or just I if the set A is clear. Hence we have for all x in A the identity function I ( )

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1 3 Functions 46 SECTON E Properties of Composite Functions the end of this section ou will be able to understand what is meant b the identit function prove properties of inverse function prove properties of composition of functions E Properties of the dentit Function What do ou think the term identit means? The word identit in everda language means being the same. What does identit function mean? The function which has the same input and output. The definition of the identit function is: Definition (3.2. Let be an set. The function f : given b f ( for all in is called the identit function on the set. The identit function on a set is normall denoted b or just if the set is clear. Hence we have for all in the identit function : satisfing ( The identit function transforms back to. This can be illustrated b f Fig is in then what is 2 2. f equal to? + is in then ( + +. What is ( provided is in set equal to? The identit function transforms back to itself. The input is same as the output. The identit function has man important properties that we use. n this section we state and prove some of these properties. Proposition (3.3. Let f be an function, f :, then f f f Let be an arbitrar element in which means it is in the domain of f because we are given f :. s it in the domain of f? f f ( Fig 32 f

2 3 Functions 47 Yes because : and so f :. We have ( f ( f ( ( [ Definition of ] ecause f f f therefore f f. Since Net we show f f. s the element in the domain of f? f f f ( Fig 33 Yes because f : and therefore f :. We have ( f ( f ( [ Definition of ] ecause f f f f f therefore f f. Since Remember both and are ident ns not change b appling these functions. Hence we have our required result, f f f. it functio so their arguments and f do Prior to this section we have used lower case letters such as f, g and h to represent functions but for the identit function we use upper case to represent this function. E2 Properties of the nverse Function Proposition (3.4. Let f : be a bijective function. Then f f and f f where is the identit function on and is the identit function on. We prove f f and leave the proof of 3(e. Since f is bijective therefore the inverse function f f as a question in Eercise f eists. Wh? ecause from Section C we have (3.5 f has an inverse f is bijective Let be an arbitrar element in the set and f (. Therefore is in because f :. f f f f f ( f [ Definition of ]

3 3 Functions 48 s in the domain of Yes because f? f : that is the domain of nc f ( ( f f ( f ( [ From bove] ec ( f f ( ( f is and f ( definition of the inverse fu tion we have and therefore. the ause the identit function Since therefore we have the required result, f f. Proposition (3.5. Let the functions f : and g: Then the function g is unique and satisf both g f and f g g f. Comment. For g f we need the function f to be bijective. Wh? ecause from Section C we have (3.5 f has an inverse f is bijective First we prove th at f is bijective and then g f. Hence first we show that f is injective and surjective. How do we prove that f is injective (one to one? (3.2 we are required to show that for all, in we have f ( f ( implies Let both, f f. Then ( [ dentit Function] g f ( [ecause we are given g f g f definition of be in and ] ( [ ] g( f ( ecause f ( f ( ( g f ( [ definition of ] ( [ ecause we are given g f ] Hence therefore b (3.2 we conclude that f is injective. Net we prove f is surjective. How? Let be in (codomain of f and then show that there is an element, call it (domain of f such that f (. We have ( [ dentit Function] f g ecause we are given f g [ f g( [ Definition of ] ], in is an injection f f (3.2 implies f

4 3 Functions 49 Since we are given g: therefore g( is in. Let g( we have found an element, g(, in, such that f ( f is surjective. ecause f is both injective and surjective therefore it is bijective. (3.5 f has an inverse f is bijective. we can sa the function f has an inverse and therefore. Hence the function f. Have we completed the proof? No because we still need to show that g is unique an d g f. How do we prove this? contradiction. Suppose the are not equal, that is g f. This means there is another function besides f with the above stated properties: g f and f g The domain of both f : and g: is the set. There must be an element sa in such that g f ecause functions g f Since the function f is injective (one to one therefore ut and ( f f f g [Not Equal] ( ( [ Definition of ] ( [ ecause we are given f g ] f g f g f ( f ( ( f f ( [ Definition of ] ecause b (3.4 we have We have f ( g( f ( f ( were not equal, that is f g( f f. Hence f f. Contradiction because earlier we stated the ( we have proved our result g f. Can ou see an relationship between the above propositions (3.4 and (3.5? n (3.5 we have proven that the inverse function f which satisfies proposition (3.4 is unique: f f and f f f is the onl function with this propert for a given function f. What does this mean? Let f : and be in the domain then ( Similarl if is in then ( f f (. f f Note that g f onl if both the following conditions are satisfied: g f and f g

5 3 Functions 50 Remember bije ctive functions are crucial in the discussion of inverse functions but the are also preserved under the composition of functions as we will show in the eercises. E3 Properti es of the Composite Functions Net we prove an etremel important propert of function composition. We prove that the composition of functions is associative. Proposition (3.6 Let h:, g: C and f : C D be functions. Then We can illustrate the functions b: ( f g h f ( g h h g f Fig 34 C D The domain of both ( f g h and f ( g h is the set. Let be an arbitrar element in. Then f g h f g h Definition of lso [ ] f g( h( Definition of [ ] ( f ( g h ( f ( g h( [ Definition of ] f g( h( Definition of [ ] Since for the arbitrar element we have f g h f g h therefore we have our required result: f g h f g h SUMMRY The identit function on a set is denoted b and has the propert ( for all in. The input and output of the function sta the same. Let f : and g: be functions which satisf both g f and f g Then g f and is unique. Composition of functions is associative: f g h f g h

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