The Category of Sets

Transcription

1 The Cateory o Sets Hans Halvorson October 16, 2016 [[Note to students: this is a irst drat. Please report typos. A revised version will be posted within the next couple o weeks.]] 1 Introduction The aim o metatheory is to theorize about theories. For simplicity, let s use M to denote this hypothetical theory about theories. Thus, M is not the object o our study, it is the tool we will use to study other other theories. And yet, it miht be helpul to ive you a sort o user s manual or M. That s the aim o this chapter. Let s think about what we hope to do with M. We want to be able to talk about theories, which are supposed to be collections o thins, or better, structured collections o thins. In the irst instance, we will think o theories as structured collections o sentences. 1 What s more, sentences themselves are structured collections o symbols. Fortunately, we won t need to press the inquiry urther into the question o the nature o symbols. It will suice to assume that there are enouh symbols, and that there is some primitive notion o identity o symbols. For example, I assume that you understand that p is the same symbol as p, and is dierent rom q. Fortunately, much o the theory we need was worked out in previous enerations. At the beinnin o the 20th century, much eort was spent on ormulatin a theory o abstract collections (i.e. sets) that would be adequate to serve as a oundation or all o mathematics. Amazinly, the theory o sets can be ormalized in predicate loic with only one symbol o non-loical vocabulary, a binary relation symbol. In the resultin irst-order theory usually called Zermelo-Frankel set theory the quantiiers can be thouht o as ranin over sets, and the relation symbol can be used to deine urther notions such as subset, Cartesian products o sets, unctions rom one set to another, etc.. 1 I don t mean to be bein the question here about what a theory is. We could just as well think o a theory as a structured collection o models. And just as sentences can be broken down into smaller components until we reach undeined primitives, so models can be broken down into smaller components until we reach undeined primitives. In both cases, metatheory bottoms out in undeined primitives. We can call these primitives symbols, or sets, or anythin else we want. But the name we choose doesn t aect the inerences we re permitted to draw. 1

2 Set theory can be presented inormally (sometimes called naive set theory ), or ormally ( axiomatic set theory ). In both cases, the relation is primitive. However, we re oin to approach thins rom a dierent anle. We re not concerned as much with what sets are, but with what we can do with them. Thus, I ll present a version o ETCS, the elementary theory o the cateory o sets. 2 Here elementary theory indicates that this theory can be ormalized in elementary (i.e. irst-order) loic. The phrase cateory o sets indicates that this theory treats the collection o sets as a structured object a cateory consistin o sets and unctions between them. Axiom 1: Sets is a cateory Sets is a cateory, i.e. it consists o two kinds o thins: objects, which we call sets, and arrows, which we call unctions. To say that Sets is a cateory means that: 1. Every unction has a domain set d 0 and a codomain set d 1. We write : Y to indicate that = d 0 and Y = d Compatible unctions can be composed. For example, i : Y and : Y Z are unctions, then : Z is a unction. (We requently abbreviate as.) 3. Composition o unctions is associative: h ( ) = (h ) when all these compositions are deined. 4. For each set, there is a unction 1 : that acts as a let and riht identity relative to composition. Discussion. I our oal was to ormalize ETCS riorously in irst-order loic, we miht use two-sorted loic, with one sort or sets, and one sort or unctions. The primitive vocabulary o this theory would include symbols, d 0, d 1, 1, but it would not include the symbol. In other words, containment is not a primitive notion o ETCS. Set theory makes requent use o bracket notation, such as: {n N n > 17}. These symbols should be read as, the set o n in N such that n > 17. Similarly, {x, y} desinates a set consistin o elements x and y. But so ar, we have no 2 All credit to William Lawvere or introducin this approach to set theory. For a entle introduction, see his Sets or Mathematics. 2

3 rules or reasonin about such sets. In the ollowin sections, we will radually add axioms until it becomes clear which rules o inerence are permitted vis-a-vis sets. Suppose or a moment that we understand the bracket notation, and suppose that and Y are sets. Then iven an element x, and an element y Y, we can take the set {x, {x, y}} as an ordered pair consistin o x and y. The pair is ordered because x and y play asymmetric roles: the element x occurs by itsel, as well as with the element y. I we could then ather toether these ordered pairs into a sinle set, we would desinate it by Y, which we call the Cartesian product o and Y. The Cartesian product construction should be amiliar rom hih school mathematics. For example, the plane (with x and y coordinates) is the Cartesian product o two copies o the real number line. In typical presentations o set theory, the existence o product sets is derived rom other axioms. Here we will proceed in the opposite direction: we will take the notion o a product set as primitive. Axiom 2: Cartesian products For any two sets and Y, there is a set Y, and unctions π 0 : Y and π 1 : Y Y, such that: or any other set Z and unctions : Z and : Z Y, there is a unique unction, : Z Y such that π 0, = and π 1, =. Here the anle brackets, are not intended to indicate anythin about the internal structure o the denoted unction. This notation is chosen merely to indicate that, is uniquely determined by and. The deinin conditions o a product set can be visualized by means o an arrow diaram. Z, Y π 0 π 1 Here each node represents a set, and arrows between nodes represent unctions. The dashed arrow is meant to indicate that the axiom asserts the existence o such an arrow (dependent on the existence o the other arrows in the diaram). Discussion. There is a close analoy between the deinin conditions o a Cartesian product and the introduction and elimination rules or conjunction. I φ ψ is a conjunction, then there are arrows (i.e. derivations) φ ψ φ and φ ψ ψ. That s the elimination rule. 3 Moreover, or any sentence θ, i there are deriva- 3 Here I m intentionally bein ambiuous between the relation and the connective. Y 3

4 tions θ φ and θ ψ, then there is a unique derivation θ φ ψ. That s the introduction rule. Deinition. Let γ and γ be paths o arrows in a diaram that bein and end at the same node. We say that γ and γ commute just in case the composition o the unctions alon γ is equal to the composition o the unctions alon γ. We say that the diaram as a whole commutes just in case any two paths between nodes are equal. Thus, or example, the product diaram above commutes. The unctions π 0 : Y and π 1 : Y Y are typically called projections o the product. What eatures do these projections have? Beore we say more on that score, let s pause to talk about eatures o unctions. You will recall rom secondary school that unctions can be one-to-one, or onto, or continuous, etc.. For bare sets, there is no notion o continuity o unctions, per se. And, with only the irst two axioms in place, we do not yet have the means to deine what it means or a unction to be one-to-one or onto. Indeed, recall that a unction : Y is typically said to be one-to-one just in case (x) = (y) implies x = y or any two points x and y o. But we don t yet have a notion o points! Nonetheless, there are point-ree surroates or the notions o bein one-toone and onto. Deinition. A unction : Y is said to be a monomorphism just in case or any two unctions, h : Z, i = h then = h. Deinition. A unction : Y is said to be a epimorphism just in case or any two unctions, h : Y Z, i = h then = h. We will requently say,... is monic as shorthand or... is a monomorphism, and... is epi or... is an epimorphism. Deinition. A unction : Y is said to be an isomorphism just in case there is a unction : Y such that = 1 and = 1 Y. I there is an isomorphism : Y, we say that and Y are isomorphic, and we write = Y. Exercise 1.1. Show the ollowin: 1. I is monic, then is monic. 2. I is epi, then is an epi. 3. I and are monic, then is monic. 4. I and are epi, then is epi. 5. I is an isomorphism, then is epi and monic. The analoy is more clear i we use the symbol the latter. 4

5 Proposition 1.2. Suppose that both (W, p 0, p 1 ) and (Z, q 0, q 1 ) are Cartesian products o and Y. Then there is an isomorphism : W Z such that q 0 = p 0 and q 1 = p 1. Proo. Since (Z, q 0, q 1 ) is a Cartesian product, there is a unique unction p 0, p 1 : W Z such that q 0 p 0, p 1 = p 0 and q 1 p 0, p 1 = p 1. There is a similar unction q 0, q 1 : Z W. We claim that these unctions are inverse to each other. Indeed, q 0 p 0, p 1 q 0, q 1 = p 0 q 0, q 1 = q 0, and similarly, q 1 p 0, p 1 q 0, q 1 = q 1. Thus, by the uniqueness clause in the deinition o Cartesian products, A similar arument shows that p 0, p 1 q 0, q 1 = 1 Z. q 0, q 1 p 0, p 1 = 1 W. Thereore, = p 0, p 1 is the requisite isomorphism. Deinition. I is a set, we let δ : denote the unique arrow 1, 1 iven by the deinition o. We call δ the diaonal o, or the equality relation on. Note that δ is monic, since π 0 δ = 1 is monic. Deinition. Suppose that : W Y and : Z are unctions. Consider the ollowin diaram: W W q 0 q 1 Y Y Z Z π 0 π 1 We let = q 0, q 1 be the unique unction rom W to Y Z such that π 0 ( ) = q 0, π 1 ( ) = q 1. Proposition 1.3. Suppose that : A B and : B C are unctions. Then 1 ( ) = (1 ) (1 ). Proo. The ollowin diaram commutes: A A 1 1 B B 1 1 C C Thus, (1 ) (1 ) has the deinin properties o 1 ( ). 5

6 Exercise 1.4. Show that 1 1 Y = 1 Y. Deinition. Let be a ixed set. Then induces two mappins, as ollows: 1. A mappin Y Y o sets to sets. 2. A mappin 1 o unctions to unctions. That is, i : Y Z is a unction, then 1 : Y Z is a unction. By the previous results, the second mappin is compatible with the composition structure on arrows. In this case, we call the pair o mappins a unctor rom Sets to Sets. Exercise 1.5. Suppose that : Y is a unction. Show that the ollowin diaram commutes. Y δ δ Y Y Y We will now recover the idea that sets consist o points by requirin the existence o a sinle-point set 1, which plays the privileed role o determinin identity o unctions. Axiom 3: Terminal Object There is a set 1 with the ollowin two eatures: 1. For any set, there is a unique unction β 1 In this case, we say that 1 is a terminal object or Sets. 2. For any sets and Y, and unctions, : Y, i x = x or all unctions x : 1, then =. In this case, we say that 1 is a separator or Sets. Exercise 1.6. Show that i and Y are terminal objects in a cateory, then = Y. Deinition. We write x to indicate that x : 1 is a unction. In this case, we say that x is an element o, and we write x. I : Y is a unction, we sometimes write (x) or x. With this notation, the statement that 1 is a separator says: = is and only i (x) = (x), or all x. 6

7 Discussion. In ZF set theory, equality between unctions is completely determined by equality between sets. Indeed, in ZF, unctions, : Y are deined to be certain subsets o Y ; and subsets o Y are deined to be equal just in case they contain the same elements. In the ETCS approach to set theory, equality between unctions is primitive, and Axiom 3 stipulates that this equality can be detected by checkin elements. Some miht see this dierence as aruin in avor o ZF: it is more parsimonious, because it derives = rom somethin more undamental. However, the deender o ETCS miht claim in reply that her theory deines x y rom somethin more undamental. Which is really more undamental, equality between arrows (unctions), or containment o objects (sets)? We ll leave that or other philosophers to think about. Exercise 1.7. Show that any unction x : 1 is monic. Proposition 1.8. A set has exactly one element i and only i = 1. Proo. The terminal object 1 has exactly one element, since there is a unique unction 1 1. Suppose now that has exactly one element x : 1. We will show that is a terminal object. First, or any set Y, there is a unction x β Y rom Y to. Now suppose that, are unctions rom Y to such that. By Axiom 3, there is an element y Y such that y y. But then has more than one element, a contradiction. Thereore, there is a unique unction rom Y to, and is a terminal object. Proposition 1.9. In any cateory with Cartesian products, i 1 is a terminal object then 1 =, or any object. Proo. Let π 0 : 1 and π 1 : 1 1 be the projections. Then 1, β is a unction rom to 1. We claim that π 0 is a let and riht inverse or 1, β. Since π 0 1, β = 1, π 0 is a let inverse or 1, β. To show that π 0 is a riht inverse or 1, β, we use the act that 1 1 is the unique unction rom 1 to itsel such that π = π 0, and π = π 1. But we also have, π 0 ( 1, β π 0 ) = π 0, and since π 1 is the unique unction rom 1 to 1, π 1 ( 1, β π 0 ) = π 1. Thus, 1, β π 0 = 1 1, and π 0 is riht inverse to 1, β. Proposition Let a and b be elements o Y. Then a = b i and only i π 0 (a) = π 0 (b) and π 1 (a) = π 1 (b). 7

8 Proo. Suppose that π 0 (a) = π 0 (b) and π 1 (a) = π 1 (b). By the uniqueness property o the product, there is a unique unction c : 1 Y such that π 0 (c) = π 0 (a) and π 1 (c) = π 1 (a). Since a and b both satisy this property, a = b. Note. The previous proposition justiies the use o the notation Y = { x, y x, y Y }. Here the identity condition or ordered pairs is iven by x, y = x, y i x = x and y = y. Proposition Let ( Y, π 0, π 1 ) be the Cartesian product o and Y. I Y is non-empty, then π 0 is an epimorphism. Proo. Suppose that Y is non-empty, and that y : 1 Y is an element. Let β : 1 be the unique map, and let = y β. Then 1, : Y such that π 0 1, = 1. Since 1 is epi, π 0 is epi. Deinition. We say that : Y is injective just in case: or any x, y i (x) = (y), then x = y. Written more ormally: x y[(x) = (y) x = y] Note. Injective is synonymous with one-to-one. Exercise Let : Y be a unction. Show that i is monic, then is injective. Proposition Let : Y be a unction. I is injective then is monic. Proo. Suppose that is injective, and let, h : A be unctions such that = h. Then or any a A, we have ((a)) = (h(a)). Since is injective, (a) = h(a). Since a was an arbitrary element o A, Axiom 3 entails that = h. Thereore, is monic. Deinition. Let : Y be a unction. We say that is surjective just in case: or each y Y, there is an x such that (x) = y. Written ormally: And in diarammatic orm: y x[(x) = y] x 1 y Y Note. Surjective is synonymous with onto. 8

9 Exercise Show that i : Y is surjective then is an epimorphism. We will eventually establish that all epimorphisms are surjective. However, irst we need a couple more axioms. Given a set, and some deinable condition φ on, we would like to be able to construct a subset consistin o those elements in that satisy φ. The usual notation here is {x φ(x)}, which we read as, the x in such that φ(x). But the important question is: which eatures φ do we allow? As an example o a deinable condition φ, consider the condition o, havin the same value under the unctions and, that is, φ(x) just in case (x) = (x). We call the subset {x (x) = (x)} the equalizer o and. Axiom 4: Equalizers Suppose that, : Y are unctions. Then there is a set E and a unction m : E with the ollowin property: m = m, and or any other set F and unction h : F, i h = h, then there is a unique unction k : F E such that mk = h. k E Y We call (E, m) an equalizer o and. F m h Exercise Suppose that (E, m) and (E, m ) are both equalizers o and. Show that there is an isomorphism k : E E. Deinition. Let A, B, C be sets, and let : A C and : B C be unctions. We say that actors throuh just in case there is a unction h : B A such that h =. Exercise Let, : Y, and let m : E be the equalizer o and. Let x. Show that x actors throuh m i and only i (x) = (x). Proposition In any cateory, i (E, m) is the equalizer o and. Then m is a monomorphism. Proo. Let x, y : Z E such that mx = my. Since mx = mx, there is a unique arrow z : Z E such that mz = mx. Since both mx = mx and my = mx, it ollows that x = y. Thereore, m is monic. Deinition. Let : Y be a unction. We say that is a reular monomorphism just in case is the equalizer o a pair o arrows, h : Y Z. 9

10 Exercise Show that i is an epimorphism and a reular monomorphism, then is an isomorphism. In other approaches to set theory, one uses to deine a relation o inclusion between sets: Y x(x x Y ). We cannot deine this exact notion in our approach since, or us, elements are attached to some particular set. However, or typical applications, every set under consideration will come equipped with a canonical monomorphism m : U, where U is some ixed set. Thus, it will suice to consider a relativized notion. Deinition. A subobject or subset o a set is a set B and a monomorphism m : B, called the inclusion o B in. Given two subsets m : B and n : A, we say that B is a subset o A (relative to ), written B A just in case there is a unction k : B A such that nk = m. When no conusion can result, we omit and write B A. Let m : B Y be monic, and let : Y. diaram: 1 p k 0 (B) B Y mp 1 Consider the ollowin where 1 (B) is deined as the equalizer o π 0 and mp 1. Intuitively, we have 1 (B) = { x, y B (x) = y} = { x, y Y (x) = y and y B} = {x (x) B}. Now we veriy that 1 (B) is a subset o. Proposition The unction p 0 k : 1 (B) is monic. Proo. To simpliy notation, let E = 1 (B). Let x, y : Z E such that p 0 kx = p 0 ky. Then p 0 kx = p 0 ky, and hence mp 1 kx = mp 1 ky. Since m is monic, p 1 kx = p 1 ky. Thus, kx = ky. (The identity o a unction into B is determined by the identity o its projections onto and B.) Since k is monic, x = y. Thereore, p 0 k is monic. Deinition. Let m : B be a subobject, and let x : 1. We say that x B just in case x actors throuh m as ollows: B 1 Proposition Let A B. I x A then x B. x m 10

11 Proo. A k B 1 x 1 Recall that x 1 (B) means: x : 1 actors throuh the inclusion o 1 (B) in. Consider the ollowin diaram: 1 x 1 (B) p B m Y m First look just at the lower-riht square. This square commutes, in the sense that ollowin the arrows rom 1 (B) clockwise ives the same answer as ollowin the arrows rom 1 (B) counterclockwise. The square has another property: or any set Z, and unctions : Z and h : Z B, there is a unique unction k : Z 1 (B) such that m k = and pk = h. [To understand better, draw a picture!] When a commutin square has this property, then it s said to be a pullback. Proposition Let : Y, and let B Y. Then x 1 (B) i and only i (x) B. Proo. I x 1 (B), then there is an arrow ˆx : 1 1 (B) such that m ˆx = x. Thus, x = mpˆx, which entails that the element (x) Y actors throuh B, i.e. (x) B. Conversely, i (x) B, then since the square is a pullback, x : 1 actors throuh 1 (B), i.e. x 1 (B). Deinition. Given unctions : Z and : Y Z, we deine Z Y = { x, y Y (x) = (y)}. In other words, Z Y is the equalizer o π 0 and π 1. The set Z Y, toether with the unctions π 0 : Z Y and π 1 : Z Y Y is called the pullback o and, alternatively the ibered product o and. The pullback o and has the ollowin distinuishin property: or any set A, and unctions h : A and k : A Y such that h = k, there is a 11

12 unique unction j : A Z Y such that π 0 j = h and π 1 j = k. A k h Z Y Y π 0 π 1 Z The ollowin is an interestin special case o a pullback. Deinition. Let : Y be a unction. Then the kernel pair o is the pullback Y, with projections p 0 : Y and p 1 : Y. Intuitively, Y is the relation, havin the same imae under. Written in terms o braces, Y = { x, x (x) = (x )}. In particular, is injective i and only i, havin the same imae under is coextensive with the equality relation on. That is, Y = { x, x x }, which is the diaonal o. Exercise Let : Y be a unction, and let p 0, p 1 : Y be the kernel pair o. Show that the ollowin are equivalent: 1. is a monomorphism. 2. p 0 and p 1 are isomorphisms. 3. p 0 = p 1. 2 Truth values and subsets Axiom 5: Truth-value object There is a set Ω with the ollowin eatures: 1. Ω has exactly two elements, which we denote by t : 1 Ω and : 1 Ω. 2. For any set, and subobject m : B, there is a unique unction χ B : Ω such that the ollowin diaram is a pullback: m B 1 χ B Ω t 12

13 In other words, B = {x χ B (x) = t}. Intuitively speakin, the irst part o Axiom 5 says that Ω is a two-element set, say Ω = {, t}. The second part o Axiom 5 says that Ω classiies the subobjects o a set. That is, each subobject m : B corresponds to a unique characteristic unction χ B : {, t} such that χ B (x) = t i and only i x B. The terminal object 1 is a set with one element. Thus, it should be the case that 1 has two subsets, the empty set and 1 itsel. Proposition 2.1. The terminal object 1 has exactly two subobjects. Proo. By Axiom 5, subobjects o 1 correspond to unctions 1 Ω, that is, to elements o Ω. By Axiom 5, Ω has exactly two elements. Thereore, 1 has exactly two subobjects. Obviously the unction t : 1 Ω corresponds to the subobject id 1 : 1 1. Can we say more about the subobject m : A 1 correspondin to the unction : 1 Ω? Intuitively, we should have A = {x 1 t = }, in other words, the empty set. To conirm this intuition, consider the pullback diaram: 1 x m k A 1 1 Ω Note that m and k must both be the unique unction rom A to 1, that is m = k = β A. Suppose that A is nonempty, i.e. there is a unction x : 1 A. Then β A x is the identity 1 1, and since the square commutes, t =, a contradiction. Thereore, A has no elements. Exercise 2.2. Show that Ω Ω has exactly our elements. We now use the existence o a truth-value object in Sets to demonstrate urther properties o unctions. Deinition. Let : Y be a unction. Then is said to be a reular monomorphism just in case there are unctions, h : Y Z such that is the equalizer o and h. Exercise 2.3. Show that in any cateory, i : Y is a reular monomorphism, then is monic. Proposition 2.4. Every monomorphism between sets is reular, i.e. an equalizer o a pair o parallel arrows. t 13

14 Proo. Let m : B be monic. By Axiom 5, the ollowin is a pullback diaram: B 1 m χ B A straihtorward veriication shows that m is the equalizer o β 1 t Ω and χ B : Ω. Thereore, m is reular monic. Students with some backround in mathematics miht assume that i a unction : Y is both a monomorphism and an epimorphism, then it is an isomorphism. However, that isn t true in all cateories! [For example, it s not true in the cateory o monoids.] Nonetheless, Sets is a special cateory, and in this case we have the result: Proposition 2.5. In Sets, i a unction is both a monomorphism and an epimorphism, then it is an isomorphism. Proo. In any cateory, i m is reular monic and epi, then m is an isomorphism (Exercise 1.18). Deinition. Let : Y be a unction, and let y Y. The iber over y is the subset 1 {y} o iven by the ollowin pullback: Ω 1 {y} 1 t y Y Proposition 2.6. Let p : Y. I p is not a surjection, then there is a y 0 Y such that the iber p 1 {y 0 } is empty. Proo. Since p is not a surjection, there is a y 0 Y such that or all x, p(x) y 0. Now consider the pullback: 1 z p 1 {y 0 } 1 m y 0 I there were a morphism z : 1 p 1 {y 0 }, then we would have p(m(z)) = y 0, a contradiction. Thereore, p 1 {y 0 } is empty. Proposition 2.7. In Sets, epimorphisms are surjective. p Y 14

15 Proo. Suppose that p : Y is not a surjection. Then there is a y 0 Y such that or all x, p(x) y 0. Since 1 is terminal, the morphism y 0 : 1 Y is monic. Consider the ollowin diaram: 1 x p 1 {y 0 } 1 1 p y 0 t Y Ω Here is the characteristic unction o {y 0 }; by Axiom 5, is the unique unction that makes the riht hand square a pullback. Let x be arbitrary. I we had (p(x)) = t, then there would be an element x p 1 {y 0 }, in contradiction with the act that the latter is empty (Proposition 2.6). By Axiom 5, either (p(x)) = t or (p(x)) = ; thereore, (p(x)) =. Now let h be the composite Y 1 Ω. Then, or any x, we have h(p(x)) =. Since p and h p aree on arbitrary x, we have p = h p. Since h, it ollows that p is not an epimorphism. In a eneral cateory, there is no uarantee that an epimorphism pulls back to an epimorphism. However, in Sets, we have the ollowin: Proposition 2.8. In Sets, the pullback o an epimorphism is an epimorphism. Proo. Suppose that : Y Z is epi, and let x. Consider the pullback diaram: 1 y x Y q 0 q 1 By Proposition 2.7, is surjective. In particular, there is a y Y such that (y) = (x). Since the diaram is a pullback, there is a unique x, y : 1 such that q 0 x, y = x and q 1 x, y = y. Thereore, q 0 is surjective, and hence epi. Proposition 2.9. I : Y and : W Z are epimorphisms, then so is : W Y Z. Proo. Since = ( 1) (1 ), it will suice to show that 1 is epi Z 15

16 when is epi. Now, the ollowin diaram is a pullback: W Y W p 0 1 By Proposition 2.8, i is epi, then 1 is epi. Suppose that : Y is a unction, and that p 0, p 1 : Y is the kernel pair o. Suppose also that h : E Y is a unction, that q 0, q 1 : E Y E E is the kernel pair o h, and that : E is an epimorphism. Then there is a unique unction b : Y E Y E, such that q 0 b = p 0 and q 1 b = p 1. p 0 Y Y b E Y E p 0 p 1 q 0 q 1 An arument similar to the one above shows that b is an epimorphism. We will use this act below to describe the properties o epimorphisms in Sets. 3 Relations 3.1 Equivalence relations and equivalence classes A relation R on a set is a subset o ; i.e. a set o ordered-pairs. A relation is said to be an equivalence relation just in case it is relexive, symmetric, and transitive. One particular way that equivalence relations on arise is rom unctions with as domain: iven a unction : Y, let say that x, y R just in case (x) = (y). [Sometimes we say that, x and y lie in the same iber over Y. ] Then R is an equivalence relation on. Given an equivalence relation R on, and some element x, let [x] = {y x, y R} denote the set o all elements o that are equivalent to. We say that [x] is the equivalence class o x. It s straihtorward to show that or any x, y, either [x] = [y] or [x] [y] =. Moreover, or any x, we have x [x]. Thus the equivalence classes orm a partition o into disjoint subsets. We d like now to be able to talk about the set o these equivalence classes, i.e. somethin that miht intuitively be written as {[x] x }. The ollowin axiom uarantees the existence o such a set, called /R, and a canonical mappin q : /R that takes each element x to its equivalence class [x] /R. Our next axiom uarantees the existence o the set o equivalence classes. E Y h 16

17 Axiom 6: Equivalence classes Let R be an equivalence relation on. Then there is a set /R, and a unction q : /R with the properties: 1. x, y R i and only i q(x) = q(y). 2. For any set Y and unction : Y that is constant on equivalence classes, there is a unique unction : /R Y such that q =. q /R Y Here is constant on equivalence classes just in case (x) = (y) whenever x, y R. An equivalence relation R can be thouht o as a subobject o, i.e. a subset o ordered pairs. Accordinly, there are two unctions p 0 : R and p 1 : R iven by: p 0 x, y = x and p 1 x, y = y. Then condition (1) in the above axiom says that q p 0 = q p 1. And condition (2) says that or any unction : Y such that p 0 = p 1, there is a unique unction : /R Y such that q =. In this case, we say that q is a coequalizer o p 0 and p 1. Exercise 3.1. Show that in any cateory, coequalizers are unique up to isomorphism. Exercise 3.2. Show that in any cateory, a coequalizer is an epimorphism. Exercise 3.3. For a unction : Y, let R = { x, y (x) = (y)}. That is, R is the kernel pair o. Show that R is an equivalence relation. Deinition. A unction : Y is said to be a reular epimorphism just in case is a coequalizer. Exercise 3.4. Show that in any cateory, i : Y is both a monomorphism and a reular epimorphism, then is an isomorphism. Proposition 3.5. Every epimorphism in Sets is reular. In particular, every epimorphism is the coequalizer o its kernel pair. Proo. Let : Y be an epimorphism. Let p 0, p 1 : Y be the kernel pair o. By Axiom 6, the coequalizer : E o p 0 and p 1 exists; and since also coequalizes p 0 and p 1, there is a unique unction m : E Y 17

18 such that = m. Y Y b E Y E p 0 p 1 q 0 q 1 Here E Y E is the kernel pair o m. Since mp 0 = p 0 = p 1 = mp 1, there is a unique unction b : Y E Y E such that p 0 = q 0 b and p 1 = q 1 b. By the considerations at the end o the previous section, b is an epimorphism. Furthermore, q 0 b = p 0 = p 1 = q 1 b, and thereore q 0 = q 1. By Exercise 1.22, m is a monomorphism. Since = m, and is epi, m is also epi. Thereore, by Proposition 2.5, m is an isomorphism. E m This last proposition actually shows that Sets is what is known as a reular cateory. In eneral, a cateory C is said to be reular just in case it has all inite limits, i coequalizers o kernel pairs exist, and i reular epimorphisms are stable under pullback. Now, it s known that i a cateory has products and equalizers, then it has all inite limits. Thus Sets has all inite limits. Our most recent axiom says that Sets has coequalizers o kernel pairs. And inally, all epimorphisms in Sets are reular, and epimorphisms in Sets are stable under pullback; thereore, reular epimorphisms are stable under pullback. Reular cateories have several nice eatures that will prove quite useul. In the remainder o this section, we will discuss one such eatures: actorization o unctions into a reular epimorphism ollowed by a monomorphism. 3.2 The epi-monic actorization Let : Y be a unction, and let p 0, p 1 : Y be the kernel pair o. By Axiom 6, the kernel pair has a coequalizer : E. Since also coequalizes p 0 and p 1, there is a unique unction m : E Y such that = m. p 0 Y Y p 1 E m An arument similar to the one in Proposition 3.5 shows that m is a monomorphism. Thus, (E, m) is a subobject o Y, which we call the imae o under, and we write E = (). The pair (, m) is called the epi-monic actorization o. Since epis are surjections, and monics are injections, (, m) can also be called the surjective-injective actorization. 18

19 Deinition. Suppose that A is a subset o, in particular, n : A is monic. Then n : A Y, and we let (A) denote the imae o A under n. A (A) n We also use the suestive notation (A) = (A) = {y Y x A.(x) = y}. Proposition 3.6. Let : Y be a unction, and let A be a subobject o. The imae (A) is the smallest subobject o Y throuh which actors. Proo. Let e : Q and m : Q Y be the epi-monic actorization o. Suppose that n : B Y is a subobject, and that actors throuh n, say = n. Consider the ollowin diaram. p 0 Y E Y p 1 e Q Then np 0 = p 0 = p 1 = np 1, since p 0, p 1 is the kernel pair o. Since n is monic, p 0 = p 1, i.e. coequalizes p 0 and p 1. Since e : Q is the coequalizer o p 0 and p 1, there is a unique unction k : Q B such that ke =. By uniqueness o the epi-monic actorization, nk = m. Thereore, Q B. Proposition 3.7. For any A and B Y, we have A 1 (B) i and only i (A) B. Proo. Suppose irst that A 1 (B), in particular that k : A 1 (B). Consider the ollowin diaram: k A 1 (B) m e k (A) By deinition, je is the epi-mono actorization o m k. Since m k also actors throuh m : B Y, we have (A) B, by Proposition 3.6. Suppose now that (A) B. Usin the act that the lower square in the diaram is a pullback, we see that there is an arrow k : A 1 (B) such that m k is the inclusion o A in. That is, A 1 (B). Exercise 3.8. Use the previous result to show that A 1 ( (A)), or any subset A o. B Y m B n j 19

20 3.3 Functional relations Deinition. A relation R Y is said to be unctional just in case or each x there is a unique y Y such that x, y R. Deinition. Suppose that : Y is a unction. We let raph() = { x, y (x) = y}. Exercise 3.9. Show that raph() is a unctional relation. The ollowin result is helpul or establishin the existence o arrows : Y. Proposition Let R Y be a unctional relation. Then there is a unique unction : Y such that R = raph(). The proo o this result is somewhat complicated, and we omit it (or the time bein). 4 Colimits Axiom 7: Coproducts For any two sets, Y, there is a set Y and unctions i 0 : Y and i 1 : Y Y with the eature that or any set Z, and unctions : Z and : Y Z, there is a unique unction : Y Z such that ( ) i 0 = and ( ) i 1 =. Z Y i 0 i 1 Y We call Y the coproduct o and Y. We call i 0 and i 1 the coprojections o the coproduct. Intuitively speakin, the coproduct Y is the disjoint union o the sets and Y. What we mean by disjoint here is that i and Y share elements in common (which doesn t make sense in our ramework, but does in some rameworks), then these elements are dis-identiied beore the union is taken. For example, in terms o elements, we could think o Y as consistin o elements o the orm x, 0, with x, and elements o the orm y, 1, with y Y. Thus, i x is contained in both and Y, then Y contains two separate copies o x, namely x, 0 and x, 1. 20

21 We now show that that the inclusions i 0 : Y and i 1 : Y Y do, in act, have disjoint imaes. Proposition 4.1. Coproducts in Sets are disjoint. In other words, i i 0 : Y and i 1 : Y Y are the coprojections, then i 0 (x) i 1 (y) or all x and y Y. Proo. Suppose or reductio ad absurdum that i 0 (x) = i 1 (y). Let : Ω be the unique map that actors throuh t : 1 Ω. Let h : Y Ω be the unique map that actors throuh : 1 Ω. By the universal property o the coproduct, there is a unique unction h : Y Ω such that ( h)i 0 = and ( h)i 1 = h. Thus, we have t = (x) = ( h)i 0 x = ( h)i 1 y = h(y) =, a contradiction. Thereore, i 0 (x) i 1 (y), and the ranes o i 0 and i 1 are disjoint. Proposition 4.2. The coprojections i 0 : Y and i 1 : Y Y are monomorphisms. Proo. We will show that i 0 is monic; the result then ollows by symmetry. Suppose irst that has no elements. Then i 0 is trivially injective, hence monic by Proposition Suppose now that has an element x : 1. Let = x β Y, where β Y : Y 1. Then (1 )i 0 = 1, and Exercise 1.1 entails that i 0 is monic. Proposition 4.3. The coprojections are jointly surjective. That is, or each z Y, either there is an x such that z = i 0 (x), or there is a y Y such that z = i 1 (y). Proo. Suppose or reductio ad absurdum that z is neither in the imae o i 0 nor in the imae o i 1. Let : ( Y ) Ω be the characteristic unction o {z 0 }. Then or all x, (i 0 (x)) =. And or all y Y, (i 1 (y)) =. Now let h : ( Y ) Ω be the constant unction, i.e. h(z) = or all z Y. Then i 0 = hi 0 and i 1 = hi 1. Since unctions rom Y are determined by their coprojections, = h, a contradiction. Thereore, all z Y are either in the rane o i 0 or in the rane o i 1. Proposition 4.4. The unction t : 1 1 Ω is an isomorphism. Proo. Consider the diaram: Ω t t 1 1 i 0 i

22 Then t is monic since every element o 1 1 actors throuh either i 0 or i 1 (Proposition 4.3), and since t. Furthermore, t is epi since t and are the only elements o Ω. By Proposition 2.5, t is an isomorphism. Proposition 4.5. Let be a set, and let B be a subset o. Then the inclusion B \B is an isomorphism. Proo. Usin the act that Ω is Boolean, or every x, either x B or x \B. Thus the inclusion B \B is a bijection, hence an isomorphism. Axiom 8: Empty set There is a set with the ollowin properties: 1. For any set, there is a unique unction α In this case, we say that is an initial object in Sets. 2. is empty, i.e. there is no unction x : 1. Exercise 4.6. Show that in any cateory with coproducts, i A is an initial object, then A =, or any object. Proposition 4.7. Any unction : is an isomorphism. Proo. Since has no elements, is trivially surjective. We now claim that has no elements. Indeed, i x : 1 is an element o, then (x) is an element o. Since has no elements, is trivially injective. By Proposition 2.5, is an isomorphism. Proposition 4.8. A set has no elements i and only i =. Proo. By Axiom 8, the set has no elements. Thus i =, then has no elements. Suppose now that has no elements. Since is an initial object, there is a unique arrow α :. Since has no elements, α is trivially surjective. Since has no elements, α is trivially injective. By Proposition 2.5, is an isomorphism. 22

23 5 Sets o unctions and sets o subsets [[Note to students. This section is hihly technical, and you are not required to read it. Please simply amiliarize yoursel with the deinition o an exponential object, and the deinition o the powerset o a set.]] One distinctive eature o the cateory o sets is its ability to model almost any mathematical construction. One such construction is atherin toether old thins into a new set. For example, iven two sets A and, can we orm a set A o all unctions rom A to? Similarly, iven a set, can we orm a set P o all subsets o? As usual, we won t be interested in hard questions about what it takes to be a set. Rather, we re interested in hypothetical questions: i such a set existed, what would it be like? The crucial eatures o A seem to be captured by the ollowin axiom: Axiom 9: Exponential objects Suppose that A and are sets. Then there is a set A, and a unction e : A A such that or any set Z, and unction : A Z, there is a unique unction : Z A such that e (1 A ) =. A A e 1 A Z The set A is called an exponential object, and the unction : Z A is called the transpose o : A Z. The way to remember this axiom is to think o Y as the set o unctions rom to Y, and to think o e : Y Y as a meta-unction that takes an element Y, and an element x, and returns the value e(, x) = (x). For this reason, e : Y Y is sometimes called the evaluation unction. Note urther that i : Z Y is a unction, then or each z Z, (, z) is a unction rom Y. In other words, corresponds uniquely to a unction rom Z to unctions rom Y to. This latter unction is the transpose : Z Y o. We have written Axiom 9 in irst-order ashion, but it miht help to think o it as statin that there is a one-to-one correspondence between two sets: hom( Z, Y ) = hom(z, Y ), where hom(a, B) is thouh o as the set o unctions rom A to B. As a particular 23

24 case, when Z = 1, the terminal object, we have hom(, Y ) = hom(1, Y ). In other words, elements o Y in the internal sense correspond to elements o hom(, Y ) in the external sense. Consider now the ollowin special case o the above construction: A A e A 1 e e A A Thus, e = 1 A. Deinition. Suppose that : Y Z is a unction. We let A : A Y A denote the transpose o the unction: A Y A e Y Y Z That is, A = ( e Y ), and the ollowin diaram commutes: A Z A e Z 1 A e Y Z A Y A Y Proposition 5.1. Let : A Y and : Y Z be unctions. Then ( ) = A. Proo. Consider the ollowin diaram: A Z A e Z Z 1 ( ) 1 A 1 A Y A e Y A Y The bottom trianle commutes by the deinition o. The upper riht trianle commutes by the deinition o A. And the outer square commutes by the deinition o ( ). It ollows that e Z (1 ( A )) =, and hence A = ( ). 24

25 Consider now the ollowin particular case: A (A ) A 1 p A 1 e A Here p = 1 is the unique unction such that e(1 A p) = 1 A. Intuitively, we can think o p as the unction that takes an element x, and returns the unction p x : A A such that p x (a) = a, x. Thus, (1 p) a, x = a, p x, and e(1 p) a, x = p x (a) = a, x. Deinition. Suppose that : Z A is a unction. We deine : Z A to be the ollowin composite unction: 1 A Z A A e Proposition 5.2. Let : Y and : Y Z A be unctions. Then ( ) = (1 A ). Proo. By deinition, ( ) = e (1 ( )) = e (1 ) (1 ) = (1 ). Proposition 5.3. For any unction : A Z, we have ( ) =. Proo. By the deinitions, we have ( ) = e (1 ) =. Proposition 5.4. For any unction : Z A, we have ( ) =. Proo. By deinition, ( ) is the unique unction such that e (1 ( ) ) =. But also e (1 ) =. Thereore, ( ) =. Proposition 5.5. For any set, we have 1 =. Proo. Let e : 1 1 be the evaluation unction rom Axiom 9. We claim that e is a bijection. Recall that there is a natural isomorphism i : Consider the ollowin diaram: x i e That is, or any element x : 1, there is a unique element x o 1 such that e(1 x ) = x. Thus, e is a bijection, and = 1 1 is isomorphic to. x 25

26 Proposition 5.6. For any set, we have = 1. Proo. Elements o correspond unctions. There is exactly one such unction, hence has exactly one element x : 1. Thus, x is a bijection, and = 1. Proposition 5.7. For any sets A,, Y, we have ( Y ) A = A Y A. Proo. An eleant proo o this proposition would note that ( ) A is a unctor, and is riht adjoint to the unctor A ( ). Since riht adjoints preserve products, ( Y ) A = A Y A. Nonetheless, we will o into urther detail. By uniqueness o Cartesian products, it will suice to show that ( Y ) A is a Cartesian product o A and Y A, with projections π0 A and π1 A. Let Z be an arbitrary set, and let : Z A and : Z Y A be unctions. Now take γ =,, where : A Z and : A Z Y. Z γ ( Y ) A A π A 0 π A 1 Y A We claim that π A 0 γ = and π A 1 γ =. Indeed, π A 0 γ = π A 0, = (π 0, ) = ( ) =. Thus, π A 0 γ =, and similarly, π A 1 γ =. Suppose now that h : Z ( Y ) A such that π A 0 h = and π A 1 h =. Then = π A 0 (h ) = (π 0 h ). Hence, π 0 h =, and similarly, π 1 h =. That is, h =,, and h =, = γ. Proposition 5.8. For any sets A,, Y, we have A ( Y ) = (A ) (A Y ). Proo. Even without Axiom 9, there is always a canonical unction rom (A ) (A Y ) to A ( Y ), namely φ := (1 A i 0 ) (1 A i 1 ), where i 0 and i 1 are the coproduct inclusions o Y. That is, φ j 0 = 1 A i 0, and φ j 1 = 1 A i 1, where j 0 and j 1 are the coproduct inclusions o (A ) (A Y ). 26

27 Y i 0 p1 i 1 A ( Y ) Y q 1 1 A i 0 φ 1 A i 1 r1 A (A ) (A Y ) A Y j 0 We will show that Axiom 9 entails that φ is invertible. Let : A ( Y ) A ( Y ) be the identity, i.e. = 1 A ( Y ). Then : Y (A ( Y )) A is the unique unction such that e(1 A ) =. By Proposition 5.2, Similarly, ( i 1 ) = 1 A i 1. Thus, ( i 0 ) = (1 A i 0 ) = 1 A i 0. j 1 = (1 A i 0 ) (1 A i 1 ). We also have (1 A i 0 ) = (φ j 0 ) = φ A j 0, and (1A i 1 ) = φ A j 1. Hence = (φ A j 0 ) (φa j 1 ) = φa (j 0 j 1 ). Now or the inverse o φ, we take ψ = (j 0 j 1 ). ((A ) (A Y )) A j 0 j 1 j 0 j 1 Y It then ollows that i 0 i 1 (φ ψ) = φ A (j 0 j 1 ) =, and thereore φ ψ = 1 A ( Y ). Similarly, Y (ψ φ j 0 ) = ψ A (φ j 0 ) = ψ A i 0 = ψ i 0 = j 0. Thus, ψ φ j 0 = j 0, and a similar calculation shows that ψ φ j 1 = j 1. It ollows that ψ φ = 1 (A ) (A Y ). Thus, ψ is a two-sided inverse or φ, and A ( Y ) is isomorphic to (A ) (A Y ). Deinition (Powerset). I is a set, we let P = Ω. Intuitively speakin, P is the set o all subsets o. For example, i = {a, b}, then P = {, {a}, {b}, {a, b}}. More riorously, each element o Ω corresponds to a unction 1 Ω, which in turn corresponds to a unction = 1 Ω, which corresponds to a subobject o. Thus, we can think o P as another name or Sub(), althouh Sub() is not really an object in Sets. 27

28 6 Cardinality Summary: When mathematics was riorized in the 19th century, one o the important advances was a riorous deinition o ininite set. It came as somethin o a suprise that there are dierent sizes o ininity, and that some ininite sets (e.. the real numbers) are strictly larer than the natural numbers. In this section, we deine inite and ininite. We then add an axiom which says there is a speciic set N that behaves like the natural numbers; in particular, N is ininite. Finally, we show that the powerset P o a set is always larer than. Deinition. A set is said to be inite i and only i or any unction m :, i m is monic, then m is an isomorphism. A set is said to be ininite i and only i there is a unction m : that is monic and not surjective. We are already uaranteed the existence o inite sets: or example, the terminal object 1 is inite, as is the subobject classiier Ω. But the axioms we have stated thus ar do not uarantee the existence o any ininite sets. We won t know that there are ininite sets until we add the natural number object axiom below. Deinition. We say that Y is at least as lare as, written Y, just in case there is a monomorphism m : Y. Proposition 6.1. Y. Proo. Proposition 4.2 shows that i 0 : Y is monic. Proposition 6.2. I Y is non-empty, then Y. Proo. Consider the unction 1, : Y, where : 1 Y. Axiom 10: Natural Number Object There is an object N, and unctions z : 1 N and s : N N such that or any other set with unctions q : 1 and :, there is a unique unction u : N such that the ollowin diaram commutes: 1 N N The set N is called a natural number object. q z u s u 28

29 Exercise 6.3. Let N be a set, and let z : 1 N and s : N N be unctions that satisy the conditions in the axiom above. Show that N is isomorphic to N. Proposition 6.4. z s : 1 N N is an isomorphism. Proo. Let i 0 : 1 1 N and i 1 : N 1 N be the coproduct inclusions. Usin the NNO axiom, there is a unique unction : N 1 N such that the ollowin diaram commutes: z 1 N N i 0 s 1 N 1 N i 1z i 1s We will show that is a two-sided inverse o z s. To this end, we irst establish that s = i 1. Consider the ollowin diaram: z N s N 1 N N i 1z sz s s i 1z i 1s 1 N 1 N The lower trianle commutes because o the commutativity o the previous diaram. Thus, the entire diaram commutes. The outer trianle and square would also commute with i 1 in place o s. By the NNO axiom, s = i 1. Now, to see that (z s) = 1 N, note irst that (z s) z = (z s) i 0 = z. s Furthermore, (z s) s = (z s) i 1 = s. Thus the NNO axiom entails that (z s) = id N. (z s) = id 1 N, we calculate: Finally, to see that Furthermore, (z s) i 0 = z = i 0. (z s) i 1 = s = i 1. Thereore, (z s) = id 1 N. This establishes that is a two-sided inverse o z s, and 1 N is isomorphic to N. Proposition 6.5. The unction s : N N is injective, but not surjective. Thus, N is ininite. 29

30 Proo. By Proposition 4.2, the unction i 1 : N 1 N is monic. Since the imaes o i 0 and i 1 are disjoint, i 0 is not surjective. Since z s is an isomorphism, (z s) i 1 = s is monic, but not surjective. Thereore, N is ininite. Proposition 6.6. I m : B is a nonempty subobject, then there is an epimorphism : B. Proo. Since B is nonempty, there is a unction : \B B. By Proposition 4.5, B = B \B. Finally, 1 B : B \B B is an epimorphism, since 1 B is an epimorphism. Deinition. We say that a set is countable just in case there is an epimorphism : N, where N is the natural numbers. Proposition 6.7. N N is countably ininite. Sketch o proo. We will ive two aruments: one quick, and one slow (but hopeully more illuminatin). For the quick arument, deine a unction : N N N by (x, y) = 2 x 3 y. I x, y x, y, then either x x or y y. In either case, unique actorizability o inteers ives 2 x 3 y 2 x 3 y. Thereore, : N N N is monic. Since N N is nonempty, Proposition 6.6 entails that there is an epimorphism : N N N. Thereore, N N is countable. Now or the slow arument. Imaine writin down all elements in N N in an ininite table, whose irst ew elements look like this: 0, 0 1, 0 2, 0 0, 1 1, 1 2, 1 0, 2 1, 2 2, Now imaine runnin a thread diaonally throuh the numbers: bein with 0, 0, then move down to 1, 0 and up to 0, 1, then over to 2, 0 and down its diaonal, etc.. This process deines a unction : N N N whose irst ew values are (0) = 0, 0 (1) = 0, 1 (2) = 1, 0. It is not diicult to show that is surjective, and so N N is countable. Exercise 6.8. Show that i A and B are countable then A B is countable. We re now oin to show that exponentiation allows us to construct sets o larer and larer size. In the case o inite sets A and, it s easy to see that the ollowin equation holds: A = A, 30

31 where denotes the number o elements in. In particular, Ω can be thouht o as the set o binary sequences indexed by. We re now oin to show that or any set, the set Ω is larer than. Deinition. Let : A A be a unction. We say that a A is a ixed point o just in case (a) = a. We say that A has the ixed point property just in case any unction : A A has a ixed point. Proposition 6.9. Let A and be sets. I there is a surjective unction p : A, then A has the ixed point property. Proo. Suppose that p : A is surjective. That is, or any unction : A, there is an x such that = p(x ). Let ϕ = p, so that = ϕ(x, ). Now let : A A be any unction. We need to show that has a ixed point. Consider the unction : A deined by = ϕ δ, where δ : is the diaonal map. Then we have ϕ(x, x) = (x) = ϕ(x, x), or all x. In particular, ϕ(x, x ) = ϕ(x, x ), which means that a = ϕ(x, x ) is a ixed point o. Since : A A was arbitrary, it ollows that A has the ixed point property. Proposition There is no surjective unction Ω. Proo. The unction Ω Ω that permutes t and has no ixed points. The result then ollows rom Proposition 6.9. Exercise Show that there is an injective unction Ω. [The proo is easy i you simply think o Ω as unctions rom to {t, }. For a bier challene, try to prove that it s true usin the deinition o the exponential set Ω.] Corollary For any set, the set P o its subsets is strictly larer than. There are several other acts about cardinality that are important or certain parts o mathematics in our case, they will be important or the study o topoloy. For example, i is an ininite set, then the set F o all inite subsets o a set has the same cardinality as. Similarly, a countable coproduct o countable sets is countable. However, these acts well known rom ZF set theory are not obviously provable in ETCS. Discussion. Intuitively speakin, N is the set o all sequences with values in. Thus, we should have somethin like N = However, we don t have any axiom tellin us that Sets has ininite products such as the one on the riht hand side above. Can it be proven that N satisies the deinition o an ininite product? In other words, are there projections π i : N which satisy an appropriate universal property? 31

32 7 The Axiom o Choice Deinition. Let : Y be a unction. We say that is a split epimorphism just in case there is a unction s : Y such that s = 1 Y. In this case, we say that s is a section o. Exercise 7.1. Prove that i is a split epimorphism, then is a reular epimorphism. Exercise 7.2. Prove that i s is a section, then s is a reular monomorphism. Axiom 11: Axiom o choice Every epimorphism in Sets has a section. The name axiom o choice comes rom a dierent ormulation o this axiom, which says (rouhly speakin) that or any set-indexed collection o sets, say { i i I}, we can choose one element rom each set, say x i i, and orm a new set with these elements. To translate that version o the axiom o choice into our version, suppose that the sets i are stacked side by side, and that is the map that projects each x i to the value i. Then a section s o would be a map with domain I that returns an element s(i) i or each i I. Further readin C. Butz, Reular cateories and reular loic LS/98/2/BRICS-LS-98-2.pd W. Lawvere, Sets or Mathematics. Cambride University Press, W. Lawvere, An elementary theory o the cateory o sets tac.mta.ca/tac/reprints/articles/11/tr11.pd T. Leinster, Rethinkin set theory ETCS in nlab 32