Functions, High-School Edition

Transcription

1 Functions

2 What is a function?

3 Functions, High-School Edition

4 f(x) = x4 5x2 + 4 source:

5 source:

6 Functions, High-School Edition In high school, functions are usually given as objects of the form x3 +3 x2 +15 x+7 f (x) = 1 x137 What does a function do? Takes in as input a real number. Outputs a real number. except when there are vertical asymptotes or other discontinuities, in which case the function doesn't output anything.

7 Functions, CS Edition

8 int int flipuntil(int flipuntil(int n) n) {{ int int numheads numheads == 0; 0; int int numtries numtries == 0; 0; while while (numheads (numheads << n) n) {{ if if (randomboolean()) (randomboolean()) numheads++; numheads++; }} }} numtries++; numtries++; return return numtries; numtries;

9 Functions, CS Edition In programming, functions might take in inputs, might return values, might have side effects, might never return anything, might crash, and might return different values when called multiple times.

10 What's Common? Although high-school math functions and CS functions are pretty different, they have two key aspects in common: They take in inputs. They produce outputs. In math, we like to keep things easy, so that's pretty much how we're going to define a function.

11 Rough Idea of a Function: A function is an object f that takes in an input and produces exactly one output. x f f(x) (This is not a complete definition we'll revisit this in a bit.)

12 High School versus CS Functions In high school, functions usually were given by a rule: f(x) = 4x + 15 In CS, functions are usually given by code: int factorial(int n) { int result = 1; for (int i = 1; i <= n; i++) { result *= i; } return result; } What sorts of functions are we going to allow from a mathematical perspective?

13 Dikdik Nubian Ibex Sloth

14

15 but also

16 f(x) = x2 + 3x 15

17 n/2 if n is even f (n)= (n+1)/2 otherwise { Functions Functionslike likethese these are arecalled calledpiecewise piecewise functions. functions.

18 To define a function, you will typically either draw a picture, or give a rule for determining the output.

19 In mathematics, functions are deterministic. That is, given the same input, a function must always produce the same output. The following is a perfectly valid piece of C++ code, but it s not a valid function under our definition: int randomnumber(int numoutcomes) { return rand() % numoutcomes; }

20 One Challenge

21 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

22 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

23 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

24 f(x) = x2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

25 f( )= f(137 ) =?

26 We need to make sure we can't apply functions to meaningless inputs.

27 Domains and Codomains Every function f has two sets associated with it: its domain and its codomain. A function f can only be applied to elements of its domain. For any x in the domain, f(x) belongs to the codomain. The Thefunction function must be must bedefined defined for every element for every element ofofthe thedomain. domain. Domain Codomain The Theoutput outputofofthe the function must function must always alwaysbe beininthe the codomain, but codomain, but not notall allelements elements ofofthe codomain the codomain must mustbe be produced producedas as outputs. outputs.

28 Domains and Codomains Every function f has two sets associated with it: its domain and its codomain. A function f can only be applied to elements of its domain. For any x in the domain, f(x) belongs to the codomain. The Thecodomain codomainofofthis thisfunction functionisis ℝ. ℝ.Everything Everythingproduced producedisisaareal real number, but not all real numbers number, but not all real numbers can canbe beproduced. produced. The Thedomain domainofofthis thisfunction function isisℝ. Any real number ℝ. Any real numbercan canbe be provided as input. provided as input. private double absolutevalueof(double x) { if (x >= 0) { return x; } else { return -x; } }

29 Domains and Codomains If f is a function whose domain is A and whose codomain is B, we write f : A B. This notation just says what the domain and codomain of the function are. It doesn't say how the function is evaluated. Think of it like a function prototype in C or C++. The notation f : ArgType RetType is like writing RetType f(argtype argument); We know that f takes in an ArgType and returns a RetType, but we don't know exactly which RetType it's going to return for a given ArgType.

30 The Official Rules for Functions Formally speaking, we say that f : A B if the following two rules hold: The function must be obey its domain/codomain rules: a A. b B. f(a) = b ( Every input in A maps to some output in B. ) The function must be deterministic: a₁ A. a₂ A. (a₁ = a₂ f(a₁) = f(a₂)) ( Equal inputs produce equal outputs. ) If you re ever curious about whether something is a function, look back at these rules and check! For example: Can a function have an empty domain? Can a function with a nonempty domain have an empty codomain?

31 Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain to some element of the codomain. Examples: f(n) = n + 1, where f : ℤ ℤ f(x) = sin x, where f : ℝ ℝ f(x) = x, where f : ℝ ℤ Notice that we're giving both a rule and the domain/codomain.

32 Defining Functions Typically, we specify a function by describing a rule that maps every element This function of the domain to some element of the This isis the the ceiling ceiling function the the smallest smallest integer integer greater greater codomain. Examples: than than or or equal equal to to x. x. For For example, example, 1 1 == 1,1, == 2, 2, f(n) = n + 1, where f : ℤ ℤ and and π π == f(x) = sin x, where f : ℝ ℝ f(x) = x, where f : ℝ ℤ Notice that we're giving both a rule and the domain/codomain.

33 Is This a Function from A to B? Stanford Cardinal Berkeley Blue Michigan Gold Arkansas White A B

34 Is This a Function from A to B? California Dover New York Sacramento Delaware Albany Washington DC A B

35 Is This a Function from A to B? Wish Funshine A B Love-a-Lot Tenderheart Friend

36 Combining Functions

37 f : People Places Keith g : Places Prices Mountain View Far Too Much Anna San Francisco King's Ransom Shloka Redding, CA A Modest Amount Dilsher Barrow, AK Pocket Change Shalom People Palo Alto Places h : People Prices h(x) = g(f(x)) Prices

38 Keith Far Too Much Anna King's Ransom Shloka A Modest Amount Dilsher Pocket Change Shalom People Prices h : People Prices h(x) = g(f(x))

39 Function Composition Suppose that we have two functions f : A B and g : B C. Notice that the codomain of f is the domain of g. This means that we can use outputs from f as inputs to g. x f f(x) g g(f(x))

40 Function Composition Suppose that we have two functions f : A B and g : B C. The composition of f and g, denoted g f, is a function where g f : A C, and (g f)(x) = g(f(x)). A few things to notice: The Thename nameofofthe thefunction functionisisgg f.f. When Whenwe weapply applyititto toan aninput inputx,x, we wewrite write(g (g f)(x). f)(x).i Idon't don'tknow know why, but that's what we do. why, but that's what we do. The domain of g f is the domain of f. Its codomain is the codomain of g. Even though the composition is written g f, when evaluating (g f)(x), the function f is evaluated first.

41 Function Composition Let f : ℕ ℕ be defined as f(n) = 2n + 1 and g : ℕ ℕ be defined as g(n) = n2. What is g f? (g f)(n) = g(f(n)) (g f)(n) = g(2n + 1) (g f)(n) = (2n + 1)2 = 4n2 + 4n + 1 What is f g? (f g)(n) = f(g(n)) (f g)(n) = f(n2) (f g)(n) = 2n2 + 1 In general, if they exist, the functions g f and f g are usually not the same function. Order matters in function composition!

42 Time-Out for Announcements!

43 Problem Set Three Problem Set Two was due at 3:00PM today. Problem Set Three goes out right now. Reminder: You have three 24-hour late days to use throughout the quarter. Checkpoint due on Monday at the start of class. Remaining problems due Friday at the start of class. As always, feel free to ask questions on Piazza or to stop by office hours with questions!

44 First Midterm Exam The first midterm exam is on Tuesday, May 2nd from 7:00PM 10:00PM, location TBA. We ll be releasing a ton of practice problems next week, and we ll talk about policies on Monday. We will be holding a practice midterm exam next Tuesday, April 25th from 7:00PM 10:00PM, location TBA. You are highly encouraged to attend. This is an excellent way to practice and prepare for the midterm. More details next week.

45 WiCS Casual CS Dinner Stanford WiCS (Women in Computer Science) is holding a Casual CS Dinner this upcoming Monday, April 24th at 6:00PM at the Women's Community Center. All are welcome. Highly recommended! RSVP using

46 Your Questions

47 I feel like I'll never catch up to my peers who are so ahead in CS. What advice do you have for changing that, especially for someone from FLI background. AA few few things things to to keep keep inin mind: mind: 1.1. The The overwhelming overwhelming majority majority of of CS CS majors majors have have no no prior prior CS CS experience experience before before coming coming here here most most schools schools don t don t offer offer anything. anything Make Make sure sure you you have have the the right right mental mental model model for for where where you you stand stand relative relative to to everyone everyone else else there s there s aa huge huge sampling sampling bias! bias! The The gap gap between between you you and and everyone everyone else else isis largest largest when when you you get get started started and and rapidly rapidly gets gets washed washed out out by by coursework coursework here. here. You d You d be be amazed amazed how how quickly quickly we we move move here! here! A A little little slope slope makes makes up up for for aa lot lot of of y-intercept y-intercept Never Never confuse confuse talent talent for for experience. experience.

48 Is there any hope for a just world under capitalism? II think think so, so, though though that that might might just just be be because because II have have aa different different conception conception of of the the question question than than what what was was intended. intended. I ll I ll take take this this one one inin class. class.

49 Back to CS103!

50 Special Types of Functions

51 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

52 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

53 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

54 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

55 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

56 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

57 Injective Functions A function f : A B is called injective (or one-to-one) if the following statement is true about f: a₁ A. a₂ A. (a₁ a₂ f(a₁) f(a₂)) ( If the inputs are different, the outputs are different. ) The following first-order definition is equivalent and is often useful in proofs. a₁ A. a₂ A. (f(a₁) = f(a₂) a₁ = a₂) ( If the outputs are the same, the inputs are the same. ) A function with this property is called an injection. How does this compare to our second rule for functions?

58 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. Since f(n₀) = f(n₁), we see that 2n₀ + 7 = 2n₁ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

59 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. Since f(n₀) = f(n₁), we see that 2n₀ + 7 = 2n₁ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

60 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₀ ℕ. n₁ ℕ. n₀ ℕ. n₁ ℕ.((f(n₀) f(n₀)==f(n₁) f(n₁) n₀ n₀==n₁ n₁)) 2n₀ = 2n₁, n₀ n₀ ℕ. ℕ. n₁ n₁ ℕ. ℕ.((n₀ n₀ n₁ n₁ f(n₀) f(n₀) f(n₁) f(n₁))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₀, n₀,n₁ n₁ ℕ ℕ where where f(n₀) f(n₀)==f(n₁), f(n₁), then then prove prove that that n₀ n₀==n₁. n₁.

61 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ ℕ. n₂ ℕ. n₁ ℕ. n₂ ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₀, n₀,n₁ n₁ ℕ ℕ where where f(n₀) f(n₀)==f(n₁), f(n₁), then then prove prove that that n₀ n₀==n₁. n₁.

62 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ ℕ. n₂ ℕ. n₁ ℕ. n₂ ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₀, n₀,n₁ n₁ ℕ ℕ where where f(n₀) f(n₀)==f(n₁), f(n₁), then then prove prove that that n₀ n₀==n₁. n₁.

63 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ ℕ. n₂ ℕ. n₁ ℕ. n₂ ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

64 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ ℕ. n₂ ℕ. n₁ ℕ. n₂ ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

65 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ ℕ. n₂ ℕ. n₁ ℕ. n₂ ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

66 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₀, n₁ ℕ where f(n₀) = f(n₁). We will prove that n₀ = n₁. What does itit mean for the function ff to be What does mean for the function to be Since f(n₀) = f(n₁), we see that injective? injective? 2n₀ + 7 = 2n₁ + 7. This in turn means that n₁ ℕ. n₂ ℕ. n₁ ℕ. n₂ ℕ.((f(n₁) f(n₁)==f(n₂) f(n₂) n₁ n₁==n₂ n₂)) 2n₀ = 2n₁, n₁ n₁ ℕ. ℕ. n₂ n₂ ℕ. ℕ.((n₁ n₁ n₂ n₂ f(n₁) f(n₁) f(n₂) f(n₂))) so n₀ = n₁, as required. Therefore, Therefore, we'll we'll pick pick arbitrary arbitrary n₁, n₁,n₂ n₂ ℕ ℕ where where f(n₁) f(n₁)==f(n₂), f(n₂), then then prove prove that that n₁ n₁==n₂. n₂.

67 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₀) = f(n₁), we see that 2n₀ + 7 = 2n₁ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

68 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₀ = 2n₁, so n₀ = n₁, as required.

69 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₁ = 2n₂ so n₀ = n₁, as required.

70 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₁ = 2n₂ so n₁ = n₂, as required.

71 Injective Functions Theorem: Let f : ℕ ℕ be defined as f(n) = 2n + 7. Then f is injective. Proof: Consider any n₁, n₂ ℕ where f(n₁) = f(n₂). We will prove that n₁ = n₂. Since f(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that 2n₁ = 2n₂ so n₁ = n₂, as required.

72 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). Let x₀ = -1 and x₁ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

73 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). Let x₀ = -1 and x₁ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

74 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₀ ℤ. x₁ ℤ. (x₀ x₀ ℤ. x₁ ℤ. (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) x₀ ℤ. x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₁)) f(x₁) = f(1) =x₁ 14 =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ ℤ. ℤ. (x₀ f(x₁))) x₀ ℤ. x₁ x₁ ℤ. (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

75 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) x₀ ℤ. x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₁)) f(x₁) = f(1) =x₁ 14 =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ ℤ. ℤ. (x₀ f(x₁))) x₀ ℤ. x₁ x₁ ℤ. (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

76 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) x₀ ℤ. x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ f(x₀) f(x₁)) f(x₁) = f(1) =x₁ 14 =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ ℤ. ℤ. (x₀ f(x₁))) x₀ ℤ. x₁ x₁ ℤ. (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

77 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₀ ℤ. = x₁ 4 f(x₀) x₀ ℤ. ℤ. x₁ x₁ ℤ. (x₀ x₁1 f(x₁)) f(x₁) =(x₀ f(1) =f(x₀) 1, f(x₁)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ ℤ. f(x₁))) x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

78 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 f(x₂)) f(x₁) =(x₁ f(1) =f(x₁) 1, f(x₂)) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ ℤ. f(x₁))) x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

79 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 f(x₂)) f(x₁) =(x₁ f(1) =f(x₁) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₀ ℤ. f(x₁))) x₀ ℤ. ℤ. x₁ x₁ ℤ.(x₀ (x₀ x₁ x₁ (f(x₀) (f(x₀) f(x₁))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

80 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 f(x₂)) f(x₁) =(x₁ f(1) =f(x₁) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) x₀ x₀ ℤ. ℤ. x₁ x₁ ℤ. ℤ.(x₀ (x₀ x₁ x₁ f(x₀) f(x₀)==f(x₁)) f(x₁)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

81 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 f(x₂)) f(x₁) =(x₁ f(1) =f(x₁) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but Therefore, we need to find x₀, x₁ ℤ such that x₀ x₁, but f(x₀) = f(x₁). Can we do that? f(x₀) = f(x₁). Can we do that?

82 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₀ and x₁ such that x₀ x₁, but f(x₀) = f(x₁). What does it mean for f to be injective? What does it mean for f to be injective? Let x₀ = -1 and x₁ = +1. Then x₁ ℤ. x₂ ℤ. (x₁ x₁ ℤ. x₂ ℤ. (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) 4 f(x₀) f(-1) = (-1) =1 What is the negation of this = statement? What is the negation of this statement? and x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) x₁ ℤ. = x₂ 4 f(x₁) x₁ ℤ. ℤ. x₂ x₂ ℤ. (x₁ x₂1 f(x₂)) f(x₁) =(x₁ f(1) =f(x₁) 1, f(x₂)) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) so f(x₀) = f(x₁) even though x₀ x₁, as required. x₁ ℤ. f(x₂))) x₁ ℤ. ℤ. x₂ x₂ ℤ.(x₁ (x₁ x₂ x₂ (f(x₁) (f(x₁) f(x₂))) x₁ x₁ ℤ. ℤ. x₂ x₂ ℤ. ℤ.(x₁ (x₁ x₂ x₂ f(x₁) f(x₁)==f(x₂)) f(x₂)) Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Therefore, we need to find x₁, x₂ ℤ such that x₁ x₂, but f(x₁) = f(x₂). Can we do that? Can we do that?

83 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₀ = -1 and x₁ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

84 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. Then f(x₀) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

85 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₁) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

86 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

87 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

88 Injective Functions Theorem: Let f : ℤ ℕ be defined as f(x) = x4. Then f is not injective. Proof: We will prove that there exist integers x₁ and x₂ such that x₁ x₂, but f(x₁) = f(x₂). Let x₁ = -1 and x₂ = +1. f(x₁) = f(-1) = (-1)4 = 1 and f(x₂) = f(1) = 14 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

89 Injections and Composition

90 Injections and Composition Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is an injection. Our goal will be to prove this result. To do so, we're going to have to call back to the formal definitions of injectivity and function composition.

91 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

92 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

93 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

94 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

95 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). There There are are two two definitions definitions of of injectivity injectivity that that we we can can use use here: here: Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = g(f(a₁)) g(f(a₂)), required. a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) a₁ a₁ A. A. a₂ a₂ A. A.(a₁ (a₁ a₂ a₂ (g (g f)(a₁) f)(a₁) (g (g f)(a₂)) f)(a₂)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁,a₂ a₂ AA where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

96 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). There There are are two two definitions definitions of of injectivity injectivity that that we we can can use use here: here: Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = g(f(a₁)) g(f(a₂)), required. a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) a₁ a₁ A. A. a₂ a₂ A. A.(a₁ (a₁ a₂ a₂ (g (g f)(a₁) f)(a₁) (g (g f)(a₂)) f)(a₂)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁,a₂ a₂ AA where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

97 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). There There are are two two definitions definitions of of injectivity injectivity that that we we can can use use here: here: Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that a₁ A. A. ((g f)(a₁) = g(f(a₁)) g(f(a₂)), required. a₁ A. a₂ a₂ A.as ((g f)(a₁) =(g (g f)(a₂) f)(a₂) a₁ a₁= = a₂) a₂) a₁ a₁ A. A. a₂ a₂ A. A.(a₁ (a₁ a₂ a₂ (g (g f)(a₁) f)(a₁) (g (g f)(a₂)) f)(a₂)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁,a₂ a₂ AA where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

98 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

99 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required.

100 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that How isis (g defined? Howas (g f)(x) f)(x) g(f(a₁)) g(f(a₂)), required. defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

101 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that How isis (g defined? Howas (g f)(x) f)(x) g(f(a₁)) g(f(a₂)), required. defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

102 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that How isis (g defined? Howas (g f)(x) f)(x) g(f(a₁)) g(f(a₂)), required. defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

103 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that How isis (g defined? Howas (g f)(x) f)(x) g(f(a₁)) g(f(a₂)), required. defined? (g (g f)(x) f)(x)==g(f(x)) g(f(x)) So So we we need need to to prove prove that that g(f(a₁)) g(f(a₁)) g(f(a₂)). g(f(a₂)).

104 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

105 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

106 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

107 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ g(f(a₂)) f(a₂) A B C

108 Theorem: If f : A B is an injection and g : B C is an injection, then the function g f : A C is also an injection. Proof: Let f : A B and g : B C be arbitrary injections. We will prove that the function g f : A C is also injective. To do so, consider any a₁, a₂ A where a₁ a₂. We will prove that (g f)(a₁) (g f)(a₂). Equivalently, we need to show that g(f(a₁)) g(f(a₂)). Since f is injective and a₁ a₂, we see that f(a₁) f(a₂). Then, since g is injective and f(a₁) f(a₂), we see that g(f(a₁)) g(f(a₂)), as required. g(f(a₁)) f(a₁) a₁ a₂ Great Great exercise: exercise: Repeat Repeat this this proof proof using using the the other other g(f(a₂)) f(a₂) A B definition definition of of injectivity. injectivity. C

109 Another Class of Functions

110 Mt. Lassen California Mt. Hood Washington Mt. St. Helens Oregon Mt. Shasta

111 Surjective Functions A function f : A B is called surjective (or onto) if this first-order logic statement is true about f: b B. a A. f(a) = b ( For every possible output, there's at least one possible input that produces it ) A function with this property is called a surjection. How does this compare to our first rule of functions?

112 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

113 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

114 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. does mean for Let What xwhat = 2y. Thenititwe see that does mean for ff to to be be surjective? surjective? f(x) = f(2y) = 2y / 2 = y. y ℝ. x ℝ. f(x) ==yy y ℝ. x ℝ. f(x) So f(x) = y, as required. Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

115 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. does mean for Let What xwhat = 2y. Thenititwe see that does mean for ff to to be be surjective? surjective? f(x) = f(2y) = 2y / 2 = y. y ℝ. x ℝ. f(x) ==yy y ℝ. x ℝ. f(x) So f(x) = y, as required. Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

116 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. does mean for Let What xwhat = 2y. Thenititwe see that does mean for ff to to be be surjective? surjective? f(x) = f(2y) = 2y / 2 = y. y ℝ. x ℝ. f(x) ==yy y ℝ. x ℝ. f(x) So f(x) = y, as required. Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

117 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. does mean for Let What xwhat = 2y. Thenititwe see that does mean for ff to to be be surjective? surjective? f(x) = f(2y) = 2y / 2 = y. y ℝ. x ℝ. f(x) ==yy y ℝ. x ℝ. f(x) So f(x) = y, as required. Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

118 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. does mean for Let What xwhat = 2y. Thenititwe see that does mean for ff to to be be surjective? surjective? f(x) = f(2y) = 2y / 2 = y. y ℝ. x ℝ. f(x) ==yy y ℝ. x ℝ. f(x) So f(x) = y, as required. Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

119 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. does mean for Let What xwhat = 2y. Thenititwe see that does mean for ff to to be be surjective? surjective? f(x) = f(2y) = 2y / 2 = y. y ℝ. x ℝ. f(x) ==yy y ℝ. x ℝ. f(x) So f(x) = y, as required. Therefore, Therefore, we'll we'll choose choose an an arbitrary arbitrary yy ℝ, ℝ, then then prove prove that that there there isis some some xx ℝℝ where where f(x) f(x)==y. y.

120 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

121 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

122 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

123 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

124 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

125 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

126 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

127 Surjective Functions Theorem: Let f : ℝ ℝ be defined as f(x) = x / 2. Then f(x) is surjective. Proof: Consider any y ℝ. We will prove that there is a choice of x ℝ such that f(x) = y. Let x = 2y. Then we see that f(x) = f(2y) = 2y / 2 = y. So f(x) = y, as required.

128 Composing Surjections

129 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

130 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

131 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

132 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

133 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) =does c. itit mean What What does mean for for gg ff :: AA CC to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is c C. = A. f)(a) = some b B such g(b) Similarly, c that C. a a A.c.(g (g f)(a)since = cc f : A B is surjective, there is some a A such that f(a) = b. This means thatwe'll there is some a A csuch Therefore, choose arbitrary C that and prove that there Therefore, we'll choose arbitrary c C and prove that there isis some AA such that (g g(b) = c,f)(a) some aa g(f(a)) such= that (g f)(a) == c. c. which is what we needed to show.

134 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) =does c. itit mean What What does mean for for gg ff :: AA CC to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is c C. = A. f)(a) = some b B such g(b) Similarly, c that C. a a A.c.(g (g f)(a)since = cc f : A B is surjective, there is some a A such that f(a) = b. This means thatwe'll there is some a A csuch Therefore, choose arbitrary C that and prove that there Therefore, we'll choose arbitrary c C and prove that there isis some AA such that (g g(b) = c,f)(a) some aa g(f(a)) such= that (g f)(a) == c. c. which is what we needed to show.

135 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) =does c. itit mean What What does mean for for gg ff :: AA CC to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is c C. = A. f)(a) = some b B such g(b) Similarly, c that C. a a A.c.(g (g f)(a)since = cc f : A B is surjective, there is some a A such that f(a) = b. This means thatwe'll there is some a A csuch Therefore, choose arbitrary C that and prove that there Therefore, we'll choose arbitrary c C and prove that there isis some AA such that (g g(b) = c,f)(a) some aa g(f(a)) such= that (g f)(a) == c. c. which is what we needed to show.

136 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

137 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

138 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A B C

139 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A f B C

140 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A f B g C

141 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is c surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A f B g C

142 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. b Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is c surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A f B g C

143 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. b Consider any c C. Since g : B C is surjective, there is a b B such that g(b) = c. Similarly, since f : A B is some c surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A f B g C

144 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. b Consider any c C. Since g : B C is surjective, there is a b B such that g(b) = c. Similarly, since f : A B is some c surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show. A f B g C

145 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

146 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

147 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

148 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

149 Theorem: If f : A B is surjective and g : B C is surjective, then g f : A C is also surjective. Proof: Let f : A B and g : B C be arbitrary surjections. We will prove that the function g f : A C is also surjective. To do so, we will prove that for any c C, there is some a A such that (g f)(a) = c. Equivalently, we will prove that for any c C, there is some a A such that g(f(a)) = c. Consider any c C. Since g : B C is surjective, there is some b B such that g(b) = c. Similarly, since f : A B is surjective, there is some a A such that f(a) = b. This means that there is some a A such that g(f(a)) = g(b) = c, which is what we needed to show.

150 Injections and Surjections An injective function associates at most one element of the domain with each element of the codomain. A surjective function associates at least one element of the domain with each element of the codomain. What about functions that associate exactly one element of the domain with each element of the codomain?

151 Katniss Everdeen Elsa Hermione Granger

152 Bijections A function that associates each element of the codomain with a unique element of the domain is called bijective. Such a function is a bijection. Formally, a bijection is a function that is both injective and surjective. Bijections are sometimes called one-toone correspondences. Not to be confused with one-to-one functions.

153 Bijections and Composition Suppose that f : A B and g : B C are bijections. Is g f necessarily a bijection? Yes! Since both f and g are injective, we know that g f is injective. Since both f and g are surjective, we know that g f is surjective. Therefore, g f is a bijection.

154 Inverse Functions

155 Katniss Everdeen Elsa Hermione Granger

156 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

157 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

158 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

159 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune