Analysis with an introduction to proof 5th by steven lay ch02

When there are no extra elements in B, the function is given a special name... A surjective function is also referred to as a surjection.. The question of whether or not a function is su

Chapter 2

Sets and Functions

Section 2.3 Functions

Definition 2.3.1

Let A and B be sets A function from A to B is a nonempty relation f  A  B that

satisfies the following two conditions:

1 Existence: For all a in A, there exists a b in B such that (a, b)  f

2 Uniqueness: If (a, b)  f and (a, c)  f , then b = c

That is, given any element a in A, there is one and only one element b in B such that (a, b) f

Set A is called the domain of f and is denoted by dom f

Set B is referred to as the codomain of f

We write f : A  B to indicate f has domain A and codomain B

R = {(1,6), (2,4)}

This is not a function It violates the existence condition

The number 3 in set A is not related to anything in set B.

R = {(1,2), (2,8), (3, 6), (1, 4)}

This is not a function It violates the uniqueness condition

The number 1 in set A is related to two numbers in set B.

R = {(1,6), (2,8), (3,2)}

This is a function from A to B.

It is OK that nothing in set A is related to number 4 in set B.

R = {(1,4), (2,6), (3,4)}

This is a function from A to B.

It is OK that both 1 and 3 in set A are related to 4 in set B.

In both of these functions, there is at least one element in B that is not related to

anything in A

When there are no extra elements in B, the function is given a special name.

Definition 2.3.4

A function f : A  B is called surjective (or is said to map A onto B) if B = rng f

A surjective function is also referred to as a surjection

The question of whether or not a function is surjective depends on the choice of codomain

A function can always be made surjective by restricting the codomain to being equal to the range, but sometimes this is not convenient

If it happens that no member of the codomain appears more than once as a second

element in one of the ordered pairs, then we have another important type of function

Definition 2.3.5

A function f : A  B is called injective (or one-to-one) if, for all a and a in A,

f (a) = f(a) implies that a = a An injective function is also referred to as an injection

If a function is both surjective and injective, then it is particularly well behaved

Definition 2.3.6

A function f : A  B is called bijective or a bijection if it is both surjective and injective

Properties of Functions

If we take for both the domain and codomain

so that f :  , then f is not surjective because there is no real number that maps onto –1

If we limit the codomain to be the set [0,), then the function g:  [0,) such that g(x) = x2 is surjective

Since g(1) = g(1), we see that g is not injective when defined on all of

But restricting g to be defined on only [0,), it becomes injective

Thus h: [0,)  [0,) such that h(x) = x2 is bijective

y = f (x)

y = h (x)

y = –1

This function is injective This function is neither.

This function is surjective This function is bijective

When thinking of a function as transforming its domain into its range, we may wish to

consider what happens to certain subsets of the domain Or we may wish to identify the

set of all points in the domain that are mapped into a particular subset of the range

To do this we use the following notation:

Functions Acting on Sets

Notation 2.3.13

Suppose that f : A  B If C  A, we let f (C) represent the subset {f (x): x  C} of B

The set f (C) is called the image of C in B.

If D  B, we let f – 1(D) represent the subset {x  A: f (x)  D} of A The set f – 1(D) is

called the pre-image of D in A or “ f inverse of D.”

Note: At this time, f – 1 is only applied to sets, not points

Given a function f : A  B, there are many relationships that hold between the images

and pre-images of subsets of A and B.

Several of these are included in the next theorem The proofs are left to the practice

problems and the exercises

Theorem 2.3.16

Suppose that f : A  B Let C, C1, and C2 be subsets of A and let D, D1, and D2 be

subsets of B Then the following hold:

(a) C  f –1[ f (C)]

(b) f [ f –1(D)]  D

(c) f (C1  C2)  f(C1)  f (C2) (d) f (C1  C2) = f (C1)  f (C2) (e) f –1(D1  D2) = f –1(D1)  f –1(D2) (f) f –1(D1  D2) = f –1(D1)  f –1(D2)

We see that f – 1[ f (C1)] = C1  C2, and this is larger than C1.

This shows that equality may not hold in part (a): C  f –1[ f (C)]

Also, since C1  C2 = , we have f (C1  C2 ) =  But f (C1)  f (C2) = f (C1)  ,

This shows that equality may not hold in part (c): f (C1  C2)  f (C1)  f (C2)

While Theorem 2.3.16 states the strongest results that hold in general, if we apply

certain restrictions on the functions involved, then the containment symbols in

parts (a), (b), and (c) may be replaced by equality

Theorem 2.3.18

Suppose that f : A  B Let C, C1, and C2 be subsets of A and let D be a subset of B

Then the following hold:

(a) If f is injective, then f – 1[ f (C)] = C

(b) If f is surjective, then f [ f – 1(D)] = D

(c) If f is injective, then f (C1  C2) = f (C1)  f (C2)

converse inclusion is Theorem 2.3.16(c)

To this end, let y  f (C1)  f (C2) Then y  f (C1) and y  f (C2)

It follows that there exists a point x1 in C1 such that f (x1) = y

Similarly, there exists a point x2 in C2 such that f (x2) = y

Since f is injective and f (x1) = y = f (x2), we must have x1 = x2

That is, x1  C1  C2 But then y = f (x1)  f (C1  C2) 

C B

then for any a  A, f (a)  B

But B is the domain of g, so g can be applied to f (a) This yields g( f (a)), an element of C

This establishes a correspondence between a in A and g( f (a)) in C

This correspondence is called the composition function of f and g and is denoted

by g ○ f (read “g of f ”) It defines a function g ○ f : A  C given by

(g ○ f )(a) = g (f (a)) for all a  A.

Our next theorem tells us that composition of functions preserves the properties of

being surjective or injective The proof is in the book and the exercises

Theorem 2.3.20

Let f : A  B and g : B  C Then

(a) If f and g are surjective, then g ○ f is surjective

(b) If f and g are injective, then g ○ f is injective

(c) If f and g are bijective, then g ○ f is bijective

Given a function f: A  B, we have seen how f determines a relationship between

subsets of B and subsets of A That is, given D  B, we have the pre-image f – 1(D) in A.

We would like to be able to extend this idea so that f – 1 can be applied to a point in B

to obtain a point in A That is, suppose D = { y}, where y  B.

There are two things that can prevent f – 1(D) from being a point in A:

It may be that f – 1(D) is empty,

and it may be that f – 1(D) contains several points instead of just one.

Inverse Functions

Given f : A  B and y  B, under what conditions on f can we

assert that there exists an x in A such that f (x) = y?

Practice 2.3.21

Given f : A  B and y  B, under what conditions on f can we

assert that there exists a unique x in A such that f(x) = y?

Practice 2.3.22

Given a bijection f : A  B, we see that each y in B corresponds to exactly one x in A,

the unique x such that f (x) = y.

This correspondence defines a function from B into A called the inverse of f and denoted f –1 Thus x = f –1(y)

Let f : A  B be bijective The inverse function of f is the function given by

If f : A  B is bijective, then it follows that f – 1:B  A is also bijective.

Indeed, since dom f = A and rng f = B, we have dom f – 1 = B and rng f – 1 =A.

Thus f –1 is a mapping from B onto A.

Since f is a function, a given x in A can correspond to only one y in B.

This means that f – 1 is injective, and hence bijective

When f is followed by f – 1, the effect is to map x in A onto f (x) in B and then back to x in A That is, ( f – 1 ○ f )(x) = x, for every x  A.

A function defined on a set A that maps each element in A onto itself is called the

○ f = iA

Furthermore, if f (x) = y, then x = f –1(y), so that ( f ○ f – 1)(y) = f ( f – 1 (y)) = f (x) = y.

Thus f ○ f – 1 = iB We summarize these results in the following theorem

Let f : A  B be bijective Then

(a) f – 1 : B  A is bijective (b) f – 1 ○ f = i A and f ○ f – 1 = i B

Theorem 2.3.24

Our final theorem relates inverse functions and composition.

Let f : A  B and g : B  C be bijective Then the composition g ○ f : A  C is bijective

and (g ○ f ) 1 = f – 1

○ g 1

Theorem 2.3.28

denoted by (g ○ f ) 1, and this inverse maps C onto A.

To show (g ○ f ) 1  f – 1 ○ g 1, we suppose (c, a)  ( g ○ f ) 1.

By the definition of an inverse function, this means (a, c)  g ○ f

The definition of composition implies that

 b  B such that (a, b)  f and (b, c)  g.

Since f and g are bijective, this means that