Math 278x - Categorical Logic

Taught by Jacob LurieNotes by Dongryul Kim

Spring 2018

This course was taught by Jacob Lurie, on Mondays, Wednesdays, and Fri-days at 3-4pm. I stopped going to lectures, and thus the notes are incom-plete. A complete set of lecture notes can be found on the instructor’s website,http://math.harvard.edu/~lurie/278x.html.

Contents

1 January 22, 2018 31.1 First-order logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 January 24, 2018 62.1 First-order logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 The weak syntactic category . . . . . . . . . . . . . . . . . . . . . 7

3 January 26, 2018 93.1 Fiber products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.2 Factorization of morphisms . . . . . . . . . . . . . . . . . . . . . 9

4 January 29, 2018 124.1 Criteria for representability . . . . . . . . . . . . . . . . . . . . . 124.2 Categories that are weak syntactic . . . . . . . . . . . . . . . . . 13

5 January 31, 2018 155.1 Weak pretopoi . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.2 Constructing a theory . . . . . . . . . . . . . . . . . . . . . . . . 16

6 February 2, 2018 186.1 Versions of the completeness theorem . . . . . . . . . . . . . . . . 186.2 Boolean coherent categories are syntactic . . . . . . . . . . . . . 19

7 February 5, 2018 217.1 Disjoint unions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217.2 Equivalence relations . . . . . . . . . . . . . . . . . . . . . . . . . 22

1 Last Update: August 27, 2018

8 February 7, 2018 248.1 Sheaves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

9 February 9, 2018 279.1 Topoi are pretopoi . . . . . . . . . . . . . . . . . . . . . . . . . . 279.2 Constructing sheafification . . . . . . . . . . . . . . . . . . . . . . 28

10 February 12, 2018 3010.1 Giraud’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

11 February 16, 2018 3311.1 Quasi-compactness . . . . . . . . . . . . . . . . . . . . . . . . . . 33

12 February 21, 2018 3612.1 Characterizing geometric morphisms . . . . . . . . . . . . . . . . 36

13 February 23, 2018 40

14 February 26, 2018 4314.1 Locales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

15 February 28, 2018 4615.1 Points of a locale . . . . . . . . . . . . . . . . . . . . . . . . . . . 4615.2 Sober spaces and spatial locales . . . . . . . . . . . . . . . . . . . 47

16 March 2, 2018 4916.1 Classifying topoi . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

17 March 5, 2018 5217.1 Open maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

2

Math 278x Notes 3

1 January 22, 2018

This course is on first-order logic.

Definition 1.1. A linearly ordered set is a set X with a relation R ⊆ X2

satisfying

(A1) R(x, x) for all x,

(A2) R(x, y) and R(y, z) implies R(x, z),

(A3) R(x, y) and R(y, x) implies x = y.

(A4) R(x, y) or R(y, x) for all x, y.

This is an example of first-order theory.

1.1 First-order logic

Definition 1.2. A language is a set of predicate symbols {Pi} with aritiesni ≥ 0. A theory is just a language and a set of first-order sentences. Theseare the axioms of the theory.

Definition 1.3. Let T be a theory. A model of T is a set M , with for eachpredicate Pi, a subset MPi ⊆ Mni , such that for every ϕ ∈ T , ϕ is true in M .(This is written as M � ϕ.)

There are theories that are not first-order. For instance, torsion groups arenot first-order. Finite fields are not first-order.

Theorem 1.4 (Completeness). Let T be a theory and ϕ be a sentence. Thefollowing are equivalent:

(1) There is a proof of ϕ from T . (Written as T ` ϕ.)

(2) ϕ is true in every model of T . (Written as T � ϕ.)

This gives relation between syntax and semantics. But to define this rigor-ously, we need to consider formulas. These are going to be sentences, such asϕ(x) = ∀yR(x, y). For ϕ a formula, let us define

Mϕ = {(a1, . . . , an) ∈Mn : M � ϕ(a1, . . . , an)}.

Theorem 1.5 (Completeness). Let T be a theory and ϕ(x1, . . . , xn) and ϕ′(x1, . . . , xn)be formulas. The following are equivalent:

(1) There is a proof (from T ) that ϕ⇔ ϕ′.

(2) For every model M , we have Mϕ = Mϕ′ .

Question. Say that for each M � T , we have subset M ⊆M . When does thereexist ϕ(x) such that M = Mϕ for all M?

Math 278x Notes 4

Definition 1.6. LetM andM ′ be models of T . We say that a function f : M →M ′ is an elementary embedding if for every ϕ(x1, . . . , xn) and a1, . . . , an ∈M ,

M � ϕ(a1, . . . , an)⇔M � ϕ(f(a1), . . . , f(an)).

If f : M → M ′, then we should have M = f−1(M ′). This is a necessarycondition, but not sufficient.

Example 1.7. Let T be the theory of fields. Inside every F , consider

F ⊇ F = {roots of unity in F}.

But the claim is that there is no single formula defining F . One way to showthis is to use the theory of ultrafilers. Let U be a non-principal ultrafilter onZ>0. We want to construct a field

F =

(∏n>0

C)/U.

But for any ϕ(x), and any sequence {zn}, we have F � ϕ({zn}) if and only if{n : C � ϕ(zn)} ∈ U . Let’s suppose that ϕ(x) was satisfied by exactly the rootsof unity. Consider

z = (e2πi, e2πi/2, e2πi/3, . . .).

Here, zk = 1 is not true for any k, but it should be true that M � ϕ(z)

Theorem 1.8 (Makkai definablity). Assume that M = f−1(M ′) for any ele-mentary embedding f : M →M ′, and is compatible with ultraproducts, i.e.,

M =

(∏i

Mi

)/U.

This should be considered as a companion for Godel’s completeness theorem.This gives the existence result, and Godel’s completeness gives uniqueness.

Theorem 1.9 (Makkai). Let T be a first-order theory. Then T can be re-covered (up to equivalence) from Mod(T ), the category with objects models andmorphisms elementary embeddings. But this Mod(T ) is a “category with ultra-products”.

We need to define a lot of stuff to make this rigorous. Of course, if everyaxiom of T can be proven from T ′ and vice versa, we would want to call themequivalent. But this is more subtle.

Example 1.10. Consider P2(K) for a field K. There are points and lines inP2(K), and incidence relations. An abstract projective plane is a set of pointsP and a set of lines L. The axioms will be:

(A1) Every pair of distinct points lies on a unique line.

Math 278x Notes 5

(A2) Every pair of distinct lines intersects on a unique point.

Actually K can be recovered from P2(K). But does every projective plane comefrom a field? This requires Papuus’ theorem.

(A3) Pappus’ theorem

So you can go back and forth between these theories, although a priori theylook different.

So for each theory T , we are going to define a category Syn(T ), called thesyntactic category of T . Then we can define

Definition 1.11. T and T ′ are equivalent if Syn(T ) ' Syn(T ′) as categories.

Theorem 1.12 (Makkai). Let T be a first-order theory. Then Syn(T )∼−→

Funultra(Mod(T ),Set) is an equivalence. Conversely, Mod(T ) is a subcategoryof Fun(Syn(T ),Set).

This syntactic category Syn(T ) is a pretopos. But to make a one-to-onedictionary between classical first-order logic, we will need to give up the law ofexcluded middle. This is because of a pragmatic reason.

Example 1.13. Consider T the first-order theory of groups. The morphismsof Mod(T ) are elementary embeddings. But there is a notion of group homo-morphism, which is strictly weaker than being an elementary embedding. For agroup homomorphism, we have z = xy implies f(z) = f(x)f(y), but we don’tconclude z 6= xy implies f(z) 6= f(x)f(y). This is what we mean by giving upthe law of excluded middle.

Math 278x Notes 6

2 January 24, 2018

We’ll begin with review of first-order logic.

2.1 First-order logic

Let L be a language with predicate symbols {Pi} and an infinite set V of freevariables.

Definition 2.1. A formula with free variables in V0 ⊆ V consist of

(1) x = y for x, y ∈ V0,

(2) Pi(x1, . . . , xni) for x1, x2, . . . ∈ V0,

(3) ϕ ∨ ψ for ϕ and ψ already defined,

(4) ¬ϕ for ϕ already defined,

(5) ∃xψ for ψ a formula with free variables in V0 q {x}.

Recall that an L-structure is a set M with M [Pi] ⊆Mni . Given any formulaϕ with variables in V0 = {x1, x2, . . . , xn} and c1, . . . , cn ∈ M , we define M �ϕ(c1, . . . , cn) inductively as the following:

(1) for ϕ = ”xi = xj”, M � ϕ(~c) if and only if ci = cj .

(2) for ϕ = ”Pi(xj1 , . . . , xjni )”, M � ϕ(~c) if and only if ~c ∈M [Pi],

(3) . . .

With these primitive notions, we can build others like

(1) ϕ ∧ ψ means ¬(¬ϕ ∨ ¬ψ)

(2) ϕ⇒ ψ means ¬ϕ ∨ ψ(3) ∃~xϕ(~x, ~y) means ∃x1∃x2 · · · ∃xnϕ(~x, ~y)

(4) ∀~xϕ(~x, ~y) means ¬∃~x¬ϕ(~x, ~y).

A sentence is a formula with no free variables, and a theory T of a languageis a set of sentences. A model is a L-structure M such that M � ϕ for all ϕ ∈ T .If M is a model and ϕ is a formula, we write

M [ϕ] = {(c1, . . . , cn) : M � ϕ(c1, . . . , cn)}.

Suppose we have a map V0 = {x1, . . . , xn} → V1 with xi 7→ yi. Given aformula ϕ(x1, . . . , xn), we would like to substitute every xi with yi. But thereis a caveat, when one of the variables we want to use as free is already bound.But life is too short to worry about such problems.

Math 278x Notes 7

2.2 The weak syntactic category

Given a first order theory T , we define a category Syn0(T ), called the weaksyntactic category of T . The objects are formulas ϕ(~x) in the language of T .(I’m going to write [ϕ(~x)] for the object.) Let [ϕ(~x)] and [ψ(~y)] be objects ofSyn0(T ). A morphism f : [ϕ]→ [ψ] is a collection of maps

fM : M [ϕ]→M [ψ]

for each model M , such that there exists a formula θ(~x, ~y) such that for all M ,M [θ] is the graph in M [ϕ]×M [ψ]. We need to check that this is a category.

Proposition 2.2. If f : [ϕ] → [ϕ′] and g : [γ′] → [γ′′] are morphisms inSyn0(T ), so is g ◦ f .

Proof. Suppose that M [θ(~x, ~y)] = Γ(fM ) and M [θ′(~y, ~z)] = Γ(gM ). Then thefirst-order formula

ρ(~x, ~z) = ∃~yθ(~x, ~y) ∧ θ(~y, ~z)cuts out the graph of g ◦ f .

Here is another way to describe this category. The morphisms from [ϕ(~x)]to [ϕ(~y)] are formulas θ(~x, ~y) such that

T � ∀~x, ~y, (θ(~x, ~y)⇒ ϕ(~x) ∧ ψ(~y)) ∧ (∀~x[ϕ(~x)⇒ ∃!~yθ(~x, ~y)]),

modulo the equivalence relation

T � ∀~x, ~yθ(~x, ~y)⇔ θ′(~x, ~y).

If we really want to be syntactic, we should replace it with `, but then we wouldbe using Godel’s completeness theorem, which we are supposed to prove. So weleave it as �.

Let X = [ϕ(~x)] ∈ Syn0(T ). For each M , we have a subset M [X] = M [ϕ] ⊆Mn. A morphism f : X → Y induces a map M [X]→M [Y ] by definition. Thatis, X 7→M [X] is a functor Syn0(T )→ Set.

Proposition 2.3. Let M and N be models of T . The following are equivalent:

(1) an elementary embedding f0 : M → N

(2) natural transformations M [−]→ N [−]

Proof. First, if f0 is elementary, for any ϕ and c1, . . . , cn ∈M , we know that

M � ϕ(c1, . . . , cn) ⇔ N � ϕ(f0(c1), . . . , f0(cn)).

So f : Mn → Nn restricts to a map fϕ : M [ϕ] → N [ϕ]. The claim is that thisdata gives a natural transformation. We need to check that the diagram

M [ψ] N [ψ]

M [ϕ] N [ϕ]

fψ

gM fN

fϕ

Math 278x Notes 8

commutes. This follows from the fact that f is elementary used on formula θ.For the other direction, we first use on ϕ(x) = (x = x). Then we get a map

f0 : M → N . We need to show that f0 is an elementary embedding, which justmeans that f0 takes M [ϕ] to N [ϕ]. This follows from functoriality.

So we have shown that the functor

Mod(T ) ↪→ Fun(Syn0(T ),Set)

is fully faithful. There is a natural question, which we will postpone this questionuntil later.

Question. What is the image? Given F : Syn0(T ) → Set, when does it arrisefrom a model?

In the abstract projective plane, we had lines and points.

Definition 2.4. A typed language is a set Typ of types, predicate symbols{Pi} with arity (t1, . . . , tn). Now every variable has a type (e.g., {p, q, r} forpoints and {`, `′, `′′} for lines). Formulas are as before, except x = y is onlyallowed when x and y have the same type and Pi(x1, . . . , xn) is allowed onlywhen the type of xj is tj . We also add the formula ⊥, which is always false. Amodel is a collection of sets M [t] for every t ∈ Typ, and M [Pi] ⊆M [t1]× · · · ×M [tn].

If there is only one type, everything reduces to the previous definition. Ifthere are only finitely many types, this is pointless. When there are no typesat all, this is propositional logic. Here, the formulas are ⊥, Pi, and theirdisjunctions and negations. In particular, every formula is a sentence.

What does Syn0(T ) look like for propositional logic? We have

HomSyn0(T )([ϕ], [ψ]) =

{∗ T � ϕ⇒ ψ,

∅ otherwise.

So this is a poset of sentences up to provable equivalences.

Math 278x Notes 9

3 January 26, 2018

Our goal is to find a criterion for when a functor Syn0(T )→ Set comes from amodel.

3.1 Fiber products

Proposition 3.1. The category Syn0(T ) have fiber products, and M [−] : Syn0(T )→Set preserves them, where M � T .

Proof. First we show that there are fiber products. For f : X → Z givenby θ(~x, ~z) and g : Y → Z given by θ′(~y, ~z), the fiber product is going to be∃~z(θ(~x′, ~z′) ∧ θ′(~y, ~z)). The projection maps are just going to be ~x′ = ~x and~y′ = ~y. To check that this is a fiber product, we first see that for every model,this will give a fiber product. If δ(~w, ~x) defines W → X and δ(~w, ~y) definesW → Y , then W → P should be defined by δ(~w, ~x′) ∧ δ(~w, ~y′).

Corollary 3.2. The corollary Syn0(T ) has finite limits, preserved by M [−] foreach M � T .

Proof. It suffices to find an initial object. Take a sentence ϕ such that T � ϕ,for instance, ∀x(x = x). If we denote [ϕ] = 1, then M [1] = {∗} and so thereis just unique map from M [ϕ] → M [1] for each ϕ. This map is a morphismX → 1 in the syntactic category.

3.2 Factorization of morphisms

Lemma 3.3. Let f : X → Y be a morphism in Syn0(T ). The following areequivalent:

(1) f is an isomorphism.

(2) Each fM : M [X]→M [Y ] is bijective.

Proof. (1) to (2) is obvious. For the other direction, just switch the variables.

Corollary 3.4. Let f : X → Y be a morphism in Syn0(T ). The following areequivalent:

(1) f is a monomorphism.

(2) fM : M [X]→M [Y ] for all M � T .

Proof. Apply the lemma to X → X ×Y X.

Example 3.5. Let ϕ(~x) and ϕ0(~x) be formulas. If T � (∀~xϕ0(~x) ⇒ ϕ(x)),then M [ϕ0] ⊆M [ϕ]. This defines a map [ϕ0]→ [ϕ], which is a monomorphism.Let’s call this a “special” monomorphism.

Math 278x Notes 10

Proposition 3.6. Let f : X → Y be any morphism in the syntactic category.

Then f has a canonical factorization as Xg−→ Y

h−→ Z where g is an epimorphismand h is a special monomorphism. Moreover, for any M � T , we would haveM [X] �M [Y ] ↪→M [Z].

Proof. Let us write X = [ϕ(~x)] and Z = [ψ(~z)] and f given by θ(~x, ~z). Nowdefine Y = [∃yθ(~x, ~z)], with g given by θ(~x, ~z) and h given by the identity map.We haven’t proven that g is an epimorphism, but let’s hold this for now.

The statement that g is an epimorphism does not characterize the factoriza-tion.

Definition 3.7. Let C be a category with fiber products, and consider

X ×Y X X T.π

π′

g

We say that g is an effective epimorphism if g is the coequalizer.

Effective epimorphisms are always epimorphisms, but the converse is nottrue. The converse is true in Set. But in CRing, effective epimorphisms aresurjections. Localizations are epimorphisms, but usually not surjective.

Proof. To show that g is an epimorphism, we show that g is an effective epi-morphism. Assume that a map U : [X] → [W ] is given by β(~x, ~w), and on thelevel of models, we should have a map [Y ] → [W ]. The formula that cuts outthis is going to be ∃~x(θ(~x, ~z) ∧ β(~x, ~w)).

In fact, the fact that g is an effective epimorphism determines this factor-ization uniquely.

Proposition 3.8. Let C be any category with fiber products, and f : X → Z

be any morphism. Suppose f factors as Xg−→ Y

h−→ Z with g an effectiveepimorphism and h a monomorphism. Then this factorization is unique.

Proof. This is just a general categorical fact.

Corollary 3.9. Let X = [ϕ(~x)]. Then every monomorphism f : X0 ↪→ X isisomorphic to a special monomorphism.

Proof. There is a factorization X0 � Y ↪→ X with Y ↪→ X a special monomor-phism, and there is also the factorization X0

∼= X0 ↪→ X.

Definition 3.10. Let C be any category and X ∈ C. Consider poset of sub-objects

Sub(X) = {monomorphisms X0 ↪→ X}/isomorphism.

Proposition 3.11. Le X = [ϕ(~x)]. Then

(1) every object of Sub(X) has the form [ϕ0(~x)] where T � (ϕ0 ⇒ ϕ).

Math 278x Notes 11

(2) Given any two formale ϕ0(~x) and ϕ(~x), [ϕ0(~x)] ⊆ [ϕ(~x)] if and only ifT � (ϕ0 ⇒ ϕ).

(3) [ϕ0(~x)] = [ϕ(~x)] in Sub(X) if an only if T � (ϕ0 ⇔ ϕ).

Corollary 3.12. For X ∈ Syn0(T ), Sub(X) is a Boolean algebra.

Proof. The least upper bound is given by ϕ0 ∨ ϕ1, the greatest lower bound isgiven by ϕ0 ∧ ϕ1, and the complement is given by ¬ϕ0 ∧ ϕ.

Moreover, for any model M � T the map Sub(X) → Sub(M [X]) is homo-morphism of Boolean algebras.

Proposition 3.13. Let F : Syn0(T ) → Set be a functor. Then F comes froma model M � T if and only if

(a) F preserves finite limits,

(b) F carries effective epimorphisms to surjections, and

(c) for any X ∈ Syn0(T ), F : Sub(X) → Sub(F (X)) preserves least upperbounds of finite subsets.

Math 278x Notes 12

4 January 29, 2018

Let T be a theory, and let C = Syn0(T ). We proved that

(A1) C has finite limits.

(A2) any f : X → Z factors as f = h ◦ g with h monic and g effective epic.

(A3) for X ∈ C, Sub(X) as least elements and joints X0 ∨X1.

Given a morphism f : X → Z in Syn0(T ), how do we check if f is aneffective epimorphism? This is true if and only if each fM : M [X] → M [Y ] isa surjection. This is because we have uniqueness of factorization. We have canalways factorize f : X → Z into M [X] � M [Y ] ↪→M [Z], and then f : X → Zis an effective epimorphism if and only if Y → Z is an isomorphism.

We actually showed that Sub(X) is a Boolean algebra, and we can deducethis from X0 ×X X1 being the meet X0 ∧X1.

4.1 Criteria for representability

Theorem 4.1. Let F : Syn0(T )→ Set be a functor. This comes from a modelif and only if

(1) F preserves finite limits,

(2) F preserves effective epimorphisms,

(3) each Sub(X)→ Sub(F (X)) is a map of upper semi-lattices.

Just as the previous discussion, (3) is equivalent to Sub(X) → Sub(F (X))being a Boolean algebra homomorphism.

The interesting direction is the converse. Assume that F satisfies (1), (2),and (3). Pick a variable e, and consider E = [e = e] ∈ Syn0(T ). Then for anymodel N , we have N [E] = N . So let us define

M = F (E).

If F came from a model, this is the right thing to do. Now let us assume wehave a finite set V0 = {x1, . . . , xn} of variables. Then

EV0 = [(x1 = x1) ∧ · · · ∧ (xn = xn)]

is always true, so N [EV0 ] = NV0 =∏x∈V0

N . This give bijection EV0 ∼=∏x∈V0

E in the syntactic category. Because F preserves finite limits, we get

F (EV0) =∏x∈V0

F (E) ∼= MV0 .

For each ϕ(~x), [ϕ(~x)] is a subobject of EV0 . So we have

F [ϕ(~x)] ⊆MV0

Math 278x Notes 13

because monomorphisms are preserved.Given any map V0 → V1 with xi 7→ yi, we have a morphism

[ϕ(~y)] EV1

[ϕ(~x)] EV0

which is a pullback square in Syn0(T ). Because F preserves finite limits, wehave F [ϕ(~y)] = F [ϕ(~x)]×MV0 M

V1 . Now for each predicate symbol Pi of arityn, choose x1, . . . , xn. We define

M [Pi] = F (Pi(x1, . . . , xn)) ⊆Mn.

This lets us define M [ϕ(~x)] ⊆MV0 for each ϕ(~x) with variables in V0.Now the claim is that for any formula ϕ(~x), we have F [ϕ(~x)] = M [ϕ(~x)]

as subsets of MV0 . This should be some kind of induction on the formation ofϕ(~x).

(i) If ϕ is x = y. You can do this.

(ii) If ϕ is Pi(y1, . . . , yn), this is tautological (unless some of the yis have thesame variables, in which case it follows from the compatibility.)

(iii) Assume ϕ = ϕ0 ∨ ϕ1. If we assume that M [ϕ0] = F [ϕ0] and M [ϕ1] =F [ϕ1], then M [ϕ0∨ϕ1] = M [ϕ1]∪M [ϕ2] = F [ϕ0]∪F [ϕ1], but F preservesjoins.

(iv) If ϕ = ¬ψ, then it follows from the fact that F : Sub(EV0) → Sub(MV0)is a Boolean algebra homomorphism.

(v) If ϕ(~x) = ∃yψ(~x, y), then [ϕ(~x, y)] ⊆ EV0∪{y}. Here, [ψ(~x, y)] → [ϕ(~x)]is an effective epimorphism, so F takes it to a surjective morphism. SoM [ψ(~x, y)] → M [ϕ(~x)] is a surjection, and it is just going to be the pro-jection.

Corollary 4.2. M is a model of T .

Proof. Let ϕ be a sentence. Then

F [ϕ] =

{{∗} M � ϕ,

∅ otherwise.

So if ϕ is a true sentence, it is a final object, and F should map it to a finalobject. So ϕ should be true in T .

4.2 Categories that are weak syntactic

Now I want to ask the following question.

Math 278x Notes 14

Question. What is special about this category? Given a category C, when is itof the form Syn0(T ) for some typed first-order theory T?

We first have the three properties (A1), (A2), and (A3). This can be thoughtof some structure on the category Syn0(T ). Here are some more compatibilityconditions.

(A4) Effective epimorphisms are stable under pullback.

This is a true statement in any Syn0(T ). This can be checked against any model,and pullback of surjective maps in Set is a surjection. There is another way toformulate this. Let’s say that C is a category with pullbacks, and consider amorphism g : Y ′ → Y . Then there is a morphism Sub(Y ) → Sub(Y ′) given byY0 7→ Y0 ×Y Y ′. Then the compatibility between (A1) and (A3) is

(A5) For any g : Y ′ → Y , the map g−1 : Sub(Y )→ Sub(Y ′) is a homomorphismof upper semi-lattices.

Again, this is true in Syn0(T ) because we can check against any model and thisis true for sets.

Proposition 4.3. Assume (A1) and (A2). Then (A4) is equivalent to for anypullback square from f : X → Y to f ′ : X ′ → Y ′, we have im(f ′) = g−1 im(f).

Proof. Exercise.

Definition 4.4. A weak pretopos is a category C satisfying these axioms(A1)–(A5).

This terminology is not standard.

Example 4.5. We know that Syn0(T ) is a weak pretopos. Also, Set is a weakpretopos.

Example 4.6. Let P be a poset. (A1) says that P has a largest element 1and meets p ∧ q. (A2) and (A4) are automatic because effective epimorphismsare identities and all morphisms are monomorphisms. (A3) states that P has asmallest element 0 and joins p∨ q. (A5) states that p∧ (q∨ r) = (p∧ q)∨ (p∧ r).So P is a weak pretopoi if and only if P is a distributive lattice.

Definition 4.7. A weak pretopos C is Boolean if for every X ∈ C, Sub(X) isa Boolean algebra.

Theorem 4.8. Let C be a small category. The following are equivalent:

(1) C ∼= Syn0(T ) of some typed theory T .

(2) C is a Boolean weak pretopos.

If you want T to be untyped, you need to add an extra axiom

(A6) There exists X ∈ C such that for all Y ∈ C, there exist monomorphismsY ↪→ Xn for some n� 0.

This object is E = [e = e] that we have used in the proof of the previoustheorem.

Math 278x Notes 15

5 January 31, 2018

We started with a first-order theory, and we extracted some of the properties:

(A1) C has finite limits.

(A2) Every morphism f : X → Y has an image.

(A3) For X ∈ C, Sub(X) is an upper semilattice.

(A4) Pullbacks of effective epimorphisms are effective epimorphisms.

(A5) For f : X → Y , f−1 : Sub(Y ) → Sub(X) preserves least elements andjoins.

I gave the name “weak pretopos” to this.

5.1 Weak pretopoi

What are morphisms between weak pretopoi? This should be a functor F : C →C′ such that

(1) F preserves finite limits,

(2) F preserves effective epimorphisms,

(3) for all X ∈ C, the induced F : Sub(X)→ Sub(F (X)) is a homomorphismof upper semilattices. (Equivalently, it is a homomorphism of lattices.)

Definition 5.1. Let C be a weak pretopos. A model of C is a morphism ofpretopoi C → Set. If C = Syn0(T ), then Mod(C) = Mod(T ) in the classicalsense.

Example 5.2. Let P and P ′ be distributive lattices. Morphisms P → P ′ asweak pretopoi is the same as lattice homomorphisms P → P ′.

Let C be a category, and let X ∈ C be an object. You can make a newcategory C/X consisting of objects (U, f : U → X) and morphisms (U, f) →(V, g) given by a map h : U → V making the diagram commute.

Proposition 5.3. If C is a weak pretopos, so is C/X.

Proof. Fiber products can be formed in C. Images also can be formed in C.Unions of subobjects are also fomred in C. Compatibility can be checked inC.

But the forgetful functor C/X → C is not going to be a morphism of weakpretopoi in most cases.

Proposition 5.4. Let f : X → Y be a morphism in C a weak pretopos. Thenwe get a functor f∗ : CY → CX given by U 7→ U ×Y X. Then f∗ a morphism ofweak pretopoi.

This is the right adjoint of the functor given by composition by f .

Math 278x Notes 16

5.2 Constructing a theory

Let C be a small weak pretopos. Our goal is to construct an associated first-ordertheory T such that models of C are models of C.

First we need to describe the language. Types are going to be objects ofC. (This is a structure that associates to each type X a set M [X].) For eachmorphism f : X → Y , we are going to consider a predicate Pf with arity (X,Y ).So we would have M [Pf ] ⊆M [X]×M [Y ]. Here are the axioms we will have:

• (for each f : X → Y ) ∀x∃!yPf (x, y) (This will give M [Pf ] the interpreta-tion of a graph of a function.)

• (for id : X → X) ∀xPid(x, x) (This will be fid the identity map on M [X].)

• (for Xf−→ Y

g−→ Z) ∀x, y, z[Pf (x, y) ∧ Pg(y, z) ⇒ Pg◦f (x, z) (This makessure that composition have as compositions.)

So far, models are simply functors M : C → Set. But we don’t want all functors.

• (for E the final obejct) ∃!e[e = e] (M preserves final objects.)

• (for f : X → Y an effective epimorphism) ∀y∃xPf (x, y) (M preserveseffective epimorphisms.)

• . . .

The defining feature of this theory is that models of T (C) are “the same” asmodels of C. There literally is a bijection between these models. But Mod(T (C))is usually not equivalent to Mod(C). Because Mod(T (C)) = Mod(Syn0(T (C))),maybe we should compare C and Syn0(T (C)).

There is always a functor

λ : C → Syn0(T (C)); X 7→ [x = x].

On the level of maps, this is going to be given by f : X → Y mapped to Pf (x, y).

Proposition 5.5. λ is a morphism of weak pretopoi.

Proof. We need to show λ(idX) = idλ(X), but this is one of the axioms. We alsoneed to show that λ(g◦f) = λ(g)◦λ(f), and it is also one of the axioms. So λ isa functor. Now we need to show more stuff like if f is an effective epimorphismthen λ(f) is an effective epimorphism. But this is also an axiom, and everythingwill be an axiom.

So we can compose the morphisms of weak pretopoi and get

Mod(T (C)) ∼= Mod(Syn0(T (C))) −◦λ−−−→ Mod(C).

This is bijective on objects, but not necessarily an equivalence.

Theorem 5.6. A weak pretopos if and only if λ : C → Syn0(T (C)) is anequivalence.

Math 278x Notes 17

One direction is obvious, because Syn0 is Boolean. If C is not Boolean, youcan think of Syn0(T (C)) as a “Booleanization” of C. Let me kind of almostprove that. Assume f : C → D is a morphism of pretopoi, and construct thefollowing:

C D

Syn0(T (C)) Syn0(T (D)).

f

λC λD

fBool

If D is Boolean, λD is an equivalence so we and construct Syn(T (C)) → D.What you need to show is that this factorization is unique.

Lemma 5.7. Let C be Boolean and let X ∈ C factor as a product X ∼= X1 ×· · · ×Xn. If ϕ(x1, . . . , xn) is a formula in T (C), and interpret

[ϕ(x1, . . . , xn)] ⊆ λ(X1)× · · · × λ(Xn) = λ(X).

Then there exists a Y ∈ Sub(X) such that λ(Y ) = [ϕ(x1, . . . , xn)] in Sub(λ(X)).

Proof. We induct on ϕ.

(i) If ϕ is xi = xj , then we can take Y = X ×Xi×Xj Xi.

(ii) If ϕ is Pf (xi, xj) then we can take Y = X×Xi×XjXi where Xi → Xi×Xj

is given by id×f .

(iii) If ϕ is ϕ0 ∨ ϕ1, with ϕ0 and ϕ1 represented by Y0 and Y1, we can takeY = Y0 ∨ Y1 in Sub(X).

(iv) If ϕ(~x) is ¬ψ(~x), we get [ψ(~x)] = λ(Z) for some Z ⊆ X. If C is Boolean,it has a complement Y in Sub(X). But if there is a morphism of uppersemilattices that happen to Boolean, it is a Boolean algebra homomor-phism.

(v) If ϕ(~x) = ∃wψ(~x,w), then we take Y to be the image of [ψ]→ [ϕ].

Example 5.8. We will see that there is a canonical (weak) pretopos C such thatMod(C) is groups with group homomorphisms. But this is not Boolean, and ifyou Booleanize it you will get Mod(Syn0(T (C))) will be groups with elementaryembeddings.

Example 5.9. LetX be a quasi-compact quasi-separated scheme. A setK ⊆ Xis constructible if it belongs to the Boolean algebra generated by the quasi-compact open sets. You can then make a new topological space Xc with basisgiven by the constructible sets in X. This is an example of a Stone space, i.e.,compact Hausdorff totally disconnected. If P is the lattice of quasi-compactopen sets in X, then Syn0(T (P )) is the lattice corresponding to Xc.

Math 278x Notes 18

6 February 2, 2018

I was informed that what I have been calling a weak pretopos is actually calleda coherent category.

6.1 Versions of the completeness theorem

Theorem 6.1 (Godel’s completeness theorem, version 0). If T is consistent,then it has a model.

Theorem 6.2 (Deligne’s completeness theorem, version 0). If C is a smallcoherent category and C is consistent, then C has a model.

Here, we are saying that C is consistent if Sub(1) has more than one ele-ment. That is, 0 and 1 are different in Sub(1). As an exercise, you can showthat C is consistent if it is not equivalent to {∗}.

Theorem 6.3 (Godel’s completeness theorem, version 1). If ϕ is a sentence ofT that is true in every model, then ϕ is provable from T . Equivalently, if T 6` ϕ,then there is a model M � ϕ.

Theorem 6.4 (Deligne’s completeness theorem, version 1). If C is small andcoherent, if U ∈ Sub(1) and U 6= 1 (in Sub(1)), then there exists a modelM : C → Set such that M [U ] = ∅.

We will take these to be the official statements. We easily see that version0 and 1 are equivalent for Godel, but it is harder to see this for Deligne.

Theorem 6.5 (Godel’s completeness theorem, version 2). If ϕ(~x) and ψ(~x) aretwo formula,s and T 6` ϕ⇒ ψ, then there exists an M such that M [ϕ] 6⊆M [ψ].

Theorem 6.6 (Deligne’s completeness theorem, version 2). If C is small andcoherent, and U ⊆ X is a subjective with U 6= X, then there exists a modelM : C → Set such that M [U ] (M [X].

It is easy to see that version 2 implies version 1. I claim that version 1implies version 2. Let U ( X be a subobject. Then we can look at the slicecategory C/X with objects Y ∈ C along with maps Y → X. The final object isX, and now U ( X is a subject of the final object. Then there exists a modelN : C/X → Set such that N [U ] = ∅. There also exists a morphism C → C/Xgiven by Y 7→ Y ×X. We compose them to get M : C → Set. Now my claim isthat

M [U ] (M [X]

which is equivalent to N [U ×X] ( N [X ×X]. But we have a pullback

N [U ×X] N [X ×X]

N [U ] N [X]

where N [U ] is empty and N [X] has one element. So it cannot be a bijection.

Math 278x Notes 19

Proposition 6.7. The following data are equivalent:

(1) models of C/X(2) models M of C with a chosen element of M [X].

You can think of this as adding a constant.

Theorem 6.8 (Deligne’s completeness theorem, version 3). Let C be a smallcoherent category and f : X → Z be a morphism in C. If f is not an isomor-phism, then there exists a model M : C → Set such that M [X] → M [Z] is notbijective.

Let’s now prove that version 2 implies version 3. Assume that f induces abijection in every model. We know that f can be factorized as

M [X]g−→M [Y ]

h−→M [Z]

where g is an effective epimorphism and h is a monomorphism. Then g and his bijective in every model, and version 2 implies that h is an isomorphism. Butwhat about g? The definition of an effective epimorphism is that

X ×Y X ⇒ Xg−→ Y

is a coequalizer. Here we can take X as a subobject X ↪→ X ×Y X, and thefact that M [X] → M [Y ] is bijective implies that M [X] = M [X ×Y X]. ThenX ∼= X ×Y X as subobjects of X × X. Then Y can be computed by thecoequalizer of X ⇒ X, which is X.

6.2 Boolean coherent categories are syntactic

Corollary 6.9. Let λ : C → D be a morphism of coherent categories. Assumethat

(1) λ is essentially surjective, (for all D ∈ D there exists an isomorphismD ∼= λ(C))

(2) for each C ∈ C, λ induces a surjection Sub(C) � Sub(λ(C)),

(3) every model of C extends to a model of D.

Then λ is an equivalence.

Example 6.10. Last time we had a morphism C → Syn0(T (C)). If C is Boolean,then (2) is satisfied. This is what we finished with last time. Then (1) is satisfiedand (3) is obvious. This will finish the proof of the theorem we stated last time.

Now let us prove the corollary. We want to know that for X,Y ∈ C, that

HomC(X,Y )→ HomD(λ(X), λ(Y ))

Math 278x Notes 20

is bijective. First, we show that it is injective. Suppose f and g are mapped tothe same morphism. If we look at the equalizer, we get

λ(Eq(f, g)) = Eq(λ(f), λ(g)) = λ(X)

as subobjects of λ(X). If f 6= g, then Eq(f, g) ( X, and then Deligne’scompleteness theorem version 2 shows that there exists a model M such thatM [Eq(f, g)] ( M [X]. That is, f, g : M [X] → M [Y ] are not equal. But everymodel of C should come from a model of D, and this gives a contradiction.

Now we show that it is surjective. Given any f : λ(X) → λ(Y ), we wantsome f : X → Y to satisfies f = λ(f). We encode this as the graph

Γ(f) ⊆ λ(X)× λ(Y ) ∼= λ(X × Y ).

Then (2) tells us that there exists a U ⊆ X × Y such that Γ(f) = λ(U) assubobjects of λ(X × Y ). Now we want to show that π : U → X is an isomor-phism. It suffices to show that π : M [U ]→ M [X] is an isomorphism for all Ma model of C. This can be checked for models of D, and then λ(U) → λ(X) isan isomorphism by definition.

Recall that for a theory T , we defined the syntactic category to be objectsX = [ϕ(~x)] and Y = [ψ(~x)] and morphisms given by θ(~x, ~y) such that

T � ∀~x, ~y(θ(~x, ~y)⇒ ϕ(~x) ∧ ψ(~y)) ∧ ∀~x(ϕ(~x)⇒ ∃!~yθ(~x, ~y)).

But this is semantic. If we really want a syntactic definition, we should replaceT � by T `. Suppose we have a reasonable definition of proof, and use this todefine Syn′0(T ). Can we show that the functor

λ : Syn′0(T )→ Syn0(T )

is an equivalence of categories? Suppose that we can justify all arguments madeso far, i.e., Syn′0(T ) is a category, and λ is a morphism of coherent categories.If we can justify that models of Syn′0(T ) are models of Syn0(T ), then we canapply our corollary to show that λ is a equivalence of categories. A corollary isGodel’s completeness theorem.

Math 278x Notes 21

7 February 5, 2018

We have been talking about coherent categories. Most categories you are famil-iar with, of abelian groups or of commutative rings, don’t have (finite) unionsof subobjects.

We saw that you can produce semantics from syntax, i.e., Syn0(T ) to Mod(T ).But we want to go in the other direction.

Example 7.1. Consider a theory T with no predicates and ∃!x(x = x). Com-pare this with the theory T ′ with a single predicate P with axioms ∃!y[P (y)]and (∃!z)[¬P (z)]. These have only one model, so are semantically equivalent,in a boring way. But Syn0(T ) and Syn0(T ′) are not equivalent. The categorySyn0(T ) is equivalent to {0 < 1} and the category Syn0(T ′) is equivalent tofinite sets.

So to produce a syntax, we need choose what to pick. We will chooseSyn0(T ′), and this has the property that we can take disjoint union.

7.1 Disjoint unions

Proposition 7.2. Let C be coherent. Consider x ∈ C and consider subobjectsX1, X2, . . . , Xn ⊆ X which are disjoint, i.e., Xi∧Xj is the least subobject. ThenX1 ∨ · · · ∨Xn ⊆ X is a coproduct of X1, . . . , Xn.

Proof. Given fi : Xi → Y , we want to a unique X1 ∨ · · · ∨Xn → Y . X does notplay a role, so assume that X = X1 ∨ · · · ∨ Xn → Y . We encode they by thegraph. We can take Γ(fi) ⊆ Xi × Y ⊆ X × Y and then take

Z = Γ(f1) ∨ · · · ∨ Γ(fn) ⊆ X × Y.

Now the claim is that Z is the graph of some morphism. This means thath : Z ↪→ X×Y → X is an isomorphism. My first claim is that h is a monomor-phism, which is that Z ↪→ Z ×X Z is an isomorphism. Applying (A5) twice, wecan describe

Z ×X Z =∨i,j

Γ(fi)×X Γ(fj) ⊆ X × Y × Y.

If i 6= j, then Γ(fi)×X Γ(fj) = ∅ because Γ(fi)×X Γ(fj)→ X factors throughXi ∧Xj . (Here, we’re using (A5) again.) So

Z ×X Z =∨i

Γ(fi)×X Γ(fi) ⊆ Z.

This shows that h is a monomorphism.Going back to the picture, we see that Z ⊆ X×Y → X is a monomorphism.

So we can take Z as a subobject of X, and it should contain each Xi. Thisshows that it is Z and so h is an isomorphism.

Example 7.3. Let us take n = 0. Then this shows that the least element∅ ∈ Sub(X) is actually the initial object in C.

Math 278x Notes 22

Definition 7.4. Let C be a category which has fiber products, and considerX,Y ∈ C such that X q Y exists. We say that the coproduct is disjoint if

(1) X ↪→ X q Y and Y ↪→ Y are monomorphisms and

(2) X ×XqY Y is the initial object in C.

Proposition 7.5. Let C be a category. The following are equivalent:

(1) C is coherent and has disjoint coproducts.

(2) C satisfies (A1), (A2), (A4), and

(A3′) C has finite coproducts and they are disjoint,

(A5′) The formation of coproducts commutes with pullback.

Proof. Let’s first check (A5′). Given f : X → Y and Y1, . . . , Yn, we want toshow that ∐

(X ×Y Yi)→ X ×Y∐Yi

is an isomorphism. Y really does’t matter, so we may replace Y by∐i Yi. But

now, it suffices to show that X ×Y Yi are disjoint subobjects of X with unionbeing X. But this is (A5).

Now let C be a category satisfying (2). Let X ∈ C and X1, . . . , Xn ⊆ X. Wewant to form X1∨· · ·∨Xn ⊆ X, and to get this, we look at X1qX2q· · ·qXn.Then we have a map X1 q · · · qXn → X, with image U . This should containall Xi, and cannot have more, and so is X1 ∨ · · · ∨Xn.

So we this is a strengthening of the original axioms that allows us to takedisjoint unions.

7.2 Equivalence relations

Let me give a loose analogy with algebraic geometry. Let T be theory andconsider [ϕ(x1, . . . , xn)]. Loosely speaking, this is like cutting out an affinevariety. But we would like to construct projective varieties as well, and this isquotienting An+1 \ {0}.

Definition 7.6. Let C be any category with finite limits, and let X ∈ C. Anequivalence relation on X is a subobject R ⊆ X ×X such that for all Y ∈ C,

Hom(Y,R) ⊆ Hom(X,R)×Hom(X,R)

is an equivalence relation.

For instance, X×Y X ⊆ X×X is an equivalence relation. In fact, an effectiveepimorphism can be thought of as a coequalizer of a equivalence relation, in thecategory of sets.

Definition 7.7. We say that an equivalence relation R ⊆ X ×X is effectiveif R = X ×Y X for some f : X → Y .

Math 278x Notes 23

Now I am going to replace axiom (A2) with

(A2′) Every equivalence relation is effective.

Proposition 7.8. Let C be a category satisfying (A1), (A4), and (A2′). Thenit also satisfies (A2).

Lemma 7.9. Let C satisfy (A1) and (A4). Suppose we have a pullback square

Y ′ X ′

Y X.

f ′

g′

f

g

If f is an effective epimorphism and f ′ is an isomorphism, then f is an iso-morphism. That is, we can test isomorphisms locally.

Proof. Because g and g′ are effective epimorphisms, we have

Y ′ ×X′ Y ′ Y ′ X ′

Y ×X Y Y X.

f ′

g′

f

g

If f ′ is an isomorphism, then the left vertical map is an isomorphism. Then fis induced by the coequalizer, and so an isomorphism.

Proof of Proposition. We need to construct an image of f : X → Z. We can

look at X×ZX and look at the coequalizer Xg−→ Y . Now we have a factorization

Xg−→ Y

h−→ Z, and need to show that h is a monomorphism. We have pullbacksquares

X ×Y X X ×Z X X ×X

Y Y ×Z Y Y × Y.

∼=

g×g

Because g is an effective epimorphism, X ×Z X → Y ×Z Y is an effectiveepimorphism. Then Y → Y ×Z Y is an isomorphism.

Definition 7.10. A pretopos is a category C such that

(A1) C has finite limits,

(A2′) every equivalence relation is effective,

(A3) C has disjoint coproducts,

(A4′) pullbacks of effective epimorphisms are effective epimorphism,

(A5′) formation of corproducts commute with pullbacks.

Example 7.11. Set and Setfin are pretopoi.

Math 278x Notes 24

8 February 7, 2018

So pretopoi are special coherent categories.

Definition 8.1. A first-order theory T eliminates imaginaries if Syn0(T ) isa pretopos.

In fact, given a first-order theory, you can construct a first-order theory T eq

that eliminates imaginaries. Here is how you do this. Given X = [ϕ(x)] and anequivalence relation R ⊆ X ×X given by [ψ(x, x′)], you might worry that X/Rdoes not exist. But you can add a new type t (interpreted as X/R) and a newpredicate P (x, y) with arity t0, t) (where t0 is the type of x) and axioms

∀~x, ~y(P (x, y)⇒ ϕ(x), ∀xϕ(x)⇒ ∃!yP (x, y),

∀y∃xP (x, y), ∀x, x′, y, y′(P (x, y) ∧ P (x′, y′)⇒ (y = y′ ⇔ ψ(x, x′))).

In fact, this new theory has the same models, because the axioms tell you howto do this.

Theorem 8.2. Let C be a small coherent category. Then there exists a pretoposCeq and a functor of coherent categories λ : C → Ceq with the following property:for any pretopos D,

Funcoh(Ceq,D)◦λ−→ Funcoh(C,D)

is an isomorphism. We call this the pretopos completion of C.

Taking D = Set, we see that C and Ceq have the same models.

Proposition 8.3. Let C and D be pretopoi and l : C → D be a functor. Thefollowing are equivalent:

(1) λ is a morphism of coherent categories.

(2) λ is a morphism of coherent categories and preserves finite coproducts.

We will also see that C → Ceq is fully faithful. Let me give you some intuitionfor what this should be. For X,Y ∈ C, we should be allowed to form X q Y .Maybe the coproduct does not exist, or maybe there is a coproduct but it is notdisjoint.

So here is a question we need to answer. What is map Z → X q Y in Ceq?It is clear that both HomC(Z,X) and HomC(Z, Y ) lies in HomC(Z,XqY ). Butthere should be more. Suppose Z = Z0 ∨ Z1 where Z0 ∧ Z1 = ∅. We have seenthat in this case, Z = Z0 q Z1. Then given Z0 → X and Z1 → Y , we can map

f : Z = Z0 q Z1 → X q Y.

This is a correct description, but we will eventually find Ceq as a full sub-category of a much larger category Shv(C).

Math 278x Notes 25

8.1 Sheaves

Definition 8.4. Let C be any category with fiber products. A Grothendiecktopology on C is a specification of a collection of maps {fi : Ui → X}i∈I thatwe will call covering families satisfying

(T1) If {fi : Ui → X} is a covering and Y → X is any map, then {Ui×XY → Y }is a covering.

(T2) If {Ui → X} is a covering and {Vij → Ui} is a covering, then {Vij → Ui →X} is a covering.

(T3) If {fi : Ui → X} such that some fi admits a section, then it is a covering.

Proposition 8.5. Any enlargement of a covering is a covering.

Example 8.6. Let X be a topological space, and let U be the collection ofopen subsets of X. We declare that {Ui → U} is a covering if U =

⋃i Ui.

This was motivated by algebraic geometry, where maps are not necessarilyinclusions.

Example 8.7. Let X be a scheme, and let C be the category of schemes Uwith an etale map U → X. We can take {Ui → U} is a covering if

∐Ui → U

is surjective.

Proposition 8.8. Let C be a coherent category. Then C has a Grothendiecktopology where we declare that {fi : Ui → X}i∈I is a covering if there is somefinite subset I0 ⊆ I such that ∨i∈I0 im(fi) = X (in Sub(X)).

Proof. (T1) follows since taking images and joins commutes with pullback. For(T2) you can replace all coverings with finite coverings. If the maps Vij → Xfactors through some X ′ ⊆ X, then all Vij → Ui factors through X ′×X Ui, andso X ′ ×X Ui ∼= Ui. Then X ′ ∼= X. You can check (T3).

Definition 8.9. Let C be a category. A presheaf of sets is just a functorF : Cop → Set. If C has a Grothendieck topology, we say that F is a sheaf iffor every covering {Ui → X}, the diagram

F (X)∏i F (Ui)

∏i,j F (Ui ×X Uj)

is an equalizer. The collection of sheaves forms a category Shv(C) ⊆ Fun(Cop,Set).

Example 8.10. If X is a topological space, U is the collection open subsets,then Shv(X) is the sheaves on X.

Let me introduce a bit of notation. If C is a category and Y ∈ C, we denotehY : Cop → Set given by

hY (X) = HomC(X,Y ).

Math 278x Notes 26

Proposition 8.11. Let C be a coherent category, and let Y ∈ C. Then hY :Cop → Set is a sheaf.

Proof. We want to check that

Hom(X,Y )∏i Hom(Ui, Y )

∏i,j Hom(Ui ×X Uj , Y )

is an equalizer. We first denote Xi = im(Ui → X) so that Ui � Xi ↪→ X. Thenby what we did last time, fi : Ui → Y uniquely factors as fi : Xi → Y .

To get X → Y , we encode the objects in the graph Γ(fi) in Sub(X × Y ).Denote Z =

∨Γ(fi). Then it suffices to show that Z → X is an isomorphism.

To show that it is a monomorphism, we look at∨i

Γ(fi, fi) ∼= Z → Z ×X Z =∨i,j

Γ(fi, fj).

In the previous lecture, the non-diagonal terms were just empty. But here, weclaim that Γ(fi, fj) ⊆ Γ(fi, fi). This follows from the fact that fi and fj agreeon Ui ×X Uj .

Note thatC → Fun(Cop,Set); Y 7→ hY

is always fully faithful, and it factors through Shv(C) if C is coherent. We willstudy categories that look like Shv(C), and if C is coherent, we are going to find

C ⊆ Cop ⊆ Shv(C).

Definition 8.12. A topos is a category of the form Shv(C) where C is a smallcategory with a Grothendieck topology.

Example 8.13. If X is a topological space, Shv(X) is a topos. If C is a co-herent category, Chv(C) is a topos. In particular, if T is a first-order theory,Shv(Syn0(T )) is a topos, called the classifying topos of T .

This Shv(Syn0(T )) is the “space of all models” of T . In general, for X atopos, the category C is not part of the data. For instance, Shv(C) and Shv(Ceq)is going to be the same topos.

Math 278x Notes 27

9 February 9, 2018

Let’s recall what we were talking about. A category X is a topos if there is anequivalence X ' Shv(C) for C is small, has limits, with a Grothendieck topology.

9.1 Topoi are pretopoi

Theorem 9.1. Every topos X is a pretopos.

Let’s assume that X = Shv(C).

Theorem 9.2. The inclusion Shv(C) ↪→ Fun(Cop,Set) has an effect adjoint L,and L preserves finite limits.

It want to show why the second theorem implies the first. Assume Theo-rem 9.2.

Lemma 9.3. X has arbitrary limits.

Proof. We can compute limits in Fun(Cop,Set). Then the limit is also going tobe a sheaf, because limits commute with limits.

Lemma 9.4. X as all colimits.

Proof. Note that this not just the colimit computed in Fun(Cop,Set), which isnot necessarily a sheaf. Nevertheless, you can still form the direct limit, andthen apply the sheafification L to it. This will do it.

Lemma 9.5. Coproducts in X are disjoint.

Proof. The product of F and G should be L(F q G ). Because L preservesfinite limits, we can compute

LF ×L(FqG ) LG ∼= L(F ×FqG G ) ∼= L(∅).

But L(∅) is the initial object because L preserves finite limits.

Lemma 9.6. Every equivalence relation in X is effective.

Proof. Let F be a sheaf and consider an equivalence relation R ⊆ F ×F . Nowif we define

E (C) = F (C)/R(C),

then E (C) is a presheaf. This can be described as R ∼= F ×E F . But then,

F ×LE F ∼= L(F ×E F ) ∼= L(R) ∼= R.

This is precisely what it means for the equivalence relation to be effective.

Lemma 9.7. Colimits in X are universal, i.e., commute with fiber products.

Math 278x Notes 28

Proof. Suppose I give you F → G and a collections of sheaves {Gα} over G .What we want to show is that

L(lim−→(F ×G Gα)) ∼= F ×G L(lim−→Gα).

But we have an isomorphism lim−→α(F ×G Gα) ∼= F ×G lim−→Gα, and then we can

apply L.

This immediately implies

Lemma 9.8 (A5). Finite coproducts in X are preserved by pullback.

Lemma 9.9. Pullbacks of effective epimorphisms in X are effective epimor-phisms.

Proof. Suppose I have a pullback of an effective epimorphism F → G to F ′ →G ′. Then the pullback of R = F×G → F is R′ = F ′×G ′F ′. By definition, Gis the coequalizer, which is a colimit, and it should pullback to the coequalizer.This shows that G ′ is the coequalizer and F ′ → G ′ is an effective epimorphism.

We didn’t use much here, except Theorem 9.2.

Proposition 9.10. Let C be a pretopos, and C0 ⊆ C be a full subcategory, whereC0 ↪→ C admits a left adjoint L. If L preserves finite limits, then C0 is a pretopos.

Proof. Exercise.

9.2 Constructing sheafification

Now let us see how to prove Theorem 9.2.

Theorem 9.11. There exists a sheafification functor L : Fun(Cop,Set)→ Shv(C)and L is left exact.

Recall that F : Cop → Set is a sheaf if, for every covering {Ui → X}i∈I , wecan write

F (X)∏i F (Ui)

∏i,j F (Ui ×X Uj).

But it is convenient to understand this in a different way. You can think of itas

F (X)→ lim←−V→X factoring through some Ui

F (V )

being an isomorphism.

Definition 9.12. A sieve on an object X ∈ C is a full subcategory C(0)/X ⊆ C/X

such that for any U → V in C/X , if V ∈ C(0)/X then U ∈ C(0)

/X .

Math 278x Notes 29

Definition 9.13. A sieve C(0)/X is a covering if it contains a covering {fi : Ui →

X}.

Example 9.14. For any collection of maps {Ui → X}, the sieve C(0)/X consisting

of maps V → X that factors through some Ui is a covering sieve.

Now we can say that F : Cop → Set is a sheaf if and only if, for every

covering sieve C(0)/X ⊆ C/X ,

F (X) = lim−→U∈C(0)

/X

F (U).

The good thing is that the space of covering sieves is partially ordered byreverse inclusion. For F any presheaf, we can construct F † : Cop → Set by

F †(X) = lim−→C(0)/X

cov.

lim←−U∈C(0)

/X

F (U).

Note that there is a canonical map F → F †. Also, note that this is naturallyequivalent to the original sheaf if F is a sheaf to begin with.

Proposition 9.15. For any presheaf F , the composite

F → F † → F ††

exhibits F †† as a sheafification of F .

Let me explain why we need the double dagger.

Definition 9.16. A presheaf F is separated if every covering {Ui → X}, themap

F (X)→∏i

F (Ui)

is injective.

Lemma 9.17. For any F , F † is separated.

Lemma 9.18. For any separated F , F † is a sheaf.

Proof. We want to check that F † is a sheaf. Suppose we are given a covering{Ui → X}, and let us assume we have sections si ∈ F †(Ui) that agree onoverlaps. Each section si comes up in a covering Vij → Ui. Then si is given byspecifying sections tij ∈ F (Vij). We can say that Vij → X is a covering. Wewould like to say that {tij} is in F †(X).

We need to check the gluing conditions in order to say this. That is, we needto check that tij and ti′j′ agree on Vij ×X Vi′j′ . Note that si and si′ agree assections of F †(Ui×X uj), and hence there exists a covering Wk → Ui×XUk, therepresentatives agree. Then you need to use the fact that F is separated.

Math 278x Notes 30

10 February 12, 2018

Last time we showed that every topos X is a pretopos, in particular, with infinitecoproducts. What I want to talk about today, is a sort of a converse. But letme introduce a semi-temporary definition.

Definition 10.1. Say X is a pretopos with infinite coproducts. We say that acollection {fi : Ui → X} is a covering if

∐Ui → X is an effective epimorphism.

10.1 Giraud’s theorem

Theorem 10.2 (Giraud). Let X be a category. The following are equivalent:

(1) X is a topos. (This means that X ' Shv(C) where C is a small categorywith finite limits and a Grothendieck topology.)

(2) There exists a fully faithful embedding i : C ↪→ Fun(Cop,Set) such that ihas a left adjoint preserving finite limits, where C is a small category.

(3) X satisfies Giraud’s axioms

(G1) X has finite limits

(G2) every equivalence relation is effective

(G3) X has arbitrary coproducts and they are disjoint

(G4) effective epimorphisms are closed under pullback

(G5) pullbacks of coproducts are coproducts

(G6) there exists a set of objects U of X such that every X ∈ X has acovering {Ui → X} where each Ui ∈ U .

We have proven (1) implies (2), and also (2) implies (3) except for (G6). Forthis, we note that there is a covering∐

X∈C,η∈F(X)

LhX � F .

The interesting direction is (3) implies (1).Using (G6) we can choose a small full subcategory C ⊆ X such that C

contains generators of X , and C is closed under finite limits in X . (Maybe takethe skeleton if you’re worried about the difference of small and essentially small.)

Theorem 10.3. (a) Say that a collection of maps {Ui → X} in C is a cov-ering if it is so in X . This defines a Grothendieck topology on C.

(b) For X ∈ X , define hX : Cop → Set. Then each hX is a sheaf.

(c) The construction X 7→ hX is a an equivalence X ' Shv(C).

Math 278x Notes 31

Proof. (a) If {Ui → X} is a covering, and Y → X is a morphism, we need toshow that {Ui×X Y → Y } is also a covering. This follows from (G4) and (G5).We also need to check that it is transitive, that if {Vi,j → Ui} and {Ui → X}is a covering then {Vi,j → X} is a covering. Here, we can imitate the prooffor checking that the coverings on a coherent category gives a Grothendiecktopology.

(b) For each {Ui → X} a covering, we need to check that

HomX (X,Y ) HomX (∐iUi, Y ) HomX (

∐i,jUi ×X Uj , Y )

is an equalizer. But this follows from that

(∐iUi)×X (

∐jUj)

∐i Ui X

is a coequalizer. Then we have the isomorphism∐i,j Ui ×X Uj ∼= (

∐i Ui) ×X

(∐j Uj) by (G5).(c) Now consider the functor h : X → Shv(C). First note that h preserves

inverse limits (in particular, monomorphisms). First let us prove that h is fullyfaithful. We want to show that

θX : HomX (X,Y )→ HomShv(C)(hX , hY )

is bijective. Fix Y , and say that X is “good” of θX is bijective.

(i) Any X ∈ C is good, by Yoneda.

(ii) If {Ui → X} is a covering, Ui are good, and Ui ×X Uj are good, then Xis good. This is because both∐

i,j Ui ×X Uj∐i Ui X

∐i,j hUi ×hX hUj

∐i hUi hX

are coequalizers by the following lemma.

(iii) If X ∈ X is a subobject of some X ∈ C, then X is good. If {Ui → X} isa covering with Ui ∈ C, then Ui ×X Uj ∼= Ui ×X Uj is in C because C isclosed under finite limits. Then we can apply (ii).

(iv) All X ∈ X are good. For {Ui → X} with Ui ∈ C, we have Ui ×X Uj ⊆Ui ∈ Uj ∈ C. So Ui ×X Uj is good, and by (ii), X is good.

This can’t be the end, because we have not used a lot of axioms. We needto show that h : X → Shv(C) is essentially surjective.

(i) h preserves initial objects. If {Ui → X} is a covering in X then {hUi →hX} is a covering in Shv(C).

Math 278x Notes 32

(ii) h preserves coproducts. Take {Xi}i∈I and X =∐iXi. Then {hXi →

hX} is a covering by the lemma, and so∐i hXi → hX is an effective

epimorphism. We need to show that this is a monomorphism. That is, wewant to show that∐

i hXi → (∐i hXi)×hX (

∐j hXj )

∼=∐i,j hXi×XXj .

This follows from (G3), because Xi×X Xj is an initial object so hXi×XXjis an initial object as well.

Now fix a sheaf F ∈ Shv(C). We want F ' hX for some X ∈ X . We knowthat there exists a covering {hCi → F} with Ci ∈ C. (We had an explicit form.)If we take U =

∐i Ci, then hU =

∐i hCi . Then hU � F .

(a) Suppose F ⊆ hX for some X ∈ X . Then we have hU � F ⊆ hX , and itcomes from U → X. Decompose this into U � X ↪→ X. Then we get

hU � hX ↪→ hX ,

and it is a fact that a decomposition is unique. So hX ' F .

(b) Let F be arbitrary. We have an effective epimorphism hU � F , and so

hR = hU ×F hU hU F

where hR is representable because it is inside hU×U . Now you can showthat R is an equivalence relation, and then write its quotient of X by(G2). Then hX ' F .

This finishes the proof.

Lemma 10.4. Let {Ui → X} be a covering of X . Then {hUi → hX} is acovering in Shv(C).

Proof. We need to show that∐i hUi → hX is an effective epimorphism. Let us

evaluate at a point η ∈ hX(C). Then we can cover Ui ×X C by Vi,j ∈ C, andthen we can map η ∈ hX(C) to hX(Vi,j), and then pull back to hUi(Vi,j).

Math 278x Notes 33

11 February 16, 2018

Let X be a topos. We will say that a collection of maps {fi : Ui → X} is acovering if the induced ∐

Ui → X

is an effective epimorphism.

11.1 Quasi-compactness

Definition 11.1. An object X ∈ X is quasi-compact if every covering {Ui →X}i∈I has a finite subcover {Ui → X}i∈I0 for I0 ⊆ I finite.

Note that {fi : Ui → X} is a covering if and only if {im(fi) → X} is acovering.

Proposition 11.2. Let f : X → Y be an effective epimorphism. If X is quasi-compact, so is Y .

Proof. If {Ui → Y } is a covering, we have {Ui ×Y X → X} a covering. Thenthere is a finite subcover, and then {Ui ×Y X → Y }, is a covering, with each ofthem factoring through Ui.

Proposition 11.3. Let {Xi}i∈I be a collection of objects with coproduct X ∈ X .If I is finite, then X is quasi-compact if and only if all X are quasi-compact.

Proof. Exercise.

For instance, the initial object is quasi-compact.

Definition 11.4. Let X be a topos. An object X ∈ X is quasi-separatedif for any quasi-compact U, V ∈ X and any U → X ← V , their fiber productU ×X V is quasi-compact.

Sometimes objects are quasi-separated for stupid reasons. In order for thisto be a useful notion, we need enough quasi-compact objects.

Example 11.5. Let X ∈ Shv(Rn). Then F ∈ X is quasi-compact if and onlyif F ∼= ∅. So everything is quasi-separated.

Definition 11.6. A topos X is coherent if there is a collection on objects Usuch that

(1) every U ∈ U is quasi-compact and quasi-separated,

(2) U generates X (every X has a covering by objects of U),

(3) U is closed under finite products.

In particular, this implies that the collection of quasi-compact objects of X areclosed under finite products.

Math 278x Notes 34

Definition 11.7. Let C be a small category with finite limits. We say that aGrothendieck topology on C is finitary if every covering {Ui → X}i∈I has afinite subcover.

Proposition 11.8. If C has a finitary Grothendieck topology, then Shv(C) iscoherent.

Proof. We need to construct a class with good finiteness property. Fr C ∈ C,denote hC : Cop → Set be hC(D) = Hom(D,C) and L : Fun(Cop,Set)→ Shv(C).I claim that U = {LhC}C∈C works.

First note that C 7→ LhC preserves finite limits because L preserves all limits.We want for C ∈ C, LhC is quasi-compact. Let {Fi → LhC} be a covering.First note that id : C → C determines an element s ∈ LhC(C). This is acovering means that there exists a covering locally, so that for some {Cj → C}can be sj ∈ LhC(Cj) can be lifted to sj ∈ Fij(Cj) for some ij ∈ I. But ourGrothendieck topology is finitary, so that we may assume that {Cj → C} isfinite. Then {Fi → LhC}i=ij is a finite covering.

Now we show that LhC is quasi-separated. Let F ,G be quasi-compact, andwe want to show that F ×LhC G is quasi-compact. There exists a covering{LhDi → F} which we assume to be finite. Here,

Hom(LhDi , LhC) = Hom(hDi , LhC) = LhC(Di).

This is a section, so each Di admits a finite cover such that each of LhDi →F → LhC comes from a map Di → C. So we may assume that F = hD, withLhD → LhC coming form a map D → C. Likewise, we can replace G with LhE .Then we have that the fiber product is LhD×CE , which is quasi-compact.

Proposition 11.9. Every coherent topos X arises in this way, for a canonicallychosen C.

Definition 11.10. Let X be a coherent topos. We say that X ∈ X is coherentif it is quasi-compact and quasi-separated.

Inside X there is a full subcategory Xcoh ⊆ X of coherent objects.

Lemma 11.11. Let X be a coherent topos. An object X ∈ X is quasi-separatedif and only if for any quasi-compact Y and any f, g : Y → X, their equalizerEq(f, g) is quasi-compact.

Proof. We first want to show that X is quasi-separated under this condition. IfU → X and V → X with U, V quasi-compact, we can write

U ×X V = Eq(U × V → X).

For the converse, assume that X is quasi-separated. For f, g : Y → X with Yquasi-compact, we want to show that Eq(f, g) is quasi-compact. We are going touse the full strength of the coherent condition. Take a finite covering {Ui → Y },and then take the fiber product. Then Eq(Ui → X) cover Eq(Y → X), and

Math 278x Notes 35

it suffices to show that Eq(Ui → X is quasi-compact. Taht is, we may assumeY ∈ U . Now we observe that

Eq(Y → X) Y

Y ×X Y Y × Y

is a pullback. Here Y ×X Y and Y are quasi-compact, and Y × Y is quasi-separated because it is in U .

Corollary 11.12. Let X be coherent and X,Y ∈ X be quasi-separated. ThenX × Y is quasi-separated.

Proof. We want to know that Eq(Z → X × Y ) is quasi-compact. But

Eq(Z → X × Y ) = Eq(Eq(Z → X)→ Y )

is quasi-compact.

Corollary 11.13. Let X be a coherent topos. The coherent objects of X areclosed under finite limits.

Proof. We want to know that if X → Y ← Z and X,Y, Z are coherent thenX ×Y Z is coherent. It is quasi-compact because Y is quasi-separated. AlsoX × Z is quasi-separated, and it is not hard to show that subobjects of quasi-separated are quasi-separated.

Lemma 11.14. Let X be a topos, and consider Xq.c. ⊆ X the full subcategoryof quasi-compact objects. Xq.c. is essentially small.

Proof. We know that there is a set U of generators for X . This means that Ican find, for every quasi-compact X, a finite covering {Ui → X}. Then thereis only a set’s worth of choices of this Ui, and a set’s worth of choices for theequivalence relations.

Theorem 11.15. Let X be a coherent topos.

(a) Xcoh has a finatary Grothendieck topology, where {Ui → X} is a coveringif it is a covering in X .

(b) The construction X 7→ hX induces an equivalence X ↪→ Shv(Xcoh).

Proof. We checked that Xcoh is a finitary Grothendieck topology. Then (b) iswhat we proved in Giraud’s theorem.

Math 278x Notes 36

12 February 21, 2018

Last time we construct from a category C with finite limits and finitary Grothendiecktopology, a topos Shv(C). We want to see what is lost in this construction.

Definition 12.1. Let X and Y be topoi. A geometric morphism from X toY is a functor f∗ : Y → X such that

(1) f∗ preserves finite limits,

(2) f∗ preserves effective epimorphisms,

(3) f∗ preserves coproducts.

Example 12.2. If f : X → Y is continuous, it induces a functor f∗ : Shv(Y )→Shv(X), and it is an example of a geometric morphism.

12.1 Characterizing geometric morphisms

Theorem 12.3. Let f∗ : Y → X be a functor preserving finite limits. Then thefollowing are equivalent:

(1) f∗ is a geometric morphism.

(2) f∗ preserves all colimits.

(3) f∗ has a right adjoint.

The equivalence between (2) and (3) is the adjoint functor theorem. From(2) to (1), we only need to show that it preserves effective epimorphisms, butthis is because coequalizers are send to coequalizers. So the main content ofthis theorem is (1) implies (2).

We want to show that f∗ preserves coequalizers. Let g, g′ : U → Y . Thenthe coequalizer can be thought of as quotienting out by an equivalence relationR = Y ×coeq Y ⊆ Y × Y . From U , this equivalence relation can be built in thefollowing way. First, without loss of generality we can assume U ⊆ Y × Y (bytaking the image U → Y ×Y ). This may not be symmetric, so we replace U byU ∨ Uop. Then for each n ≥ 0, we form the n-fold fiber product

U ×Y U ×Y · · · ×Y U ⊆ Y n+1 → Y × Y.

Take the image Un ⊆ Y ×Y of this map. Now, we can take R =∨n≥0 Un. Note

that all of these operations can be constructed using images, fiber products, andjoins.

Definition 12.4. For X ,Y topoi, we denote Fun∗(Y,X ) ⊆ Fun∗(Y,X ) the fullsubcategory consisting of geometric morphisms.

How do we compute this in practice? Let C be a small category with finitelimits and a Grothendieck topology. Then we have

j : C h−→ Fun(Cop,Set)L−→ Shv(C).

Math 278x Notes 37

Theorem 12.5. Let X be any topos. Then composition with j determines afully faithful

Fun(Shv(C),X )→ Fun(C,X )

such that the essential image is those f : C → X such that

(1) f preserves finite limits,

(2) f preserves coverings.

Example 12.6. Let C be a category with finite limits. Then C has a topologywhere {fi : Ci → C} is a covering of f where todotodo

We have the following general statement.

Theorem 12.7. Let X be any category with small colimits. Then compositionwith h determines an equivalence

LFun(Fun(Cop,Set),X )→ Fun(C,X )

where LFun is the colimit-preserving functors.

Let me only give a construction in the reverse direction. If f : C → X isany functor, define FX : Cop → Set with FX(C) = HomX (f(C), X). ThenX 7→ FX has a left adjoint.

Now let X be a topos, and let C be a category with finite limits. In this case,there is a subcategory

Fun∗(Fun(Cop,Set),X ) ⊆ LFun(Fun(Cop,Set),X ).

This should correspond to some full subcategory, and my claim is that it corre-sponds to

Funlex(C,X ) ⊆ Fun(C,X ),

the functors preserving finite limits. This means that if F : Fun(Cop,Set) → Xpreserves colimits, then F preserves finite limits if and only if f = F ◦h preservesfinite limits.

Assume that f preserves finite limits. We are going to do something similarto the proof of Giraud’s theorem. Let D ⊆ X be a small full subcategorycontaining a set of generators, images of f , and closed under finite limits. Wecan write this functor C → X as

C D X

Fun(Cop,Set) Fun(Dop,Set) Shv(D).

h

f

h′ '

f ′ L

By the universal property we have mentioned earlier, we have F ' L ◦ F ′.Because L preserves finite limits, it suffices to show that F ′ preserves finitelimits.

Math 278x Notes 38

Note that F ′ is given by the left Kan extension along fop : Cop → Dop. Thisjust means that

F ′(F )(D) = lim−→D→f(C)

F (C).

Then for any object D ∈ D, we want that the construction

F 7→ lim−→D→f(C)

F (C)

preserves finite limits. This is because the diagram is filtered.So this was when C did not have a topology. Now let C be a category with

finite limits and a Grothendieck topology. We have

Fun∗(Shv(C ),X ) Fun(C,X )

Fun∗(Fun(Cop,Set),X ).

◦j

◦L◦h

We have shown that the diagonal map is fully faithful, with essential imagebeing the left exact functors.

Let f : C → X be left exact. Then f extends to a geometric morphismF : Fun(Cop,Set)→ X . But we want to find a dotted arrow in

Fun(Cop,Set) X

Shv(C).

L

F

So when does this exist? This is easier to think in the right adjoints. Recallthat FX(C) = HomX (f(C), X). Then we are asking when there exists

Fun(Cop,Set) X

Shv(C).

X 7→FX

So when is FX a sheaf on C for all X ∈ X ?We can just check the sheaf axioms. For every covering {Ci → C}, we need

HomX (f(C), X) HomX (∐f(Ci), X) HomX (

∐i,j Ci ×C Cj , X)

to be an equalizer for all X. This just means that∐f(Ci ×C Cj)

∐f(Ci) f(C).

But because we are in a topos, we have that the left term is∐i,j

f(Ci ×C Cj) ∼=∐i,j

f(Ci)×f(C) f(Cj) ∼= (∐f(Ci))×f(C) (

∐j f(Cj)).

Math 278x Notes 39

So that being a coequalizer is just that {f(Ci)→ f(C)} is a covering.Next time, we will apply these ideas in some examples, like C a coherent

category.

Example 12.8. Consider C = {∗}. Then Shv(C) = Set. So what we proved isthat

Fun∗(Set,X ) ∼= Funlex(C,X ) ∼= {∗}

because left exact functors should take the final object to the final object. Soin the (2-)category of topoi, Set = Shv(∗) is a final object.

Definition 12.9. Let X be a topos. A point of X is a geometric morphismfrom Set to X , that is, objects of Fun∗(X ,Set).

Math 278x Notes 40

13 February 23, 2018

We introduced two lectures ago the notion of a coherent topos. Recall that Xis coherent if it has “enough” quasi-compact quasi-separated objects. In thiscase, Xcoh is closed under finite limits.

Lemma 13.1. If X is coherent, then Xcoh ⊆ X is closed under finite coproducts.

Proof. Let X =∐i∈I Xi, with I finite and Xi coherent. Because each Xi is

quasi-compact, X is quasi-compact. For U → X ← V where U and V quasi-compact, we want to show that U ×X V is quasi-compact. We can then writeUi → Xi ← Vi, and then Ui ×Xi Vi are all quasi-compact. Now we know thatU ×X V =

∐i Ui ×Xi Vi.

Lemma 13.2. Let X be coherent, and let U → X be an effective epimorphism.Let R = U ×X U . If U is coherent and R is quasi-compact, then X is coherent.

Proof. X is quasi-compact because it takes an effective epimorphism from aquasi-compact object. Now we want to prove that X is quasi-separated, andwe use the other characterization of quasi-separated objects in the context ofcoherent topoi. We need to show that if Y is quasi-compact and f, g : Y → Xthen Eq(f, g) ⊆ Y is quasi-compact. We first change Y by taking the pullback

Y U × U

Y X ×X.(f,g)

Then we have Eq(f , g) � Eq(f, g). So we may assume that f and g factorthrough U . Then we get

Eq(f, g) = Y ×U×U R.

Y and R are quasi-compact, and U × U is coherent.

So we have proven the following.

Proposition 13.3. Let X be a coherent topos. Then Xcoh is a pretopos. (Thatis, it is closed under finite limits, finite coproducts, quotients by equivalencerelations.)

Proposition 13.4. Let C and D be coherent categories, and let f : C → D be afunctor. The following are equivalent:

(1) f is a morphism of coherent categories (i.e., f preserves finite limits,effective epimorphisms, finite joins)

(2) f preserves finite limits and also sends coverings to coverings.

Proof. Exercise.

Math 278x Notes 41

Proposition 13.5. Let C be a coherent category, and let X be a topos. Then

Fun∗(Shv(C),X ) Fun(C,X )

Funcoh(C,X ).

'

Proposition 13.6. Let C be a pretopos. Then h : C → Shv(C)coh is an equiva-lence.

Proof. Because it is a Yoneda embedding, it is fully faithful. So the content isthat it is essentially surjective. First lat X be in C and assume that F ⊆ hX . IfF is coherent (equivalently quasi-compact), then we want to show that F ' hX0

for some X0 ⊆ X. Let {hUi → F}i∈I be a covering with I finite. ThenhUi → F ⊆ hX and so we get fi : Ui → X. Then define X0 =

∨i im(fi) ⊆ X.

Then hX0= F .

Now we want to show that any coherent F ∈ Shv(C) has the form hX forsome X ∈ C. The functor h : C → Shv(C) is a morphism of coherent categoriesbecause it preserves coverings. This implies h preserves finite coproducts. Nowthe strategy is the same. We have a finite covering {hUi → F} and thenhU '

∐hUi � F where U =

∐Ui. Sow we have hU ×F hU → hU → F , and

by the first part, we have hU ×F hU ' hR. So we have

hR hU F .

Because C is a pretopos, R is effective. Then we can quotient U by R and getR→ U � X. Then hX ' F .

So the upshot is that there are mutually inverse constructions{coherent

topoi

} {essentially

small pretopoi

}.

X 7→Xcoh

C7→Shv(C)

For C a coherent category and D a pretopos, we get

Funcoh(C,D) ⊆ Funcoh(C,Shv(D)) ' Fun∗(Shv(C),Shv(D)).

We can defineC ↪→ Ceq = Shv(C)coh.

This is going to be the pretopos completion of C. We have just proved that thisis an equivalence for pretopoi C.

Proposition 13.7. Ceq is a “pretopos-completion” of C, i.e., for any pretoposD,

Funcoh(Ceq,D)→ Funcoh(C,D)

is an equivalence.

Math 278x Notes 42

Proof. Without loss of generality, assume that D is small, because we can makeit larger and larger. Then D ' Xcoh for some X a coherent topos. We have

Funcoh(Ceq,Xcoh) Funcoh(C,Xcoh)

Funcoh(Ceq,X ) Funcoh(C,X ).

But landing in Xcoh can be tested on C, because we can built everything in Ceq

can be built from C by finite limits, coproducts, and equivalence relations. Sothis is a pullback square, and

Funcoh(Ceq,X ) ' Fun∗(Shv(Ceq),X ) ' Fun∗(Shv(C),X ) ' Funcoh(C,X ).

Let’s look at an example.

Definition 13.8. Let C be a category with finite limits. A group object ofC is an object G ∈ C with multiplication m : G × G → G such that for eachC ∈ C,

Hom(C,G)×Hom(C,G)m−→ Hom(C,G)

is a group structure on Hom(C,G).

Proposition 13.9. Let G be a group object C, and let Γ be a finitely presentedgroup. Then the functor

C 7→ HomGroup(Γ,HomC(C,G))

is represented by an object GΓ ∈ C.

This construction gives a functor G 7→ {GΓ} and this we get a functor

Group(C)→ Fun(Groupopf.p., C).

Proposition 13.10. This functor is fully faithful, and the image is the leftexact functor.

Definition 13.11. Xgroup = Fun(Groupf.p.,Set) is the classifying topos of groups.

This is because if Y is any topos,

Fun∗(Xgroup,Y) ∼= Funlex(Groupopf.p.,Y) ' Group(Y).

Note that Xgroup is coherent. So Xgroup = Shv(Xgroup,coh) and so its models willbe

Mod(Xgroup,coh) ' Group.

Contemplate about this pretopos.

Math 278x Notes 43

14 February 26, 2018

What I want to do today and Wednesday is to talk about a large class of topoithat come from topology.

14.1 Locales

Definition 14.1. A locale is a poset U such that

(1) every subset S ⊆ U has a least upper bound∨U∈S U (this implies that

every T has a greatest lower bound given by∨{U a lower bound of T})

(2) U ∧∨i∈I Vi =

∨i∈I(U ∧ Vi).

Example 14.2. Let X be a topological space. Then U(X) the collection ofopen subsets U ⊆ X is a locale. It has joins given by taking unions, andmeets given by the interior of the intersections. Then (2) is satisfied, becauseU ∩

⋃i∈I Vi =

⋃i∈I(U ∩ Vi).

Proposition 14.3. Let U be a poset satisfying (1). Then U is a local if andonly if it is a Heyting algebra: that is, for U, V ∈ U , there exists another object(U → V ) such that W ⊆ (U → V ) if and only if W ∩ U ⊆ V . (This is alsoequivalent to that it is cartesian closed.)

Proof. Exercise.

Example 14.4. Any complete Boolean algebra is a locale. This is because(U → V ) = V

∨U c.

Example 14.5. Let X be a topos, and let X ∈ X be an object. Then Sub(X)is a locale.

Proof. For {Ui ⊆ X}i∈I a set of subobjects of X, we have∨Ui = im(

∐Ui → X).

Then we have

V ∧∨Ui = V ×X Ui = V ×X im(

∐Ui → X) = im((

∐Ui ×X V )→ V )

= im(∐

(Ui ×X V )→ V ) =∨i

(V ∧ Ui).

So any topos gives you lots and lots of locales.

Definition 14.6. Let X be a topos and let 1 = 1X be a final object. We defineSub(1) as the underlying locale of X .

Note that Sub(1) ⊆ X is a full subcategory.

Definition 14.7. A topos X is localic if Sub(1) generates X , that is, everyobject X ∈ X has a covering {Ui → X} with Ui ↪→ 1.

Math 278x Notes 44

Proposition 14.8. Let U be any locale. Then U has a topology whose coveringsare families {Ui ≤ X} such that X =

∨Ui.

Proof. Exercise. Note that for X a topological space and U = U(X), we recoverthe usual Grothendieck topology.

So for any locale U , we can associate to it a topos Shv(U).

Proposition 14.9. Let X be a localic topos. Then

h : X → Shv(Sub(1))

is an equivalence.

Proof. This follows from the proof of Giraud’s theorem. Note that Sub(1) isclosed under finite limits.

Proposition 14.10. Let U be a locale. Then U ' Sub(1Shv(U)). More precisely,

h : U → Fun(Uop,Set)

as essential image consisting of sheaves which are subobjects of the final object.

Proof. First, let U ∈ U . Then

hU (V ) =

{{∗} V ⊆ U∅ otherwise.

Let {Vi → V } be a covering. We want

hU (V )→∏

hU (Vi) ⇒∏

hU (Vi ∧ Vj)

to be an equalizer. That is, V ≤ U if and only if Vi ≤ U . This shows that it isa sheaf, and that it is subobject of 1.

To complete the proof, suppose that F is any sheaf on U with F ⊆ 1. ThenF (U) is either empty or a singleton for each U . We want F to be representable.So we define

U =∨

F(V ) 6=∅

V.

The claim is that F ' hU , which means that F (V ) 6= ∅ if and only if V ≤ U .For the reverse direction, invoke the fact that V is a sheaf. It suffices to knowthat F (U) 6= ∅. We have {V → U}F(V ) 6=∅ a covering of U , and then we canrecover F (U) as a one-element set.

The upshot is that the following are equivalent:{localic

topos X

} {locale U

}X 7→Sub(1X )

U7→Shv(U)

I want to sharpen this a bit.

Math 278x Notes 45

Definition 14.11. Let U and V be locales. A morphisms of locales is a mapf∗ : V → V such that f∗ preserves ordering, infinite joins, finite meets. Let mewrite Fun∗(V,U) ⊆ Fun(V,U) for the full subcategory morphisms of locales.

Let U be a locale, and X be a topos. What can we say about Fun∗(Shv(U),X )?We have shown that

Fun∗(Shv(U),X )◦h−→ Fun(U ,X )

is fully faithful, with image consisting of functors preserving finite limits andcoverings. If f : U → X preserves finite limits, then f(1) ∼= 1. Then f(U) ⊆Sub(1X ). Then the condition of preserving finite limits coverings is the same aspreserving finite meets and joins. Thus the image is equivalent to Fun∗(U ,Sub(1X )).

Proposition 14.12. Geometric morphisms of topoi from X to Shv(U) areequivalent to morphisms of locales Sub(1X ) to U .

So we have a pair of adjoint functors

{topoi X

} {locales U

}.

X 7→Sub(1X )

U7→Shv(U)

Question. What is the relationship between locales and topological spaces?

I already explained the construction X 7→ U(X). This is a functor. That is,if f : X → Y is continuous, then we get a map

f−1 : U(Y )→ U(X)

of locales. But this functor

Top→ {locales}; X 7→ U(X)

is not fully faithful.

Example 14.13. Let X have the trivial topology: U(X) = {∅, X}. This doesnot depend on X at all.

It is also not essentially surjective. Many locales do not come from topolog-ical spaces. We are going to prove the following next time.

Theorem 14.14. The construction X 7→ U(X) is induces an equivalence{sobor topological

spaces

} {spatiallocales

}.'

Math 278x Notes 46

15 February 28, 2018

Last time we defined a locale, which is a poset behaving like the open sets of atopological space. We had a construction

{topological spaces} → {locales}; X 7→ U(X).

15.1 Points of a locale

Definition 15.1. Let U be a locale. A point x of U is:

(A) a morphism of locales {0, 1} → U , which is a morphism x∗ : U → {0 < 1}that preserves joins and finite meets.

(B) a subset Ux ⊆ U (the elements Ux = {U ∈ U : x∗U = 1}) such that

(a) if U ≤ V and U ∈ Ux then V ∈ Ux,

(b) if∨Ui ∈ Ux, then some Ui ∈ Ux,

(c) Ux is closed under finite limits.

Let us write Pt(U) as the set of points of U .

Note that (a) and (b) imply that there is a U(x) =∨U /∈Ux U such that

Ux = {U 6≤ U(x)}. (c) translates to the assertion that U(x) is prime: U(x) isnot the largest element of U and U(x) = V ∧W implies U(x) = V or U(x) = W .Then a point of U just a prime element.

For any U ∈ U , associate the set of points

U = {x : U ∈ Ux} ⊆ Pt(U).

That is, we have a map

U −→ {subsets of Pt(U)}.

The conditions tell us that

(a) this is order-preserving, i.e., U ≤ V implies U ⊆ V ,

(b) U 7→ U preserves joins,

(c) U 7→ U preserves finite meets.

So we they form a topology, and we can set Pt(U) to be a topological space.Then we have a construction

U −→ U(Pt(U)).

Let X be a topological space, and let U be any locale. Then a continuousmap f : X → Pt(U) is taking x ∈ X to Uf(x). This is a binary relation on X×Umeaning “f(x) ∈ U”. This relation satisfies

(a) “f(x) ∈ U” implies “f(x) ∈ V ” if U ≤ V ,

Math 278x Notes 47

(b) “f(x) ∈∨Ui” implies “f(x) ∈ Ui” for some i,

(c) “f(x) ∈ Ui” implies “f(x) ∈∧Ui” where the meet is finite

(d) for each U ∈ U , the set {x ∈ X : “f(x) ∈ U”} is open.

You can also think of this as taking U to subsets of X. Then this is just a mapfrom U(X) to U of locales. That is, we have a canonical bijection

Homcont(X,Pt(U)) ∼= HomLocales(U(X),U) = Fun∗(U(X),U).

So we have a pair of adjoint functors{topological

spaces

} {locales

}.

X 7→U(X)

U7→Pt(U)

What happens if we do this construction in succession? The points inPt(U(X)) is the open subsets U ⊆ X which are prime, that is, U 6= X andU = V ∩W implies U = V or U = W . Equivalently, if we write K = X − U ,then K is irreducible, that is, K 6= ∅ and K = K0 ∪K1 implies K = K0 orK = K1.

15.2 Sober spaces and spatial locales

Definition 15.2. A topological space X is called sober if X → Pt(U(X)) isbijective, i.e., every irreducible set in X has a unique “generic point”.

If this is satisfied, then X → Pt(U(X)) is a homeomorphism.

Example 15.3. Any Hausdorff space is sober. Any scheme is sober.

Proposition 15.4. For any locale U , the points Pt(U) is sober.

Proof. We can show that

Pt(U)fPt(U)−−−−→ Pt(U(Pt(U)))

Pt(g:U(Pt(U))→U)−−−−−−−−−−−−→ Pt(U)

is the identity. We want to show that fPt(U) is bijective. So it suffices to showthat Pt(g) is injective. But note that g∗ : U → U(Pt(U)) is surjective, so thepoints should be injective as sets. You can also prove this directly.

Definition 15.5. A locale U is called spatial if g : U(Pt(U)) → U is andisomorphism of locales.

We know that U → U(Pt(U)) is necessarily surjective. Note that U is spatialif and only if U has “enough points”, i.e., if U, V ∈ U are distinct, then thereexist a point x ∈ Pt(U) such that U ∈ Ux and V /∈ Ux or vice versa. (This justmeans U and V are different in U(Pt(U)).)

Proposition 15.6. A locale U is spatial if and only if U ∼= U(X) for sometopological space X.

Math 278x Notes 48

Proof. Only if is clear. For if, we note that for U, V ∈ U(X) these are open setsand f : X → Pt(U(X)). todotodo

Corollary 15.7. The adjunction gives an equivalence{sober

topological spaces

}'

{spatial locales

}.

The inclusion{sober spaces} ↪→ {spaces}

has a left adjoint X 7→ Pt(U(X)). The inclusion

{spatial locales} ↪→ {locales}

has a right adjoint U 7→ U(Pt(U)) = U/ ' with U ' V if U = V .

Example 15.8. If X has the trivial topology and nonempty, then this is X 7→{∗}.

Example 15.9. This doesn’t have to be surjective. If X is a variety over C withthe Zariski topology (in the classical sense). Then Pt(U(X)) is the underlyingspace of X as a scheme.

In practice, failure to be sober is a pathology. So X to Pt(U(X)) is animprovement. For locales, U 7→ U(Pt(U)) often loses interesting information.There are many interesting examples of locales with no points.

Example 15.10 (Deligne). Let U be the set of equivalence classes of measurablesets X ⊆ [0, 1], with X ∼ Y if µ((X − Y ) ∪ (Y − X)) = 0. This is a poset,and in fact a Boolean algebra. It is even a complete: every {Xi}i∈I has a join∨Xi. (This is generally not given by the union unless I is countable.) For any

collection {Xi}i∈I , we can find a countable subcollection with the same join,because the measure of the union of the segment can only increase in a countableway. So U is a locale. But U has no points. If U ⊆ [0, 1] is prime, then [0, 1]−Udoes not have measure zero and cannot decompose into two smaller sets. Butthis is impossible.

Math 278x Notes 49

16 March 2, 2018

So we finished the first part of the class. Let me give a outline of this class.We first translated a first-order theory into a Boolean pretopos. By passing toa sheaf, we showed that these are equivalent to coherent topoi, and they arecontained in topoi. In topoi, there are localic topoi, which are equivalent to lo-cales. Inside tehre are spatial locales, which were equivalent to sober topologicalspaces.

Question. Let X be a topos. How far can X be from localic topoi?

Theorem 16.1 (Joyale–Tierney). Let X be a topos. Then there is a geometricmorphism π : U → X where U is localic and

colim(U ×X U ×X U ⇒→ U ×X U ⇒ U)∼=−→ X .

Moreover, everything in this diagram (except X ) will be localic.

I haven’t explained what these are, but topoi are going to be 2-categories,and this is supposed to be quotienting U . Let me tell you the end of the proof,which is constructing U from X .

16.1 Classifying topoi

Definition 16.2. Let Setfin be the category of finite sets. Now define

XSet = Fun(Setfin,Set) ' Funfil.colim(Set,Set).

This is going to be a topos, and we are going to call this the classifying toposof sets. There is a canonical object X0 ∈ XSet, given by the inclusion functor.In other words, it is corepresented by ∗.

Proposition 16.3. Let X be any topos. Evaluation on X0 ∈ XSet induces andequivalence

Fun(XSet,X ) ' X .

Proof. Note that XSet are presheaves on Setopfin, and have finite limits. So we

haveFun∗(XSet,X ) ' Funlex(Setop

fin,X ).

But note that if S is a finite set, we have S =∐s∈S{s} in Setfin so we have

S =∏s∈S{s} in Setop

fin. So we should have f(S) = f(∗)S .

It follows that Xset is not localic. If it were localic, given any X ∈ X wecan write down a geometric morphism π : X → XSet such that π∗X0

∼= X, andthen we would get that X is covered by subobjects of 1. So this is the furthestpossible topos from being localic.

Proposition 16.4. There is a Grothendieck topology on Setopfin where a collection

of maps {S → Ti}i∈I is a covering if some S → Ti is injective.

Math 278x Notes 50

Definition 16.5. Let us define XSet 6=∅ = Shv(Setopfin) ⊆ XSet. This is the clas-

sifying topos of nonempty sets.

Proposition 16.6. F : Setfin → Set is a sheaf if and only if F (∅) → F (∗) ⇒F (∗∗) is an equalizer.

You can also check that XSet 6=∅∼= Fun(Set6=∅fin ,Set). Note that X0 ∈ XSet 6=∅ .

Proposition 16.7. Let X be any topos. Then

Fun∗(XSet6=∅ ,X )→ X

given by evaluation on X0 is fully faithful, and the essential image consists ofX ∈ X such that X → 1 is an effective epimorphism.

Proof. There is a fully faithful embedding of the left hand side to Fun(Setopfin,X )

with image the functors preserving coverings and finite limits. If f(∗) = X, thenwe have f(S) ∼= XS . When does this preserve coverings? We need every injec-tion S ↪→ T to induce an effective epimorphism XT → XS . This is automaticif S 6= ∅. If S = ∅, then XT → 1 should be an effective epimorphism.

So XSet 6=∅ and this is again some free thing generated by X0. This is alsovery very far from localic, but let us look at morphisms from a localic topos toX .

Example 16.8. Consider Equiv(Z) the set of all equivalence relations on Z.There is a topology generated by

Ui,j = {E : iEj}.

On this topological space, we have a sheaf F informally described as

FE = Z/E.

More formally, we are taking the quotient

Z× Z ⊇ R ⇒ Z � F .

Here, F → 1 is an effective epimorphism. Now we have a geometric morphism

ϕ : Shv(Equiv(Z))→ XSet 6=∅

such that F ' ϕ∗X0.

More generally, let X be any topos, and let X ∈ X be a “nonempty” object.(Again, this means that the map to 1 is an effective epimorphism.) Then thereexists a geometric morphism ψ : X → XSet 6=∅ such that X ' ψ∗X0. In the2-category of topoi, you can take fiber products.

Enum(X) Shv(Equiv(Z))

X XSet 6=∅ .

ψ′

ϕ′ ϕ

ψ

Math 278x Notes 51

Now we haveψ′∗Z � ψ′∗F ' ϕ′∗X.

So X is covered by countable copies. So here is how the proof of the theoremof Joyale–Tierney will work. Let X be any topos, and choose a set of gener-ators {Xi}. Take X =

∐iXi. Without loss of generality, assume that X is

“nonempty”. Then we have a covering

U = Enum(X)→ X .

Example 16.9. Let X = Set and X some set. Consider

Enum(X) Shv(Equiv(Z))

Set XSet 6=∅ .

Let us take points here. Then we should get

Surjection(Z→ X) Equiv(Z)

∗ Set6=∅.X

IfX was uncountable, the this the points of Enum(X) is empty, but Enum(X)→Set is still going to be an open surjection.

Math 278x Notes 52

17 March 5, 2018

We were trying to prove Joyal and Tierney’s theorem of resolving topoi.

Enum(X) Shv(Equiv(Z))

X XSet 6=∅

17.1 Open maps

Recall that that a map f : X → Y of topological spaces is open if U ⊆ Ximplies f(U) ⊆ Y open. In this case, the map U(X)→ U(Y ) has a left adjointU 7→ f(U).

Definition 17.1. Let U and V be locales. We say that f : U → V is a morphismof locales is open if

(1) f∗ has a left adjoint f!,

(2) for any U ∈ U and V ∈ V, f!(U ∧ f∗V ) = f!(U) ∧ V .

Example 17.2. Let f : X → Y be a map of topological spaces. If f is openthen the induced map U(X)→ U(Y ) is open. This is because U 7→ f(U) givesthe adjoint and f(U ∩ f−1(V )) = f(U) ∩ V .

The converse can fail even for sober spaces.

Proposition 17.3. Let f : X → Y is a continuous map and assume that eachy ∈ Y is closed. Then the following are equivalent:

(1) f is open.

(2) U(X)→ U(Y ) is open.

(3) f∗ : U(Y )→ U(X) has a left adjoint.

Proof. We only need to show (3)⇒ (1). Then for each U ⊆ X, we have f!(U) ⊆Y . This should be the smallest open subset V of Y such that U ⊆ f−1(V ),or f(U) ⊆ V . Tautologically, we have f!(U) ⊇ f(U). Suppose that y ∈ f!(U).Then f!(U) is not contained in Y −{y} which is open. This shows that f(U) isnot contained in Y − {y} and y ∈ f(U). So f!(U) = f(U) and we get (1).

Example 17.4. Consider Y = {0, η} with X = {0}. There exists an f! given byf!(X) = Y , but it is not an open morphism of locales because f!(X∩f−1({η})) =∅ while f!(X) ∩ {η} = {η}.

Definition 17.5. A map of locales f : U → V is an open surjection if f isopen and f!(1U ) = 1V .

Note that f!(1U ) = 1V implies more generally that f!(f∗V ) = V for any

V ∈ V by the projection formula.

Math 278x Notes 53

Proposition 17.6. Let f : U → V and g : V → W be maps of locales. If f andg are open, then g ◦f is open. (The corresponding statement for open surjectionalso holds.)

Example 17.7. Let X be a topos, and let f : X → Y be a map in X . Thenwe have a morphism

Sub(X)→ Sub(Y ); V 7→ X ×Y V,

and this is an open map of locales. This is because we have f!(U) = im(U → Y )and

im(U → Y ) ∧ V = im(U → Y )×Y V= im(Y ×Y V → Y ) = im(U ×X (X ×Y V )→ Y )

giving the projection formula. This is an open surjection if and only if f is aneffective epimorphism.

Now let me give the general definition.

Definition 17.8. Let f : X → Y be a geometric morphism of topoi. Then foreach Y ∈ Y, we get a map of locales Sub(f∗Y ) → Sub(Y ). We say that f isopen if each fY is open, and it is an open surjection if each fY is.

Proposition 17.9. Let f : X → Y be a geometric morphism. Suppose I giveyou Y0 ⊆ Y a set of generators. Then f is open if and only if fY is open foreach Y ∈ Y0, and similarly for open surjections.

Proof. One direction is obvious. Assume that fY is open for Y ∈ Y0. Choose anyY ∈ Y, and we want to show that fY is open. Choose a covering {Yi → Y }i∈Iwith Yi ∈ Y0. We have a morphism of locales Sub(Y ) → Sub(f∗Y ) and firstwant to show that this has a left adjoint. When is U ⊆ f∗V (in Sub(f∗Y ))?This this true if and only if U ×f∗Y f∗Yi ⊆ f∗V ×f∗Y f∗Yi = f∗(V ×Y Yi) forall i. Because Sub(f∗Yi)→ Sub(Y ) is an open surjection, this is equivalent to

fYi!(U ×f∗Y f∗Yi) ⊆ V ×Y Yi

for all i. The conclusion is that given U ⊆ f∗Y ,

fY !(U) =∨i∈I

im(fYi!(U ×f∗Y f∗Yi)→ Y )

is the smallest V ⊆ Y such that U ⊆ f∗V . This gives the left adjoint fY !

for f∗Y : Sub(Y ) → Sub(f∗Y ). To complete the proof, we need to check theprojection formula for fY !, and this is an exercise.

Proposition 17.10. Let X be a topos, and let U be a locale. Let f : X → Shv(U)be a geometric morphism, which corresponds to g : Sub(1X ) → U . Then f isopen (as a map of topoi) if and only if g is open (as a map of locales).

Math 278x Notes 54

Proof. The only if direction is clear. For the if direction, assume that g is open.We want that for any Y ∈ Shv(U) the map fY : Sub(f∗Y ) → Sub(Y ) is open.But we can check this on representable sheaves hU for U ∈ U . So we check that

fY : Sub(f∗hU )→ Sub(hU )

is open. Then we are actually proving that

{W ⊆ 1X : W ≤ g∗U} → {V ∈ U : V ≤ U}

is open. But we can construct the left adjoint of f∗Y by just restricting g! to thesubset, and also the projection formula follows immediately.

Corollary 17.11. Let f : U → V be a morphism of locales. Then f is open(open surjective) if and only if Shv(U)→ Shv(V) is open (open surjective) as amap of topoi.

Corollary 17.12. Let X be any topos. Then the unit map u : X → Shv(Sub(1X ))is an open surjection.

Proof. This is because Sub(1X )→ Sub(1X ) is an open surjection.

Let me advertise what we will do next time. We will study the map

Shv(Equiv(Z))→ XSet6=∅

is an open surjection.

Proposition 17.13. Let f : X → Y be a geometric morphism, and let Y ∈ Ybe an object. Given a map α : X → f∗Y , we get a morphism

Sub(X)→ Sub(f∗Y )→ Sub(Y )

of locales.

(1) If fY is open, then so is uX : Sub(X)→ Sub(Y ) is open.

(2) If f∗Y has a covering {Xi → f∗Y } and each UXi is open, so is fY .

Proof. For (2), we can find this smallest open object by

fY !(U) =∨i∈I

uxi!(U ×f∗Y Xi).

The projection formula follows.

Index

classifying topos, 26of nonempty sets, 50of sets, 49

coherent, 33, 34coherent category, 18consistence, 18covering, 30, 33covering families, 25covering sieve, 29

Deligne’s completeness theorem, 18disjoint coproduct, 22

effective, 22effective epimorphism, 10, 33eliminates imaginaries, 24equivalence relation, 22

finitary, 34formula, 6

Godel’s completeness theorem, 18geometric morphism, 36Giraud’s axioms, 30Giraud’s theorem, 30Grothendieck topology, 25group object, 42

irreducible, 47

language, 3linearly ordered set, 3locale, 43

morphism, 45point, 46

localic topos, 43

model, 3, 6, 15

open morphism, 52, 53open surjection, 52, 53

point, 39presheaf, 25pretopos, 23pretopos completion, 24prime, 47propositional logic, 8

quasi-compact, 33quasi-separated, 33

sentence, 6separated, 29sheaf, 25sieve, 28sober space, 47spatial locale, 47subobject, 10

theory, 3, 6topos, 26

localic, 43typed language, 8

underlying locale, 43

weak pretopos, 14, 16Boolean, 14

weak syntactic category, 7

55