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# sole4 - ANALYSIS EXERCISE 4SOLUTIONS 1(a Let x1 x2 X be...

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ANALYSIS EXERCISE 4–SOLUTIONS 1. (a) Let x 1 , x 2 X be arbitrary. Suppose that g ( f ( x 1 )) = g ( f ( x 2 )). Then f ( x 1 ) = f ( x 2 ) since g is injective. Therefore x 1 = x 2 since f is injective. This proves that g f is injective. (b) Let z Z be arbitrary. Then y Y such that g ( y ) = z since g is surjective. Also x X such that f ( x ) = y since f is surjective. Therefore we have proved ( z Z )( x X )[ g ( f ( x )) = z ] . So g f is surjective. 2. (a) Let z Z . Then ( x X )[( g f )( x ) = z ] as g f is surjective. So g ( y ) = z where Y 3 y = f ( x ), thus g : Y Z is surjective. (a) Let y Y . Put z := g ( y ) Z . As g f is surjective, ( x X )[( g f )( x ) = z ]. But then z = g ( y ) = ( g f )( x ) = g ( f ( x )) . As g : Y Z is injective, it follows that y = f ( x ). In other words, ( x X )[ f ( x ) = y ] and hence f is surjective. 3. The composition g f : R - {- 1 , 1 } → R is defined by ( g f )( x ) = 1 f ( x ) = x - 1 x + 1 . 4. (a) f is injective since ( x 1 +1 x 1 - 1 = x 2 +1 x 2 - 1 ) ( x 1 = x 2 ). Let y X be arbitrary. We have to show that the equation x +1 x - 1 = y has a solution x X . Direct computation show that

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