) : ( a , b ) f } . Solution. By de f nition, f 1 ( b ) = a if and only if b = f ( a ) . Hence, the graph of f 1 consists of all ordered pairs ( b , f 1 ( b )) = ( b , a ) = ( f ( a ), a ). 2.7 Let X and Y be sets, and let f : X Y be a function. (i) If S is a subset of X , prove that the restriction f | S is equal to the composite f i , where i : S X is the inclusion map. Solution. Both f | S and f i have domain S and target Y .If s S , then ( f i )( s ) = f ( s ) = ( f | S )( s ) . Therefore, f | S = f i ,by Proposition 2.2. (ii) If im f = A Y , prove that there exists a surjection f 0 : X A with f = j f 0 , where j : A Y is the inclusion. Solution. For each x X ,de f ne f 0 ( x ) = f ( x ) . Thus, f 0 differs from f only in its target. 2.8 If f : X Y has an inverse g , show that g is a bijection. Solution. We are told that f g = 1 Y and g f = 1 X . Therefore, g has an inverse, namely, f , and so g is a bijection. 2.9 Show that if f : X Y is a bijection, then it has exactly one inverse.
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Unformatted text preview: Solution. Let g : Y → X and h : Y → X both be inverses of f . Then h = h 1 Y = h ( f g ) = ( h f ) g = 1 X g = g . 2.10 Show that f : R → R , de f ned by f ( x ) = 3 x + 5, is a bijection, and f nd its inverse. Solution. The function g , de f ned by g ( x ) = 1 3 ( x − 5 ) , is the inverse of f , and so f is a bijection. (Alternatively, one could prove that f is a bijection by showing directly that it is injective and surjective.) 2.11 Determine whether f : Q × Q → Q , given by f ( a / b , c / d ) = ( a + c )/( b + d ) is a function. Solution. f is not a function: 1 2 = 2 4 and 2 6 = 1 3 , but f ( 1 2 , 2 6 ) = 3 8 ±= 3 7 = f ( 2 4 , 1 3 )....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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