lect12 - Todays topics Properties of the composite...

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1 Today’s topics: Properties of the composite functions. Inverse functions. Proofs with functions.
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2 Theorem 5 . Consider functions f : A B and g : B C , and the composition g f : A C. a) If both f and g are injective, then g f is injective. b) If both f and g are surjective, then g f is surjective. c) If both f and g are bijective, then g f is bijective. Proof. a) Let a 1 and a 2 be arbitrary elements of A such that ( g f )( a 1 ) = ( g f )( a 2 ). By the Theorem 4, it implies that g ( f ( a 1 )) = g ( f ( a 2 )). Since g is injective it implies that f ( a 1 ) = f ( a 2 ), and since f is injective by assumption, a 1 = a 2 . a 1 a 2 g ( f ( a 1 ))= g ( f ( a 2 )). f g f ( a 1 ) f ( a 2 ) A B C
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3 b) Take an arbitrary element c C. Since g is surjective, there exists b B such that g ( b )= c . Similarly, since f is surjective,there exists a A such that f ( a )= b . Then ( g f )( a ) = g ( f ( a )) = g ( b ) = c . So, g f is surjective. g surjective: 2200 c, 5 b ( g ( b )= c ) f surjective: 2200 b, 5 a ( f ( a )= b ) C c g B b f A a g f
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4 c) If both f and g are bijective, then both f and g are injective and surjective. By parts a) and b) g f is injective and surjective, so g f is bijective. So, injective (surjective) properties of two functions, f and g are sufficient condition for injective (surjective) property of the composite function. The question arises whether both conditions are necessary as well.
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5 Example. Let f : R R, and g : R R denote two functions, where R is the set of real numbers. Suppose g f ( x )=2 x +1 for all x R ( i. e. the composite function is both surjective and injective). a) Prove that the function f is injection. Let f ( x )= f ( y ), we need to show that x = y. Since g is a function on a set of real numbers, f ( x )= f ( y ) implies g ( f ( x ))= g ( f ( y )), i. e. 2 x +1 = 2 y +1 (by assumption g f ( x )=2 x +1). So, it yields x = y . QED. b) Prove the function g is a surjection. Take arbitrary z R, we need to show that 5 y R such that g ( y )= z . But for any z R there always exist x, y R, x = ( z - 1)/2 and y=f ( x ), so that g ( y ) = g ( f ( x )) = 2 ( z - 1)/2+1= z
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6 Example . Let f : A B and g : B C be two functions such that the composite function g f : A C is injective. Does it follow that
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lect12 - Todays topics Properties of the composite...

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