Lecture 21 21 127 Concepts of Math 10 17 2012 Administrivia Prep Questions for Exam 2 which is next Friday will be posted tomorrow No lecture on Friday Composition of Functions Definition Let A B C be sets and let f A B and g B C be functions Consider the function h A C defined by a A h a g f a We say that h is the composition of g with f and we write h g f We also shorten this terminology somewhat to say h is g composed with f It helps to read the as after to remind you of the order of operations g f means g is applied after f Notice the ordering of the notation g f and how it compares to the order in which we apply the functions f first and then g i e g f a In words we would read g f a as g of f of a Notation comments Be careful about the distinction between a function itself and the function applied to a domain element The following is correct h x g f x g f x Notice the use of parentheses The following is incorrect h x g f x because f x is an element of f s codomain whereas g is a function what does it mean to compose a function and an element Example Converting Kelvin to Celsius then to Fahrenheit Let C R R be defined by C x x 273 15 Let F R R be defined by F x 59 x 32 The function C converts from Kelvin to Celsisus The function F converts from Celsius to Fahrenheit Then F C converts from Kelvin to Fahrenheit directly x R F C x F C x F x 273 15 9 9 x 273 15 32 x 459 67 5 5 Example Let f R Z be given by f x bxc Let g Z N be given by z if z 0 g z z 1 if z 0 Find g f Is this function injective Surjective Composition is Associative Claim For any functions f A B and g B C and h C D we have h g f h g f 1 Proof WWTs that the outputs of the two functions are the same for every possible input Let x A be given Observe that by the definition of composition twice we have h g f x h g f x h g f x and h g f x h g f x h g f x Compositions and Jections Claim f A B and g B C and g f injective f injective Notice that this doesn t assume any properties of g it doesn t even have to be injective necessarily Try to find an example of f g where g f is injective amd g is injective and also an example where g f is injective but g is not injective Proof Suppose x y A Suppose f x f y WWTS x y Since g is a function g f x g f y This means g f x g f y Since g f is injective x y This was our goal Claim It is not necessarily true that f injective g f injective Proof We will exhibit a counterexample Define A 1 2 and B and C Define f by setting f 1 and f 2 Notice f is injective Define g by setting g 1 g 2 Notice g f is defined by g f 1 and g f 2 This shows g f is not injective Claim Suppose f A B and g B C are surjective Then g f A C is also surjective Note You will prove on the Prep Questions that this also holds for injections Proof Let c C Then b B g b c since g is surjective Since b B a A f a b since f is surjective Then g f a g f a g b c Since c was arbitrary we know g f is surjective Bijections and Inverses Definition A function that is both injective and surjective is said to be bijective or a bijection Definition Given a set X the identity function IdX X X is defined by z X Id X z z Definition Let f A B be a function If there is a function g B A such that f g B B satisfies f g IdB and g f A A satisfies g f IdA then we say g is the inverse of f and write g f 1 Theorem f A B is bijective f has an inverse f 1 B A Proof Assume f is bijective Let b B be arbitrary and fixed Since f surjective a A such that f a b Since f is injective that a is unique because x 6 a f x 6 f a b i e PreImf b a Define g b a This is a well defined function where g B A Also f g b f a b so f g IdB and g f a g b a so g f IdA Thus g f 1 2 Assume f has an inverse function call it f 1 To show f injective WWTS f a1 f a2 a1 a2 Let a1 a2 A be arbitrary and fixed Since f a1 f a2 B we know f a1 f a2 f 1 f a1 f 1 f a2 f 1 f a1 f 1 f a2 IdA a1 IdA a2 a1 a2 Next to show f surjective WWTS b B a A such that f a b Since f 1 is a function then given any b B a A such that f 1 b a Then f 1 b a f f 1 b f a f f 1 b f a IdB b f a b f a 3
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