Functions (Mappings)

Definitions A function (or mapping)  from a set A to a set B is a rule that assigns to each element a of A exactly one element b of B. The set A is called the domain of , and B is called the range of . If  assigns b to a, then b is called the image of a under . The subset of B comprising all the images of elements of A is called the image of A under 

Notation  A –> B means  is a mapping from set A to set B.  (a) = b or  : a –> b means that function  maps element a to element b (i.e. the image of a is b).

Example 1 Let  : R –> R be given by  (x) = sin(x). The image of  /2 under  is 1 The image of R under  is [-1,1]. domain of  is R range of  is R  (R) = [-1,1]

Example 2  : Q –> Z given by  : p/q –> p+q  (1/2) = = 3  (2/4) = = 6 Since 1/2 = 2/4, but  (1/2) ≠  (2/4)  is not a function!

To prove a rule is a function Assume x 1 = x 2 Show  (x 1 ) =  (x 2 ) In this case we say that  is well-defined.

Show  is well-defined Let  :Z –> Z be given by  (n) = n 2 mod 2 Suppose n 1 = n 2.  (n 1 )–  (n 2 ) = n 1 2 mod 2 –n 2 2 mod 2 = (n 1 – n 2 )(n 1 +n 2 ) mod 2 = 0 since (n 1 – n 2 ) = 0 So  (n 1 ) =  (n 2 ) Therefore,  is well-defined.

Composition of functions Let  : A –> B and  : B –> C. The composition  is the mapping from A to C defined by  (a) =  (  (a)). AB C a  (a)  (a)   

Order matters! When we compose  and , we must write  Unless A = C,  does not make sense. AB C a  (a)  (a)   

One to one functions A function from a set A is called one- to-one if Note: This is the converse to the well- defined condition.

Show not one-to-one Show  :R –> R given by  (x) = x 2 is not one–to–one.  (–2)= 4 =  (2), but –2 ≠ 2 So  is not one-to-one.

Show one-to-one Show  :[0,∞) –> R given by  (x) = x 2 is one– to–one. Suppose  (x 1 )=  (x 2 ). Then 0 =  (x 1 )–  (x 2 ) = x 1 2 –x 2 2 = (x 1 –x 2 )(x 1 +x 2 ) So either (x 1 –x 2 ) = 0 or (x 1 +x 2 ) = 0 If (x 1 –x 2 ) = 0, then x 1 = x 2 If (x 1 +x 2 ) = 0, then x 1 = x 2 = 0 since the domain is [0,∞) In either case, x 1 = x 2, so  is one-to-one.

Onto functions A function  from a set A to a set B is said to be onto B if each element of B is the image of at least one element of A.

Show not onto Show  :[0,∞) –> R given by  (x) = x 2 is not onto. Suppose –1 =  (x) for some x in [0,∞). Then -1 = x 2 ≥ 0 This counterexample shows  is not onto.

Show onto Show  :[0,∞) –> [0,∞) given by  (x) = x 2 is onto. Proof: Choose any number b ≥ 0. Let a = √b. Then  (a) = (√b) 2 = b. So  is onto.

Properties of functions Given  :A–>B,  :B–>C, and  :C–>D, then 1.  (  ) = (  ) . (Associativity) 2.If  and  are one-to-one, then  is one-to-one. 3.If  and  are onto, then  is onto. 4.If  is one-to-one and onto, then there is a function  -1 from B to A such that  -1  (a)=a for all a in A and  -1 (b)=b for all b in B.

Proof of Associativity Choose any a in A. Let b =  (a), c =  (b), and d =  (c). Notice that  (a) = c and  (b) = d. Then (  )  (a) =  (b) = d Also,  (  )(a) =  (c) = d Since (  )  (a) =  (  )(a) for all a in A, (  )  =  (  )

Prove one-to-one Suppose  (x 1 ) =  (x 2 ) Set y 1 =  (x 1 ) and y 2 =  (x 2 ). Then  (y 1 ) =  (y 2 ) Since  is one-to-one, y 1 =y 2 But then  (x 1 ) =  (x 2 ). Since  is one-to-one, x 1 = x 2. Therefore,  is one-to-one.

Prove onto Choose any c in C Since  is onto, there is a b in B with  (b)=c. Since  is onto, there is an a in A with  (a)=b. Then  (a) =  (b) = c Therefore,  is onto.

Proof of inverse functions The proof consists of three steps. 1.Construct the inverse function. 2.Show that the inverse is well-defined. 3.Show that the inverse function has the required cancellation properties.

1. Construction Given  :A->B is one-to-one and onto. Choose any b in B. Since  is onto, there is an a in A with  (a)=b. Set  -1 (b) = a.

2. Well-defined Suppose b 1 = b 2 in B. Let a 1 =  -1 (b 1 ) and a 2 =  -1 (b 2 ) Then  (a 1 ) = b 1 = b 2 =  (a 2 ) Since  is one-to-one, a 1 = a 2 That is,  -1 (b 1 ) =  -1 (b 2 ) Therefore,  -1 is well-defined.

3. Cancellation Choose any a in A. Set b = f(a) and note that  -1 (b) = a. Then  -1  (a) =  -1 (b) = a, and  -1 (b) =  (a) = b. Since  -1 is well-defined and the cancellation laws hold, we are done.