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Unformatted text preview: MATH 74 HOMEWORK 8 (DUE WEDNESDAY OCTOBER 31) 1(a). Let z Z be given. Since g : Y Z is surjective, there is y Y with g ( y ) = z . Since f : X Y is surjective, there is x X with f ( x ) = y . We conclude that ( g f )( x ) = g ( f ( x )) = g ( y ) = z, showing that z Im( g f ). Since z was arbitrary, g f is surjective. 1(b). Yes, g is necessarily surjective. To see this, let z Z be given. As g f : X Z is surjective, there is x X with ( g f )( x ) = z . This means that g ( f ( x )) = z , so we conclude that z Im g . Since z was arbitrary, g is surjective. 1(c). No, f is not necessarily surjective. Let X = { 1 } and Y = { 1 , 2 } and Z = { 1 } , and let f : X Y be the function given by f (1) = 1, and let g : Y Z be the function given by g (1) = 1 and g (2) = 1. Then g f : { 1 } { 1 } is the function sending 1 to 1, hence g f is surjective. But f is not surjective as 2 6 Im( f ). 2(a). Suppose x X and y X satisfy ( g f )( x ) = ( f g )( y ). This means (by definition of ) that f ( g ( x )) = f ( g ( y )). Since f is injective, we conclude that g ( x ) = g ( y ). Since g is injective, we conclude that x = y . As x and y were arbitrary we conclude that g f is injective....
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This note was uploaded on 02/10/2010 for the course MATH 74 taught by Professor Courtney during the Fall '07 term at University of California, Berkeley.
 Fall '07
 COURTNEY
 Math

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