Today s topics Proof that a given function is injective or surjective Composite functions Properties of the composite functions Inverse functions Proofs with functions 1 Let A R 1 and f A A be defined by the formula x 1 f x x 1 Prove that f is injective and surjective 1 injective Let x and y be arbitrary elements of A such that f x f y 1 We must show that 1 implies x y x 1 y 1 x 1 y 1 x 1 y 1 y 1 x 1 xy y x 1 yx x y 1 by assumption since x 1 0 and y 1 0 by algebra y x x y x y 2 2 surjective for any y there exists x A such that f x y Take arbitrary y A and find x A from f x y x 1 y x 1 y x 1 x 1 yx y x 1 x y 1 y 1 y 1 x A y 1 3 Composition of functions B A a f C g f a b g b c g f Note that in the context of function composition the order in which the functions appear is backward i e the rightmost function is applied first Theorem Let f A B and g B C be two functions The composition of f and g as relations defines a function g f A C such that g f a g f a 4 Proof We need to prove that composite relation g f is a function For this we must prove two things 1 the composition g f relates each element a A to some c C 2 an element c C assigned to a by g f is unique so we can denote it c g f a 1 existence Let b f a B Let c g b C So by the definition of composition of relations a c g f Thus c C a c g f B A a f C g f a b g f g b c 5 2 uniqueness Suppose a c1 g f and a c2 g f Then by the definition of composition b1 B such that a b1 f and b1 c1 g and b2 B such that a b2 f and b2 c2 g Since f is a function there may be only one b B such that a b f so b1 b2 Since g is a function b c1 g and b c2 g imply that c1 c2 b1 g f a c1 b2 c2 Thus g f a c g b g f a 6 Properties of the composite function How the injective surjective property of g f depends on properties of f and g g f f g f g f g 7 Theorem 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 a1 and a2 be arbitrary elements of A such that g f a1 g f a2 By the previous Theorem it implies that g f a1 g f a2 Since g is injective it implies that f a1 f a2 and since f is injective by assumption B a1 a2 g C f Aa 1 f a1 g f a1 g f a2 a2 f a2 8 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 C B A g f c a b g f f surjective b a f a b g surjective c b g b c 9 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 Let f A B and g B C be two functions and the composition g f A C is injective What properties of of f and g can be implied What properties of f and g can be implied if g f is surjective 10 Example Let f A B and g B C be two functions such that the composite function g f A C is injective Prove or disprove that both f and g are injective Injective property of g f A C implies if g f x g f y then x y Let s prove by contradiction that f must be injective For this assume that g f is injective but f is not injective Then we can find x y A such that x y and f x f y By taking composite function for these x and y we have that g f x g f y while x y But this results to contradiction with injective property of composite function g f x x g g f x g f y x y f g f y 11 y g f injective f injective What about g f injective g injective This implication can be disproved by the following counterexample g f g is not injective but g f is If f was surjective then we would not be able to disprove that g f injective g injective 12 Example f 0 0 f x x 2 g 0 0 1 g y sin y not injective g y sin y 0 0 1 g f sin x 2 0 0 1 g f sin x 2 0 0 1 is injective but g is not 13 Example Let f A B and g B C be two functions such that the composite function g f A C is surjective Prove or disprove that both f and g are surjective Let s prove that if g f is surjective then g is surjective Surjective property of g f implies that for any z C there exists x A such that g f x z It means that g f x z Since f is a function there exists a unique element y B such that y f x But g f x g y z g f x z y f x Thus for any z C there exists y B such that g y z Than means that g is surjective 14 The following example shows that g f is surjective f is surjective g f If we knew that g were injective we could prove that g f is surjective f is surjective 15 Example Let f R R and g R R denote two functions where R is the set …
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