Functions Mappings Definitions A function or mapping from

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]. (R) = [-1, 1] domain of is R range of is R

Example 2 • : Q –> Z given by : p/q –> p+q • (1/2) = 1 + 2 = 3 (2/4) = 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 12 mod 2 –n 22 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)). a A B (a) C (a)

Order matters! • When we compose and , we must write • Unless A = C, does not make sense. a A B (a) C (a)

One to one functions • A function from a set A is called one-to -one if • Note: This is the converse to the welldefined 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 12–x 22 = (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. 1. If and are onto, then is onto. • 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.