1. Chapter 2
Sets and Functions

2. Section 2.3
Functions

3. 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.
The range of f, denoted rng  f, is the set of all second elements of members of f . That is,
rng f = {b  B :  a  A (a, b)  f }.
If (x, y)  f , we often say that f maps x onto y and use the notation y = f (x).
Domain A
Codomain B  x
f (x) 
rng  f
f : A  B

4. Practice 2.3.2*
*Similar to Practice in the text.
Let A = {1, 2, 3} and B = {2, 4, 6, 8}. Which of the following relations are functions
from A to B? A 1 2 3 B 4 6 8
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. A 1 2 3 B 4 6 8
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.

5. Practice 2.3.2*
*Similar to Practice in the text.
Let A = {1, 2, 3} and B = {2, 4, 6, 8}. Which of the following relations are functions
from A to B? A 1 2 3 B 4 6 8
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. A 1 2 3 B 4 6 8
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.

6. Properties of Functions
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.

7. Example 2.3.7 Consider the function given by the formula f (x) = x2.
y = h (x) 4
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.
y = f (x)
y = g (x) 2
If we limit the codomain to be the set [0, ), then the
function g :  [0, ) such that g(x) = x2 is surjective.
– 2 2 x
y = – 1
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.

8. Practice
Consider the four functions pictured below. For each function, the domain and codomain
are sets consisting of two or three points as indicated. Classify each function as being
surjective, injective, bijective, or none of these.
 
  
This function is injective.
This function is neither.
 
  
This function is surjective.
This function is bijective.

9. Functions Acting on Sets
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:
Notation
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. f A B
f (A)
f – 1
f – 1(D) D
f (C) C f
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.

10. Example
Let f :     be given by f (x) = x2. Then the following hold.
– 2 2 x 4 y
y = x2
If C1 = [0, 2], then f (C1) = [0, 4].
f (C1) C1

11. Example
Let f :     be given by f (x) = x2. Then the following hold.
– 2 2 x 4 y
y = x2
If C1 = [0, 2], then f (C1) = [0, 4].
If C2 = [1, 2], then f (C2) = [0, 4].
f (C2) C2

12. Example
Let f :     be given by f (x) = x2. Then the following hold.
– 2 2 x 4 y
y = x2
If C1 = [0, 2], then f (C1) = [0, 4].
If C2 = [1, 2], then f (C2) = [0, 4].
If C3 = [2, 1]  [0, 1], then f (C3) = [0, 4].
f (C3) C3

13. Example
Let f :     be given by f (x) = x2. Then the following hold.
– 2 2 x 4 y
y = x2
If C1 = [0, 2], then f (C1) = [0, 4].
If C2 = [1, 2], then f (C2) = [0, 4].
If C3 = [2, 1]  [0, 1], then f (C3) = [0, 4].
If D1 = [0, 4], then f – 1(D1) = [2, 2]. D1
f – 1(D1)

14. Example
Let f :     be given by f (x) = x2. Then the following hold.
– 2 2 x 4 y
y = x2
If C1 = [0, 2], then f (C1) = [0, 4].
If C2 = [1, 2], then f (C2) = [0, 4].
If C3 = [2, 1]  [0, 1], then f (C3) = [0, 4].
If D1 = [0, 4], then f – 1(D1) = [2, 2].
If D2 = [1, 4], then f – 1(D2) = [2, 2]. D2
f – 1(D2)

15. Example
Let f :     be given by f (x) = x2. Then the following hold.
– 2 2 x 4 y
y = x2
If C1 = [0, 2], then f (C1) = [0, 4].
If C2 = [1, 2], then f (C2) = [0, 4].
If C3 = [2, 1]  [0, 1], then f (C3) = [0, 4].
If D1 = [0, 4], then f – 1(D1) = [2, 2].
If D2 = [1, 4], then f – 1(D2) = [2, 2].
If D3 = [1, 4], then f – 1(D3) = [2, 1]  [1, 2]. D3
f – 1(D3)

16. 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
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)
(g)  f  – 1(B \D) = A\ f  – 1(D)

17. Given f : A  B, suppose C1 and C2 are nonempty subsets of A such that
To see that equality may not hold in Theorem (a) and (c), consider the
following situation:
Given f : A  B, suppose C1 and C2 are nonempty subsets of A such that
C1  C2 =  and f (C1) = f (C2). A B f C1
f (C1) C2
f (C2)
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).

18. 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
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:
If f is injective, then f – 1[ f (C)] = C.
If f is surjective, then f [ f – 1(D)] = D.
If f is injective, then f (C1  C2) = f (C1)  f (C2).
Proof of (c): We only have to show that f (C1)  f (C2)  f (C1  C2), since the
converse inclusion is Theorem (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). 

19. Composition of Functions
If f and g are functions with f : A  B and g : B  C,
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. B C A f g
g ( f (a))
f (a)   a 
( g ○ f )(a)
g ○ f 
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.

20. Theorem 2.3.20 Inverse Functions
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
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.
Inverse Functions
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.

21. Practice 2.3.21 Practice 2.3.22  Definition 2.3.23
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
Ans: f must be surjective.
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
Ans: f must be bijective.
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). A B
x = f  – 1.( y)
y = f  ( x)  f 
f  – 1
Definition
Let f : A  B be bijective. The inverse function of f is the function given by
f  – 1  = {( y, x)  B  A : (x, y)  f }.

22. 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
identity function on A, and is denoted by iA.
So, f – 1 ○ 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.
Theorem
Let f : A  B be bijective. Then
(a) f – 1 : B  A is bijective (b) f – 1 ○ f = iA and f ○ f – 1 = iB.

23. 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.
Proof: We know from Theorem that g ○ f is bijective, so g ○ f has an inverse
denoted by ( g ○ f )  1, and this inverse maps C onto A. f g
f  –1
g –1 A B C
g ○ f
( g ○ f ) –1 = f –1 ○ g –1
We are asked to verify the equality of the two functions ( g ○ f )  1 and f – 1 ○ g  1,
as sets of ordered pairs.

24. 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
(b, a)  f  – 1 and (c, b)  g – 1.
That is, f – 1(b) = a and g – 1(c) = b.
But then,
( f – 1 ○ g  1)(c) = f – 1( g  1(c)) = f – 1(b) = a,
so that (c, a)  ( f – 1 ○ g  1) and ( g  ○  f )  1  ( f – 1 ○ g  1).
The reverse set inclusion is Exercise 31. 