Functions, High-School Edition

Transcription

1 Functions

2 What is a function?

3 Functions, High-School Edition

4 f(x) = x 4 5x source:

5 source:

6 Functions, High-School Edition In high school, functions are usually given as objects of the form f (x) = x3 +3 x x+7 1 x 137 What does a function do? It takes in as input a real number. It 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 flipuntil(int n) { int numheads = 0; int numtries = 0; while (numheads < n) { if (randomboolean()) numheads++; numtries++; } return 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) = x 2 + 3x 15

17 f (n)={ n/2 (n+1)/2 if n is even otherwise Functions like like these these are are called piecewise 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) = x 2 + 2x + 5 f( 3 ) = = 20 f( 0 ) = = 5 f( 3 ) =?

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

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

24 f(x) = x 2 + 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 The function function must must be be defined defined for for every every element element of of the the domain. domain. Domain Codomain The The output output of of the the function function must must always always be be in in the the codomain, codomain, but but not not all all elements elements of of the the codomain codomain must must be be produced produced as 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 The domain domain of of this this function function is is R. R. Any Any real real number number can can be be provided provided as as input. input. The The codomain codomain of of this this function function is is R. R. Everything Everything produced produced is is a a real real number, number, but but not not all all real real numbers numbers can can be be produced. produced. 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. Think of this like a function prototype in C++. Return Return type type Argument Argument type type Argument Argument type type Return Return type type B f(a arg); f : A B Function Function name name Function Function name name

30 The Official Rules for Functions Formally speaking, we say that f : A B if the following two rules hold. First, f must be obey its domain/codomain rules: a A. b B. f(a) = b ( Every input in A maps to some output in B. ) Second, f 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 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 g(x) = sin x, where g : R R 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 of the domain to some element of the codomain. Examples: f(n) = n + 1, where f : Z Z g(x) = sin x, where g : R R 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? م ح ر م ص ف ر ر بيع األو ل ر بيع الثاني ج مادى األولى ج مادى اآلخرة ر ج ب ش ع بان ر م ضان شو ال ذو القعدة عيد الفطر عيد األضحى A ذو الحجة B

36 Combining Functions

37 f : People Places g : Places Prices Keith Josh Mountain View San Francisco Far Too Much King's Ransom Divya Teresa Julian Redding, CA Barrow, AK Palo Alto A Modest Amount More Than You d Expect People Places Prices h : People Prices h(x) = g(f(x))

38 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))

39 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 The name name of of the the function function is is g g f. f. When When we we apply apply it it to to an an input input x, x, we we write write (g (g f)(x). f)(x). I I don't don't know know why, why, but but that's that's what what we we do. 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.

40 Time-Out for Announcements!

41 Applications are now open for Google s 2019 student scholarships. Selected student will each receive a $10,000 scholarship for the academic year and be invited to attend the annual Google Scholars' Retreat at the Googleplex next summer. Here is a list of open scholarships that you can share with your students: Women Techmakers Scholars Program ( is open to current undergraduate or graduate students who will be studying at a university in the United States or Canada for the academic year. We strongly encourage people who identify as female to apply. Deadline: December 6, 2018 Generation Google Scholarship ( is open to current undergraduate or graduate students who will be studying at a university in the United States or Canada for the academic year. We strongly encourage students from historically underrepresented groups, including Black/African American, Hispanic/Latino, American Indian, or Filipino/Native Hawaiian/Pacific Islander, to apply. Deadline: December 6, 2018 Google Lime Scholarship ( open to current undergraduate or graduate students with disabilities who will be studying at a university in the United States or Canada for the academic year. Deadline: December 9, 2018 Google Student Veterans of America Scholarship ( is open to current undergraduate or graduate student veterans who will be studying at a university in the United States for the academic year. Deadline: November 1, 2018 For questions and complete details on all of our scholarships, please visit

42 Problem Set Three Problem Set Two was due today at 2:30PM. Want to use late days? Submit the assignment by Sunday at 2:30PM. Problem Set Three goes out today. The checkpoint is due on Monday at 2:30PM. The remaining problems are due Friday at 2:30PM. Play around with binary relations, functions, their properties, and their applications! As usual, feel free to ask questions! Ask on Piazza! Stop by office hours!

43 Extra Practice Problems We ve posted a set of ten extra practice problems to the course website, with solutions. Feel free to use these to help solidify your understanding of concepts or to practice techniques. We re happy to discuss these questions in office hours there s no stakes on them.

44 Your Questions

45 I went to the career fair and just felt so inadequate after. How do you keep from letting those situations get to you? How should I decide between working for a smaller startup or a big company? How much do resumes matter in the tech industry? Are career fairs helpful and necessary? I didn t go because I was scared. I worry if I never attend one it will be difficult to get a job. What s do you think? I m I m going going to to give give a a meta-answer meta-answer to to this this one! one!

46 Can you answer a more diverse set of questions on here beyond career advice? This area seems to be overrepresented, and there's many interesting others! Yes, Yes, absolutely! absolutely! I ll I ll turn turn that that back back around around and and ask ask for for all all y all y all to to ask ask questions questions on on other other topics topics as as well. well.

47 How do I feel like I can belong in CS? I find it really hard when all of the CS professors I've had at Stanford are men I m I m going going to to split split this this into into two two separate separate questions questions one one about about belonging belonging and and one one about about faculty faculty diversity diversity rather rather than than take take on on the the two two together. together. Want Want to to talk talk about about the the two two in in tandem? tandem? Come Come talk talk to to me me in in office office hours, hours, or or me me and and let s let s set set up up a a time time to to chat chat oneon-one. oneon-one. I I want want to to hear hear your your thoughts! thoughts! We We have have some some stellar stellar female female faculty faculty that that we ve we ve hired hired in in the the past past few few years years and and I d I d be be happy happy to to make make introductions introductions if if you d you d like. like. You You can can also also get get involved involved in in the the hiring hiring process; process; the the department department solicits solicits input input from from students students about about faculty faculty candidates. candidates. As As for for community community the the fact fact that that CS CS is is so so big big cuts cuts both both ways. ways. The The major major is is fairly fairly heterogeneous heterogeneous and and there there are are lots lots of of communities communities to to reach reach out out to. to. Fun Fun fact fact there there are are more more women women majoring majoring in in CS CS than than in in any any other other major major at at Stanford. Stanford. Check Check out out groups groups like like WiCS, WiCS, SOLE, SOLE, SBSE, SBSE, etc. etc. if if this this is is something something you re you re interested interested in! in!

48 Back to CS103!

49 Special Types of Functions

50 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

51 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

52 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

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 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?

57 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 so n₀ = n₁, as required. 2n₀ = 2n₁,

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 so n₀ = n₁, as required. 2n₀ = 2n₁,

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. What Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₀ in turn means that n₀ N. N. n₁ n₁ N. 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, we'll we'll pick pick arbitrary arbitrary n₀, n₀, n₁ n₁ N where where f(n₀) f(n₀) = f(n₁), f(n₁), then then prove prove that that n₀ n₀ = n₁. n₁.

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 Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₁ in turn means that n₁ N. N. n₂ n₂ N. 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, we'll we'll pick pick arbitrary arbitrary n₀, n₀, n₁ n₁ 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 Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₁ in turn means that n₁ N. N. n₂ n₂ N. 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, we'll we'll pick pick arbitrary arbitrary n₀, n₀, n₁ n₁ 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 Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₁ in turn means that n₁ N. N. n₂ n₂ N. 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, we'll we'll pick pick arbitrary arbitrary n₁, n₁, n₂ n₂ N where where f(n₁) f(n₁) = f(n₂), f(n₂), then then prove prove that that n₁ n₁ = n₂. n₂.

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 Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₁ in turn means that n₁ N. N. n₂ n₂ N. 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, we'll we'll pick pick arbitrary arbitrary n₁, n₁, n₂ n₂ 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 Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₁ in turn means that n₁ N. N. n₂ n₂ N. 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, we'll we'll pick pick arbitrary arbitrary n₁, n₁, n₂ n₂ 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 Since What does f(n₀) does it = f(n₁), it mean mean for we see for the that the function function f f to to be be injective? 2n₀ + 7 = 2n₁ + 7. This n₁ in turn means that n₁ N. N. n₂ n₂ N. 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, we'll we'll pick pick arbitrary arbitrary n₁, n₁, n₂ n₂ 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 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 so n₀ = n₁, as required. 2n₀ = 2n₁,

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(n₁) = f(n₂), we see that 2n₁ + 7 = 2n₂ + 7. This in turn means that so n₀ = n₁, as required. 2n₀ = 2n₁,

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 so n₀ = n₁, as required. 2n₁ = 2n₂

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 so n₁ = n₂, as required. 2n₁ = 2n₂

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 so n₁ = n₂, as required. 2n₁ = 2n₂ Good Good exercise: exercise: Repeat Repeat this this proof proof using using the the other other definition definition of of injectivity! injectivity!

71 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 and f(x₀) = f(-1) = (-1) 4 = 1 f(x₁) = f(1) = 1 4 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

72 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 and f(x₀) = f(-1) = (-1) 4 = 1 f(x₁) = f(1) = 1 4 = 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) = x 4. 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 = x₀ -1 and = +1. Then x₀ Z. Z. x₁ x₁ Z. Z. (x₀ (x₀ x₁ x₁ f(x₀) f(x₀) f(x₁)) f(x₁)) f(x₀) = f(-1) = (-1) 4 = 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 x₀ Z. Z. x₁ x₁ f(x₁) Z. = Z. (x₀ f(1) (x₀ = x₁ x₁ 1 4 = f(x₀) f(x₀) 1, f(x₁)) 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) x₀ x₀ = f(x₁) Z. Z. x₁ x₁ even Z. Z. though (x₀ (x₀ x₁ x₀ x₁ (f(x₀) x₁, (f(x₀) as required. f(x₁))) 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?

74 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₀ Z. Z. x₁ x₁ f(x₁) Z. = Z. (x₀ f(1) (x₀ = x₁ x₁ 1 4 = f(x₀) f(x₀) 1, f(x₁)) 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) x₀ x₀ = f(x₁) Z. Z. x₁ x₁ even Z. Z. though (x₀ (x₀ x₁ x₀ x₁ (f(x₀) x₁, (f(x₀) as required. f(x₁))) 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) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₀ Z. Z. x₁ x₁ f(x₁) Z. = Z. (x₀ f(1) (x₀ = x₁ x₁ 1 4 = f(x₀) f(x₀) 1, f(x₁)) 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) x₀ x₀ = f(x₁) Z. Z. x₁ x₁ even Z. Z. though (x₀ (x₀ x₁ x₀ x₁ (f(x₀) x₁, (f(x₀) as required. f(x₁))) 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) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₀ Z. Z. x₁ x₁ f(x₁) Z. Z. = (x₀ (x₀ f(1) = x₁ x₁ 1 4 = f(x₀) f(x₀) 1, f(x₁)) 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) x₀ x₀ = f(x₁) Z. Z. x₁ x₁ even Z. though Z. (x₀ (x₀ x₁ x₀ x₁ (f(x₀) x₁, (f(x₀) as required. f(x₁))) 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) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₀ x₀ = f(x₁) Z. Z. x₁ x₁ even Z. though Z. (x₀ (x₀ x₁ x₀ x₁ (f(x₀) x₁, (f(x₀) as required. f(x₁))) 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) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₀ x₀ = f(x₁) Z. Z. x₁ x₁ even Z. though Z. (x₀ (x₀ x₁ x₀ x₁ (f(x₀) x₁, (f(x₀) as required. f(x₁))) 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) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₁ x₁ = f(x₁) Z. Z. x₂ x₂ even Z. though Z. (x₁ (x₁ x₂ x₀ x₂ (f(x₁) x₁, (f(x₁) as required. f(x₂))) 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?

80 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₁ x₁ = f(x₁) Z. Z. x₂ x₂ even Z. though Z. (x₁ (x₁ x₂ x₀ x₂ (f(x₁) x₁, (f(x₁) as required. f(x₂))) 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?

81 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₁ x₁ = f(x₁) Z. Z. x₂ x₂ even Z. though Z. (x₁ (x₁ x₂ x₀ x₂ (f(x₁) x₁, (f(x₁) as required. f(x₂))) 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?

82 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₁ x₁ = f(x₁) Z. Z. x₂ x₂ even Z. though Z. (x₁ (x₁ x₂ x₀ x₂ (f(x₁) x₁, (f(x₁) as required. f(x₂))) 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) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₁ x₁ = f(x₁) Z. Z. x₂ x₂ even Z. though Z. (x₁ (x₁ x₂ x₀ x₂ (f(x₁) x₁, (f(x₁) as required. f(x₂))) 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?

84 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 = x₁ -1 and x₁ = +1. Then x₁ Z. Z. x₂ x₂ Z. Z. (x₁ (x₁ x₂ x₂ f(x₁) f(x₁) f(x₂)) f(x₂)) f(x₀) = f(-1) = (-1) 4 = 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 x₁ Z. Z. x₂ x₂ f(x₁) Z. Z. = (x₁ (x₁ f(1) = x₂ x₂ 1 4 = f(x₁) f(x₁) 1, f(x₂)) 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) x₁ x₁ = f(x₁) Z. Z. x₂ x₂ even Z. though Z. (x₁ (x₁ x₂ x₀ x₂ (f(x₁) x₁, (f(x₁) as required. f(x₂))) 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?

85 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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 and f(x₀) = f(-1) = (-1) 4 = 1 f(x₁) = f(1) = 1 4 = 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) = x 4. 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 and f(x₀) = f(-1) = (-1) 4 = 1 f(x₁) = f(1) = 1 4 = 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) = x 4. 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. and f(x₁) = f(-1) = (-1) 4 = 1 f(x₁) = f(1) = 1 4 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

88 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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. and f(x₁) = f(-1) = (-1) 4 = 1 f(x₂) = f(1) = 1 4 = 1, so f(x₀) = f(x₁) even though x₀ x₁, as required.

89 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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. and f(x₁) = f(-1) = (-1) 4 = 1 f(x₂) = f(1) = 1 4 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

90 Injective Functions Theorem: Let f : Z N be defined as f(x) = x 4. 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. and f(x₁) = f(-1) = (-1) 4 = 1 f(x₂) = f(1) = 1 4 = 1, so f(x₁) = f(x₂) even though x₁ x₂, as required.

91 Injections and Composition

92 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.

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)). 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.

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)). 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.

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 a₁ g(f(a₁)) A. A. a₂ a₂ g(f(a₂)), A. A. as ((g ((g required. f)(a₁) 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)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁, a₂ a₂ A 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)). 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 a₁ g(f(a₁)) A. A. a₂ a₂ g(f(a₂)), A. A. as ((g ((g required. f)(a₁) 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)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁, a₂ a₂ A where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

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)). 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 a₁ g(f(a₁)) A. A. a₂ a₂ g(f(a₂)), A. A. as ((g ((g required. f)(a₁) 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)) Therefore, Therefore, we ll we ll choose choose an an arbitrary arbitrary a₁, a₁, a₂ a₂ A where where a₁ a₁ a₂, a₂, then then prove prove that that (g (g f)(a₁) f)(a₁) (g (g f)(a₂). f)(a₂).

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 g(f(a₁)) g(f(a₂)), as required.

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 g(f(a₁)) g(f(a₂)), as required.

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 g(f(a₁)) g(f(a₂)), How is as is (g required. (g f)(x) f)(x) defined? 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 g(f(a₁)) g(f(a₂)), How is as is (g required. (g f)(x) f)(x) defined? 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 How g(f(a₁)) g(f(a₂)), How is as is (g required. (g f)(x) f)(x) defined? 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)).

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 How g(f(a₁)) g(f(a₂)), How is as is (g required. (g f)(x) f)(x) defined? 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)).

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. a₁ f(a₁) g(f(a₁)) a₂ f(a₂) A B C g(f(a₂))

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. a₁ f(a₁) g(f(a₁)) a₂ f(a₂) A B C g(f(a₂))

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. a₁ f(a₁) g(f(a₁)) a₂ f(a₂) A B C g(f(a₂))

109 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. a₁ f(a₁) g(f(a₁)) a₂ f(a₂) A B C g(f(a₂))

110 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. a₁ f(a₁) g(f(a₁)) a₂ f(a₂) A B C Great Great exercise: exercise: Repeat Repeat this this proof proof using using the the other g(f(a₂)) other definition definition of of injectivity. injectivity.

111 Another Class of Functions

112 Lassen Peak California Mt. Hood Washington Mt. St. Helens Oregon Mt. Shasta

113 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 output, there's an input that produces it ) A function with this property is called a surjection. How does this compare to our first rule of functions?

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. Let x = 2y. Then we see that So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = 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. Let x = 2y. Then we see that So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = 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. Let What x What = 2y. does does Then it it we mean mean see that for for f to to be be surjective? f(x) = f(2y) = 2y / 2 = y. y So f(x) = y, as required. y R. R. x x R. R. f(x) f(x) = y Therefore, we'll we'll choose choose an an arbitrary arbitrary y R, R, then then prove prove that that there there is is some some x R 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. Let What x What = 2y. does does Then it it we mean mean see that for for f to to be be surjective? f(x) = f(2y) = 2y / 2 = y. y So f(x) = y, as required. y R. R. x x R. R. f(x) f(x) = y Therefore, we'll we'll choose choose an an arbitrary arbitrary y R, R, then then prove prove that that there there is is some some x R 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. Let What x What = 2y. does does Then it it we mean mean see that for for f to to be be surjective? f(x) = f(2y) = 2y / 2 = y. y So f(x) = y, as required. y R. R. x x R. R. f(x) f(x) = y Therefore, we'll we'll choose choose an an arbitrary arbitrary y R, R, then then prove prove that that there there is is some some x R 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. Let What x What = 2y. does does Then it it we mean mean see that for for f to to be be surjective? f(x) = f(2y) = 2y / 2 = y. y So f(x) = y, as required. y R. R. x x R. R. f(x) f(x) = y Therefore, we'll we'll choose choose an an arbitrary arbitrary y R, R, then then prove prove that that there there is is some some x R 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 What x What = 2y. does does Then it it we mean mean see that for for f to to be be surjective? f(x) = f(2y) = 2y / 2 = y. y So f(x) = y, as required. y R. R. x x R. R. f(x) f(x) = y Therefore, we'll we'll choose choose an an arbitrary arbitrary y R, R, then then prove prove that that there there is is some some x R where where f(x) f(x) = y. y.

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 What x What = 2y. does does Then it it we mean mean see that for for f to to be be surjective? f(x) = f(2y) = 2y / 2 = y. y So f(x) = y, as required. y R. R. x x R. R. f(x) f(x) = y Therefore, we'll we'll choose choose an an arbitrary arbitrary y R, R, then then prove prove that that there there is is some some x R where where f(x) f(x) = y. y.

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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

128 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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

129 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 So f(x) = y, as required. f(x) = f(2y) = 2y / 2 = y.

130 Composing Surjections

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)) = 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.

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)) = 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.

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)) = c. What What does does it it mean mean for for g f f : : A C to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is some b B such c c that C. C. a a g(b) = A. A. c. (g (g Similarly, f)(a) f)(a) since = c 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 Therefore, Therefore, we'll we'll choose choose arbitrary arbitrary c c C and and prove prove that that there there is is some some a a g(f(a)) A such such = that that g(b) (g (g = c, f)(a) 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. What What does does it it mean mean for for g f f : : A C to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is some b B such c c that C. C. a a g(b) = A. A. c. (g (g Similarly, f)(a) f)(a) since = c 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 Therefore, Therefore, we'll we'll choose choose arbitrary arbitrary c c C and and prove prove that that there there is is some some a a g(f(a)) A such such = that that g(b) (g (g = c, f)(a) f)(a) = = c. 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. What What does does it it mean mean for for g f f : : A C to to be be surjective? surjective? Consider any c C. Since g : B C is surjective, there is some b B such c c that C. C. a a g(b) = A. A. c. (g (g Similarly, f)(a) f)(a) since = c 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 Therefore, Therefore, we'll we'll choose choose arbitrary arbitrary c c C and and prove prove that that there there is is some some a a g(f(a)) A such such = that that g(b) (g (g = c, f)(a) f)(a) = = c. 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.

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. c a A B C b

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. c a A B C b

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 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. c a A B C b

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. 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. c a A B C b

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. 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. Then we see that g(f(a)) = g(b) = c, which is what we needed to show. c a A B C b

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. 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. Then we see that g(f(a)) = g(b) = c, which is what we needed to show. c a A B C b

145 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?

146 Katniss Everdeen Elsa Hermione Granger

147 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.

148 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.

149 Inverse Functions

150 Katniss Everdeen Elsa Hermione Granger

151 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

152 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

153 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

154 Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

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

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