TOPOLOGY

004C

Contents
1. Introduction 1
2. Basic notions 2
3. Hausdorff spaces 2
4. Separated maps 3
5. Bases 4
6. Submersive maps 5
7. Connected components 7
8. Irreducible components 10
9. Noetherian topological spaces 14
10. Krull dimension 15
11. Codimension and catenary spaces 16
12. Quasi-compact spaces and maps 18
13. Locally quasi-compact spaces 21
14. Limits of spaces 25
15. Constructible sets 27
16. Constructible sets and Noetherian spaces 30
17. Characterizing proper maps 31
18. Jacobson spaces 34
19. Specialization 36
20. Dimension functions 39
21. Nowhere dense sets 41
22. Profinite spaces 42
23. Spectral spaces 43
24. Limits of spectral spaces 49
25. Stone-Čech compactification 52
26. Extremally disconnected spaces 53
27. Miscellany 56
28. Partitions and stratifications 57
29. Colimits of spaces 58
30. Topological groups, rings, modules 59
31. Other chapters 62
References 63

1. Introduction
004D Basic topology will be explained in this document. A reference is [Eng77].

This is a chapter of the Stacks Project, version 2c3bdd57, compiled on Jun 18, 2024.
1
 TOPOLOGY 2

2. Basic notions
004E The following is a list of basic notions in topology. Some of these notions are
discussed in more detail in the text that follows and some are defined in the list,
but others are considered basic and will not be defined. If you are not familiar with
most of the italicized concepts, then we suggest looking at an introductory text on
topology before continuing.
004F (1) X is a topological space,
004G (2) x ∈ X is a point,
0B12 (3) E ⊂ X is a locally closed subset,
004H (4) x ∈ X is a closed point,
08ZA (5) E ⊂ X is a dense subset,
004I (6) f : X1 → X2 is continuous,
0BBW (7) an extended real function f : X → R ∪ {∞, −∞} is upper semi-continuous
if {x ∈ X | f (x) < a} is open for all a ∈ R,
0BBX (8) an extended real function f : X → R ∪ {∞, −∞} is lower semi-continuous
if {x ∈ X | f (x) > a} is open for all a ∈ R,
(9) a continuous map of spaces f : X → Y is open if f (U ) is open in Y for
U ⊂ X open,
(10) a continuous map of spaces f : X → Y is closed if f (Z) is closed in Y for
Z ⊂ X closed,
004J (11) a neighbourhood of x ∈ X is any subset E ⊂ X which contains an open
subset that contains x,
09R7 (12) the induced topology on a subset E ⊂ X,
S
004K (13) U : U = i∈I Ui is an open covering of U (note: we allow any Ui to be
empty and we even allow, in case U is empty, the empty set for I),S
0GM1 (14) a subcover of a covering as in (13) is an open covering U ′ : U = i∈I ′ Ui
where I ′ ⊂ I,
004L (15) S open covering V isSa refinement of the open covering U (if V : U =
the
j∈J Vj and U : U = i∈I Ui this means each Vj is completely contained
in one of the Ui ),
004M (16) {Ei }i∈I is a fundamental system of neighbourhoods of x in X,
004N (17) a topological space X is called Hausdorff or separated if and only if for every
distinct pair of points x, y ∈ X there exist disjoint opens U, V ⊂ X such
that x ∈ U , y ∈ V ,
08ZB (18) the product of two topological spaces,
08ZC (19) the fibre product X ×Y Z of a pair of continuous maps f : X → Y and
g :Z →Y,
0B30 (20) the discrete topology and the indiscrete topology on a set,
(21) etc.

3. Hausdorff spaces
08ZD The category of topological spaces has finite products.

08ZE Lemma 3.1. Let X be a topological space. The following are equivalent:
(1) X is Hausdorff,
(2) the diagonal ∆(X) ⊂ X × X is closed.
 TOPOLOGY 3

Proof. We suppose that X is Hausdorff. Let (x, y) ̸∈ ∆(X), i.e., x ̸= y. There

are U and V disjoint open sets of X such that x ∈ U and y ∈ V . This implies
that U × V ⊂ (X × X) \ ∆(X). This shows that (X × X) \ ∆(X) is an open set
of X × X which is equivalent to say that the diagonal ∆(X) ⊂ X × X is closed in
X × X. The converse is similar: The complement (X × X) \ ∆(X) consist precisely
of (x, y) ∈ X × X with x ̸= y. Thus, if ∆(X) ⊂ X × X is closed, then, by the
definition of the product topology, for every such (x, y), there are opens U, V ⊂ X
with (x, y) ∈ U × V and (U × V ) ∩ ∆(X) = ∅. In other words, with x ∈ U and
y ∈ V such that U ∩ V = ∅. □
08ZF Lemma 3.2. Let f : X → Y be a continuous map of topological spaces. If Y is
Hausdorff, then the graph of f is closed in X × Y .
Proof. The graph is the inverse image of the diagonal under the map X × Y →
Y × Y . Thus the lemma follows from Lemma 3.1. □
08ZG Lemma 3.3. Let f : X → Y be a continuous map of topological spaces. Let
s : Y → X be a continuous map such that f ◦ s = idY . If X is Hausdorff, then
s(Y ) is closed.
Proof. This follows from Lemma 3.1 as s(Y ) = {x ∈ X | x = s(f (x))}. □
08ZH Lemma 3.4. Let X → Z and Y → Z be continuous maps of topological spaces.
If Z is Hausdorff, then X ×Z Y is closed in X × Y .
Proof. This follows from Lemma 3.1 as X ×Z Y is the inverse image of ∆(Z) under
X × Y → Z × Z. □

4. Separated maps
0CY0 Just the definition and some simple lemmas.
0CY1 Definition 4.1. A continuous map f : X → Y of topological spaces is called
separated if and only if the diagonal ∆ : X → X ×Y X is a closed map.
0CY2 Lemma 4.2. Let f : X → Y be continuous map of topological spaces. The
following are equivalent:
(1) f is separated,
(2) ∆(X) ⊂ X ×Y X is a closed subset,
(3) given distinct points x, x′ ∈ X mapping to the same point of Y , there exist
disjoint open neighbourhoods of x and x′ .
Proof. If f is separated, by Definition 4.1, ∆ is a closed map. The fact that X is
closed in X gives us that ∆(X) is closed in X ×Y X. Thus (1) implies (2).
Assune ∆(X) ⊂ X ×Y X is a closed subset and denote U the complementary open.
This means we have an open set W ⊂ X × X such that W ∩ (X ×Y X) = U .
However, by definition of the product topology, if (x, x′ ) ∈ W ∩ (X ×Y X), we
have V and V ′ open sets of X such that x ∈ V , x′ ∈ V ′ and V × V ′ ⊂ W . If
we had V ∩ V ′ ̸= ∅, we would have z ∈ V ∩ V ′ . However, (z, z) ∈ X ×Y X, so
(z, z) ∈ (V × V ′ ) ∩ (X ×Y X) ⊂ U , which is absurd. Therefore V ∩ V ′ = ∅, and we
have two disjoint open neighborhoods for x and x′ . It proves that (2) implies (3).
Finally, we suppose that given distinct points x, x′ ∈ X mapping to the same point
of Y , there exist disjoint open neighbourhoods of x and x′ . Let F be a closed set
 TOPOLOGY 4

of X. We will show that ∆(F ) is a closed subset of X ×Y X. Let (x, x′ ) ∈ X ×Y X

be a point not contained in ∆(F ). Then either x ̸= x′ or x ̸∈ F . In the first
case, we choose disjoint open neighbourhoods V, V ′ ⊂ X of x, x′ and we see that
(V × V ′ ) ∩ X ×Y X is an open neighbourhood of (x, x′ ) not meeting ∆(F ). In the
second case, we see that ((X \ F ) × X) ∩ X ×Y X is an open neighbourhood of
(x, x′ ) not meeting ∆(F ). We have shown that (3) implies (1). □
0CY3 Lemma 4.3. Let f : X → Y be continuous map of topological spaces. If X is
Hausdorff, then f is separated.
Proof. Clear from Lemmas 4.2 and 3.1 as ∆(X) closed in X × X implies ∆(X)
closed in X ×Y X. □
0CY4 Lemma 4.4. Let f : X → Y and Y ′ → Y be continuous maps of topological
spaces. If f is separated, then f ′ : Y ′ ×Y X → Y ′ is separated.
Proof. Follows from characterization (2) of Lemma 4.2. Namely, with X ′ = Y ′ ×Y
X the diagonal ∆(X ′ ) in the fibre product X ′ ×Y ′ X ′ is the inverse image of ∆(X)
in X ×Y X. □

5. Bases
004O Basic material on bases for topological spaces.
004P Definition 5.1. Let X be a topological space. A collection of subsets B of X is
called a base for the topology on X or a basis for the topology on X if the following
conditions hold:
(1) Every element B ∈ B is open in X.
(2) For every open U ⊂ X and every x ∈ U , there exists an element B ∈ B
such that x ∈ B ⊂ U .
The following lemma is sometimes used to define a topology.
0D5P Lemma S 5.2. Let X be a set and let B be a collection of subsets. Assume that
X = B∈B B and that given x ∈ B1 ∩ B2 with B1 , B2 ∈ B there is a B3 ∈ B with
x ∈ B3 ⊂ B1 ∩ B2 . Then there is a unique topology on X such that B is a basis for
this topology.
Proof. Let σ(B) be the set of subsets of X which can be written as unions of
elements of B. We claim σ(B) is a topology. Namely, the empty set is an element of
σ(B) (as an empty union) and X is an element of σ(B) (as the union of all elements
of B). It isSclear that σ(B) isSpreserved under unions. Finally, if U, V ∈ σ(B) then
write U = i∈I Ui and V = j∈J Vj with Ui , Vj ∈ B for all i ∈ I and j ∈ J. Then
[
U ∩V = Ui ∩ Vj
i∈I,j∈J

The assumption in the lemma tells us that Ui ∩Vj ∈ σ(B) hence we see that U ∩V is
too. Thus σ(B is a topology. Properties (1) and (2) of Definition 5.1 are immediate
for this topology. To prove the uniqueness of this topology let T be a topology on
X such that B is a base for it. Then of course every element of B is in T by (1)
of Definition 5.1 and hence σ(B ⊂ T . Conversely, part (2) of Definition 5.1 tells us
that every element of T is a union of elements of B, i.e., T ⊂ σ(B). This finishes
the proof. □
 TOPOLOGY 5

004Q Lemma 5.3.S Let X be a topological space. Let B be a basis for the topology on X.
Let US: U = i Ui be an open covering of U ⊂ X. There exists an open covering
U = Vj which is a refinement of U such that each Vj is an element of the basis
B.
S
Proof. If x ∈ U = i∈I Ui , there is an ix ∈ I such that x ∈ Uix . Thus we have
a Bix ∈ B verifying x ∈ Bix ⊂ Uix . Set J = {ix |x ∈ U } and for j = ix ∈ J set
Vj = Bix . This gives the desired open covering of U by {Vj }j∈J . □
08ZI Definition 5.4. Let X be a topological space. A collection of subsets B of X is
called a subbase for the topology on X or a subbasis for the topology on X if the
finite intersections of elements of B form a basis for the topology on X.
In particular every element of B is open.
08ZJ Lemma 5.5. Let X be a set. Given any collection B of subsets of X there is a
unique topology on X such that B is a subbase for this topology.
T
Proof. By convention ∅ B = X. Thus we can apply Lemma 5.2 to the set of
finite intersections of elements from B. □
0D5Q Lemma 5.6.S Let X be a topological space. Let B be a collection
S of opens of X.
Assume X = U ∈B U and for U, V ∈ B we have U ∩ V = W ∈B,W ⊂U ∩V W . Then
there is a continuous map f : X → Y of topological spaces such that
(1) for U ∈ B the image f (U ) is open,
(2) for U ∈ B we have f −1 (f (U )) = U , and
(3) the opens f (U ), U ∈ B form a basis for the topology on Y .
Proof. Define an equivalence relation ∼ on points of X by the rule
x ∼ y ⇔ (∀U ∈ B : x ∈ U ⇔ y ∈ U )
Let Y be the set of equivalence classes and f : X → Y the natural map. Part (2)
holds by construction. The assumptions on B exactly mirror the assumptions in
Lemma 5.2 on the set of subsets f (U ), U ∈ B. Hence there is a unique topology on
Y such that (3) holds. Then (1) is clear as well. □

6. Submersive maps
0405 If X is a topological space and E ⊂ X is a subset, then we usually endow E with
the induced topology.
09R8 Lemma 6.1. Let X be a topological space. Let Y be a set and let f : Y → X be
an injective map of sets. The induced topology on Y is the topology characterized
by each of the following statements:
(1) it is the weakest topology on Y such that f is continuous,
(2) the open subsets of Y are f −1 (U ) for U ⊂ X open,
(3) the closed subsets of Y are the sets f −1 (Z) for Z ⊂ X closed.
Proof. The set T = {f −1 (U )|U ⊂ X open} is a topology on Y . Firstly, ∅ = f −1 (∅)
and f −1 (X) = Y . So T contains ∅ and Y .
Now let {Vi }i∈I be a collection of open subsets where Vi ∈ T and write Vi = f −1 (Ui )
where Ui is an open subset of X, then
[ [ [ 
Vi = f −1 (Ui ) = f −1 Ui
i∈I i∈I i∈I
 TOPOLOGY 6

S S
So i∈I Vi ∈ T as i∈I Ui is open in X. Now let V1 , V2 ∈ T . We have U1 , U2 open
in X such that V1 = f −1 (U1 ) and V2 = f −1 (U2 ). Then
V1 ∩ V2 = f −1 (U1 ) ∩ f −1 (U2 ) = f −1 (U1 ∩ U2 )
So V1 ∩ V2 ∈ T because U1 ∩ U2 is open in X.
Any topology on Y such that f is continuous contains T according to the definition
of a continuous map. Thus T is indeed the weakest topology on Y such that f is
continuous. This proves that (1) and (2) are equivalent.
The equivalence of (2) and (3) follows from the equality f −1 (X \ E) = Y \ f −1 (E)
for all subsets E ⊂ X. □
Dually, if X is a topological space and X → Y is a surjection of sets, then Y can
be endowed with the quotient topology.
08ZK Lemma 6.2. Let X be a topological space. Let Y be a set and let f : X → Y be
a surjective map of sets. The quotient topology on Y is the topology characterized
by each of the following statements:
(1) it is the strongest topology on Y such that f is continuous,
(2) a subset V of Y is open if and only if f −1 (V ) is open,
(3) a subset Z of Y is closed if and only if f −1 (Z) is closed.
Proof. The set T = {V ⊂ Y |f −1 (V ) is open} is a topology on Y . Firstly ∅ =
f −1 (∅) and f −1 (Y ) = X. So T contains ∅ and Y .
Let (Vi )i∈I be a family of elements Vi ∈ T . Then
[ [ 
f −1 (Vi ) = f −1 Vi
i∈I i∈I
−1
S S
Thus i∈I Vi ∈ T as i∈I f (Vi ) is open in X. Furthermore if V1 , V2 ∈ T then
f −1 (V1 ) ∩ f −1 (V2 ) = f −1 (V1 ∩ V2 )
So V1 ∩ V2 ∈ T because f −1 (V1 ) ∩ f −1 (V2 ) is open in X.
Finally a topology on Y such that f is continuous is included in T according to the
definition of a continuous function, so T is the strongest topology on Y such that
f is continuous. It proves that (1) and (2) are equivalent.
Finally, (2) and (3) equivalence follows from f −1 (X \E) = Y \f −1 (E) for all subsets
E ⊂ X. □
Let f : X → Y be a continuous map of topological spaces. In this case we obtain
a factorization X → f (X) → Y of maps of sets. We can endow f (X) with the
quotient topology coming from the surjection X → f (X) or with the induced
topology coming from the injection f (X) → Y . The map
(f (X), quotient topology) −→ (f (X), induced topology)
is continuous.
0406 Definition 6.3. Let f : X → Y be a continuous map of topological spaces.
(1) We say f is a strict map of topological spaces if the induced topology and
the quotient topology on f (X) agree (see discussion above).
(2) We say f is submersive1 if f is surjective and strict.
1This is very different from the notion of a submersion between differential manifolds! It is
probably a good idea to use “strict and surjective” in stead of “submersive”.
 TOPOLOGY 7

Thus a continuous map f : X → Y is submersive if f is a surjection and for any

T ⊂ Y we have T is open or closed if and only if f −1 (T ) is so. In other words, Y
has the quotient topology relative to the surjection X → Y .
02YB Lemma 6.4. Let f : X → Y be surjective, open, continuous map of topological
spaces. Let T ⊂ Y be a subset. Then
(1) f −1 (T ) = f −1 (T ),
(2) T ⊂ Y is closed if and only if f −1 (T ) is closed,
(3) T ⊂ Y is open if and only if f −1 (T ) is open, and
(4) T ⊂ Y is locally closed if and only if f −1 (T ) is locally closed.
In particular we see that f is submersive.
Proof. It is clear that f −1 (T ) ⊂ f −1 (T ). If x ∈ X, and x ̸∈ f −1 (T ), then there
exists an open neighbourhood x ∈ U ⊂ X with U ∩ f −1 (T ) = ∅. Since f is
open we see that f (U ) is an open neighbourhood of f (x) not meeting T . Hence
x ̸∈ f −1 (T ). This proves (1). Part (2) is an easy consequence of (1). Part (3) is
obvious from the fact that f is open and surjective. For (4), if f −1 (T ) is locally
closed, then f −1 (T ) ⊂ f −1 (T ) = f −1 (T ) is open, and hence by (3) applied to the
map f −1 (T ) → T we see that T is open in T , i.e., T is locally closed. □
0AAU Lemma 6.5. Let f : X → Y be surjective, closed, continuous map of topological
spaces. Let T ⊂ Y be a subset. Then
(1) T = f (f −1 (T )),
(2) T ⊂ Y is closed if and only if f −1 (T ) is closed,
(3) T ⊂ Y is open if and only if f −1 (T ) is open, and
(4) T ⊂ Y is locally closed if and only if f −1 (T ) is locally closed.
In particular we see that f is submersive.
Proof. It is clear that f −1 (T ) ⊂ f −1 (T ). Then T ⊂ f (f −1 (T )) ⊂ T is a closed
subset, hence we get (1). Part (2) is obvious from the fact that f is closed and
surjective. Part (3) follows from (2) applied to the complement of T . For (4), if
f −1 (T ) is locally closed, then f −1 (T ) ⊂ f −1 (T ) is open. Since the map f −1 (T ) → T
is surjective by (1) we can apply part (3) to the map f −1 (T ) → T induced by f to
conclude that T is open in T , i.e., T is locally closed. □

7. Connected components
004R
004S Definition 7.1. Let X be a topological space.
(1) We say X is connected if X is not empty and whenever X = T1 ⨿ T2 with
Ti ⊂ X open and closed, then either T1 = ∅ or T2 = ∅.
(2) We say T ⊂ X is a connected component of X if T is a maximal connected
subset of X.
The empty space is not connected.
0376 Lemma 7.2. Let f : X → Y be a continuous map of topological spaces. If E ⊂ X
is a connected subset, then f (E) ⊂ Y is connected as well.
Proof. Let A ⊂ f (E) an open and closed subset of f (E). Because f is continuous,
f −1 (A) is an open and closed subset of E. As E is connected, f −1 (A) = ∅ or
f −1 (A) = E. However, A ⊂ f (E) implies that A = f (f −1 (A)). Indeed, if x ∈
 TOPOLOGY 8

f (f −1 (A)) then there is y ∈ f −1 (A) such that f (y) = x and because y ∈ f −1 (A),
we have f (y) ∈ A i.e. x ∈ A. Reciprocally, if x ∈ A, A ⊂ f (E) implies that there is
y ∈ E such that f (y) = x. Therefore y ∈ f −1 (A), and then x ∈ f (f −1 (A)). Thus
A = ∅ or A = f (E) proving that f (E) is connected. □
004T Lemma 7.3. Let X be a topological space.
(1) If T ⊂ X is connected, then so is its closure.
(2) Any connected component of X is closed (but not necessarily open).
(3) Every connected subset of X is contained in a unique connected component
of X.
(4) Every point of X is contained in a unique connected component, in other
words, X is the disjoint union of its connected components.
Proof. Let T be the closure of the connected subset T . Suppose T = T1 ⨿ T2 with
Ti ⊂ T open and closed. Then T = (T ∩ T1 ) ⨿ (T ∩ T2 ). Hence T equals one of the
two, say T = T1 ∩ T . Thus T ⊂ T1 . This implies (1) and (2).
T
Let A be a nonempty set ofSconnected subsets of X such that Ω = T ∈A T is
nonempty. We claim E = T ∈A T is connected. Namely, E is nonempty as it
contains Ω. Say E = E1 ⨿ E2 with Ei closed in E. We may assume E1 meets Ω
(after renumbering). Then each T ∈ A meets E1 and hence must be contained in
E1 as T is connected. Hence E ⊂ E1 which proves the claim.
Let W ⊂ X be a nonempty connected subset. If we apply the result of the previous
paragraph to the set of all connected subsets of X containing W , then we see that
E is a connected component of X. This implies existence and uniqueness in (3).
Let x ∈ X. Taking W = {x} in the previous paragraph we see that x is contained
in a unique connected component of X. Any two distinct connected components
must be disjoint (by the result of the second paragraph).
To get anQexample where connected components are not open, just take an infinite
product n∈N {0, 1} with the product topology. Its connected components are
singletons, which are not open. □
0FIY Remark 7.4. Let X be a topological space and x ∈ X. Let Z ⊂ X be the [Eng77, Example
connected component of X passing through x. Consider the intersection E of all 6.1.24]
open and closed subsets of X containing x. It is clear that Z ⊂ E. In general
Z ̸= E. For example, let X = {x, y, z1 , z2 , . . .} with the topology with the following
basis of opens, {zn }, {x, zn , zn+1 , . . .}, and {y, zn , zn+1 , . . .} for all n. Then Z = {x}
and E = {x, y}. We omit the details.
0377 Lemma 7.5. Let f : X → Y be a continuous map of topological spaces. Assume
that
(1) all fibres of f are connected, and
(2) a set T ⊂ Y is closed if and only if f −1 (T ) is closed.
Then f induces a bijection between the sets of connected components of X and Y .
Proof. Let T ⊂ Y be a connected component. Note that T is closed, see Lemma
7.3. The lemma follows if we show that f −1 (T ) is connected because any connected
subset of X maps into a connected component of Y by Lemma 7.2. Suppose that
f −1 (T ) = Z1 ⨿ Z2 with Z1 , Z2 closed. For any t ∈ T we see that f −1 ({t}) =
Z1 ∩ f −1 ({t}) ⨿ Z2 ∩ f −1 ({t}). By (1) we see f −1 ({t}) is connected we conclude
 TOPOLOGY 9

that either f −1 ({t}) ⊂ Z1 or f −1 ({t}) ⊂ Z2 . In other words T = T1 ⨿ T2 with

f −1 (Ti ) = Zi . By (2) we conclude that Ti is closed in Y . Hence either T1 = ∅ or
T2 = ∅ as desired. □

0378 Lemma 7.6. Let f : X → Y be a continuous map of topological spaces. Assume

that (a) f is open, (b) all fibres of f are connected. Then f induces a bijection
between the sets of connected components of X and Y .
Proof. This is a special case of Lemma 7.5. □

07VB Lemma 7.7. Let f : X → Y be a continuous map of nonempty topological spaces.

Assume that (a) Y is connected, (b) f is open and closed, and (c) there is a point
y ∈ Y such that the fiber f −1 (y) is a finite set. Then X has at most |f −1 (y)|
connected components. Hence any connected component T of X is open and closed,
and f (T ) is a nonempty open and closed subset of Y , which is therefore equal to Y .
Proof. If the topological space X has at least N connected components for some
N ∈ N, we find by induction a decomposition X = X1 ⨿ . . . ⨿ XN of X as a disjoint
union of N nonempty open and closed subsets X1 , . . . , XN of X. As f is open and
closed, each f (Xi ) is a nonempty open and closed subset of Y and is hence equal
to Y . In particular the intersection Xi ∩ f −1 (y) is nonempty for each 1 ≤ i ≤ N .
Hence f −1 (y) has at least N elements. □

04MC Definition 7.8. A topological space is totally disconnected if the connected com-
ponents are all singletons.
A discrete space is totally disconnected. A totally disconnected space need not be
discrete, for example Q ⊂ R is totally disconnected but not discrete.
08ZL Lemma 7.9. Let X be a topological space. Let π0 (X) be the set of connected
components of X. Let X → π0 (X) be the map which sends x ∈ X to the connected
component of X passing through x. Endow π0 (X) with the quotient topology. Then
π0 (X) is a totally disconnected space and any continuous map X → Y from X to
a totally disconnected space Y factors through π0 (X).
Proof. By Lemma 7.5 the connected components of π0 (X) are the singletons. We
omit the proof of the second statement. □

04MD Definition 7.10. A topological space X is called locally connected if every point
x ∈ X has a fundamental system of connected neighbourhoods.
04ME Lemma 7.11. Let X be a topological space. If X is locally connected, then
(1) any open subset of X is locally connected, and
(2) the connected components of X are open.
So also the connected components of open subsets of X are open. In particular,
every point has a fundamental system of open connected neighbourhoods.
Proof. For all x ∈ X let write N (x) the fundamental system of connected neigh-
bourhoods of x and let U ⊂ X be an open subset of X. Then for all x ∈ U , U is
a neighbourhood of x, so the set {V ∈ N (x)|V ⊂ U } is not empty and is a funda-
mental system of connected neighbourhoods of x in U . Thus U is locally connected
and it proves (1).
 TOPOLOGY 10

Let x ∈ C ⊂ X where C is the connected component of x. Because X is locally

connected, there exists N a connected neighbourhood of x. Therefore by the def-
inition of a connected component, we have N ⊂ C and then C is a neighbourhood
of x. It implies that C is a neighbourhood of each of his point, in other words C is
open and (2) is proven. □

8. Irreducible components
004U
004V Definition 8.1. Let X be a topological space.
(1) We say X is irreducible, if X is not empty, and whenever X = Z1 ∪ Z2 with
Zi closed, we have X = Z1 or X = Z2 .
(2) We say Z ⊂ X is an irreducible component of X if Z is a maximal irreducible
subset of X.
An irreducible space is obviously connected.
0379 Lemma 8.2. Let f : X → Y be a continuous map of topological spaces. If E ⊂ X
is an irreducible subset, then f (E) ⊂ Y is irreducible as well.
Proof. Clearly we may assume E = X (i.e., X irreducible) and f (E) = Y (i.e., f
surjective). First, Y ̸= ∅ since X ̸= ∅. Next, assume Y = Y1 ∪Y2 with Y1 , Y2 closed.
Then X = X1 ∪ X2 with Xi = f −1 (Yi ) closed in X. By assumption on X, we must
have X = X1 or X = X2 , hence Y = Y1 or Y = Y2 since f is surjective. □

004W Lemma 8.3. Let X be a topological space.

(1) If T ⊂ X is irreducible so is its closure in X.
(2) Any irreducible component of X is closed.
(3) Any irreducible subset of X is contained in an irreducible component of X.
(4) Every point of X is contained in some irreducible component of X, in other
words, X is the union of its irreducible components.
Proof. Let T be the closure of the irreducible subset T . If T = Z1 ∪ Z2 with
Zi ⊂ T closed, then T = (T ∩ Z1 ) ∪ (T ∩ Z2 ) and hence T equals one of the two, say
T = Z1 ∩ T . Thus clearly T ⊂ Z1 . This proves (1). Part (2) follows immediately
from (1) and the definition of irreducible components.
Let T ⊂ X be irreducible. Consider the set A of irreducible subsets T ⊂ Tα ⊂ X.
Note that A is nonempty since T ∈ A. There is a partial ordering on A coming
′
from inclusion: α ≤ α S ⇔ Tα ⊂ Tα′ . Choose a ′ maximal totally ordered subset
′ ′
A ⊂ A, and let T = α∈A′ Tα . We claim that T is irreducible. Namely, suppose
that T ′ = Z1 ∪ Z2 is a union of two closed subsets of T ′ . For each α ∈ A′ we have
either Tα ⊂ Z1 or Tα ⊂ Z2 , by irreducibility of Tα . Suppose that for some α0 ∈ A′
we have Tα0 ̸⊂ Z1 (say, if not we’re done anyway). Then, since A′ is totally ordered
we see immediately that Tα ⊂ Z2 for all α ∈ A′ . Hence T ′ = Z2 . This proves (3).
Part (4) is an immediate consequence of (3) as a singleton space is irreducible. □
S
0G2Y Lemma 8.4. Let X be a topological space and suppose X = i=1,...,n Xi where
each Xi is an irreducible closed subset of X and no Xi is contained in the union
of the other members. Then each Xi is an irreducible component of X and each
irreducible component of X is one of the Xi .
 TOPOLOGY 11

S
Proof. Let Y be an irreducible component of X. Write Y = i=1,...,n (Y ∩ Xi )
and note that each Y ∩ Xi is closed in Y since Xi is closed in X. By irreducibility
of Y we see that Y is equal to one of the Y ∩ Xi , i.e., Y ⊂ Xi . By maximality of
irreducible components we get Y = Xi .
Conversely, take one of the Xi and expand it to an irreducible component Y , which
we have already shown is one of the Xj . So Xi ⊂ Xj and since the original union
does not have redundant members, Xi = Xj , which is an irreducible component. □
0GM2 Lemma 8.5. Let f : X → Y be a surjective, continuous map of topological
spaces. If X has a finite number, say n, of irreducible components, then Y has ≤ n
irreducible components.
Proof. Say X1 , . . . , Xn are the irreducible components of X. By Lemmas 8.2 and
8.3 the closure Yi ⊂ Y of f (Xi ) is irreducible. Since f is surjective, we see that Y
is theSunion of the Yi . We may choose a minimal subset I ⊂ {1, . . . , n} such that
Y = i∈I Yi . Then we may apply Lemma 8.4 to see that the Yi for i ∈ I are the
irreducible components of Y . □
A singleton is irreducible. Thus if x ∈ X is a point then the closure {x} is an
irreducible closed subset of X.
004X Definition 8.6. Let X be a topological space.
(1) Let Z ⊂ X be an irreducible closed subset. A generic point of Z is a point
ξ ∈ Z such that Z = {ξ}.
(2) The space X is called Kolmogorov, if for every x, x′ ∈ X, x ̸= x′ there exists
a closed subset of X which contains exactly one of the two points.
(3) The space X is called quasi-sober if every irreducible closed subset has a
generic point.
(4) The space X is called sober if every irreducible closed subset has a unique
generic point.
A topological space X is Kolmogorov, quasi-sober, resp. sober if and only if the
map x 7→ {x} from X to the set of irreducible closed subsets of X is injective,
surjective, resp. bijective. Hence we see that a topological space is sober if and only
if it is quasi-sober and Kolmogorov.
0B31 Lemma 8.7. Let X be a topological space and let Y ⊂ X.
(1) If X is Kolmogorov then so is Y .
(2) Suppose Y is locally closed in X. If X is quasi-sober then so is Y .
(3) Suppose Y is locally closed in X. If X is sober then so is Y .
Proof. Proof of (1). Suppose X is Kolmogorov. Let x, y ∈ Y with x ̸= y. Then
{x} ∩ Y = {x} ̸= {y} = {y} ∩ Y . Hence {x} ∩ Y ̸= {y} ∩ Y . This shows that Y is
Kolmogorov.
Proof of (2). Suppose X is quasi-sober. It suffices to consider the cases Y is closed
and Y is open. First, suppose Y is closed. Let Z be an irreducible closed subset
of Y . Then Z is an irreducible closed subset of X. Hence there exists x ∈ Z with
{x} = Z. It follows {x} ∩ Y = Z. This shows Y is quasi-sober. Second, suppose
Y is open. Let Z be an irreducible closed subset of Y . Then Z is an irreducible
closed subset of X. Hence there exists x ∈ Z with {x} = Z. If x ∈ / Y we get the
 TOPOLOGY 12

contradiction Z = Z ∩ Y ⊂ Z ∩ Y = {x} ∩ Y = ∅. Therefore x ∈ Y . It follows

Z = Z ∩ Y = {x} ∩ Y . This shows Y is quasi-sober.
Proof of (3). Immediately from (1) and (2). □

06N9 Lemma 8.8. Let X be a topological space and let (Xi )i∈I be a covering of X.
(1) Suppose Xi is locally closed in X for every i ∈ I. Then, X is Kolmogorov
if and only if Xi is Kolmogorov for every i ∈ I.
(2) Suppose Xi is open in X for every i ∈ I. Then, X is quasi-sober if and
only if Xi is quasi-sober for every i ∈ I.
(3) Suppose Xi is open in X for every i ∈ I. Then, X is sober if and only if
Xi is sober for every i ∈ I.
Proof. Proof of (1). If X is Kolmogorov then so is Xi for every i ∈ I by Lemma
8.7. Suppose Xi is Kolmogorov for every i ∈ I. Let x, y ∈ X with {x} = {y}.
There exists i ∈ I with x ∈ Xi . There exists an open subset U ⊂ X such that Xi is
a closed subset of U . If y ∈
/ U we get the contradiction x ∈ {x} ∩ U = {y} ∩ U = ∅.
Hence y ∈ U . It follows y ∈ {y} ∩ U = {x} ∩ U ⊂ Xi . This shows y ∈ Xi . It
follows {x} ∩ Xi = {y} ∩ Xi . Since Xi is Kolmogorov we get x = y. This shows X
is Kolmogorov.
Proof of (2). If X is quasi-sober then so is Xi for every i ∈ I by Lemma 8.7.
Suppose Xi is quasi-sober for every i ∈ I. Let Y be an irreducible closed subset of
X. As Y ̸= ∅ there exists i ∈ I with Xi ∩ Y ̸= ∅. As Xi is open in X it follows
Xi ∩ Y is non-empty and open in Y , hence irreducible and dense in Y . Thus Xi ∩ Y
is an irreducible closed subset of Xi . As Xi is quasi-sober there exists x ∈ Xi ∩ Y
with Xi ∩ Y = {x} ∩ Xi ⊂ {x}. Since Xi ∩ Y is dense in Y and Y is closed in X it
follows Y = Xi ∩ Y ∩ Y ⊂ Xi ∩ Y ⊂ {x} ⊂ Y . Therefore Y = {x}. This shows X
is quasi-sober.
Proof of (3). Immediately from (1) and (2). □

0B32 Example 8.9. Let X be an indiscrete space of cardinality at least 2. Then X is

quasi-sober but not Kolmogorov. Moreover, the family of its singletons is a covering
of X by discrete and hence Kolmogorov spaces.
0B33 Example 8.10. Let Y be an infinite set, furnished with the topology whose closed
sets are Y and the finite subsets of Y . Then Y is Kolmogorov but not quasi-sober.
However, the family of its singletons (which are its irreducible components) is a
covering by discrete and hence sober spaces.
0B34 Example 8.11. Let X and Y be as in Example 8.9 and Example 8.10. Then,
X ⨿ Y is neither Kolmogorov nor quasi-sober.
0B35 Example 8.12. Let Z be an infinite set and let z ∈ Z. We furnish Z with the
topology whose closed sets are Z and the finite subsets of Z \ {z}. Then Z is sober
but its subspace Z \ {z} is not quasi-sober.
004Y Example 8.13. Recall that a topological space X is Hausdorff iff for every distinct
pair of points x, y ∈ X there exist disjoint opens U, V ⊂ X such that x ∈ U , y ∈ V .
In this case X is irreducible if and only if X is a singleton. Similarly, any subset of
X is irreducible if and only if it is a singleton. Hence a Hausdorff space is sober.
 TOPOLOGY 13

004Z Lemma 8.14. Let f : X → Y be a continuous map of topological spaces. Assume

that (a) Y is irreducible, (b) f is open, and (c) there exists a dense collection of
points y ∈ Y such that f −1 (y) is irreducible. Then X is irreducible.
Proof. Suppose X = Z1 ∪Z2 with Zi closed. Consider the open sets U1 = Z1 \Z2 =
X \ Z2 and U2 = Z2 \ Z1 = X \ Z1 . To get a contradiction assume that U1 and U2
are both nonempty. By (b) we see that f (Ui ) is open. By (a) we have Y irreducible
and hence f (U1 )∩f (U2 ) ̸= ∅. By (c) there is a point y which corresponds to a point
of this intersection such that the fibre Xy = f −1 (y) is irreducible. Then Xy ∩ U1
and Xy ∩ U2 are nonempty disjoint open subsets of Xy which is a contradiction. □
037A Lemma 8.15. Let f : X → Y be a continuous map of topological spaces. Assume
that (a) f is open, and (b) for every y ∈ Y the fibre f −1 (y) is irreducible. Then f
induces a bijection between irreducible components.
Proof. We point out that assumption (b) implies that f is surjective (see Definition
8.1). Let T ⊂ Y be an irreducible component. Note that T is closed, see Lemma 8.3.
The lemma follows if we show that f −1 (T ) is irreducible because any irreducible
subset of X maps into an irreducible component of Y by Lemma 8.2. Note that
f −1 (T ) → T satisfies the assumptions of Lemma 8.14. Hence we win. □
The construction of the following lemma is sometimes called the “soberification”.
0A2N Lemma 8.16. Let X be a topological space. There is a canonical continuous map [GD71, Page 68 ff]
′
c : X −→ X
from X to a sober topological space X ′ which is universal among continuous maps
from X to sober topological spaces. Moreover, the assignment U ′ 7→ c−1 (U ′ ) is a
bijection between opens of X ′ and X which commutes with finite intersections and
arbitrary unions. The image c(X) is a Kolmogorov topological space and the map
c : X → c(X) is universal for maps of X into Kolmogorov spaces.
Proof. Let X ′ be the set of irreducible closed subsets of X and let
c : X → X ′, x 7→ {x}.
′ ′
For U ⊂ X open, let U ⊂ X denote the set of irreducible closed subsets of X
which meet U . Then c−1 (U ′ ) = U . In particular, if U1 ̸= U2 are open in X, then
U1′ ̸= U2′ . Hence c induces a bijection between the subsets of X ′ of the form U ′ and
the opens of X.
Let U1 , U2 be open in X. Suppose that Z ∈ U1′ and Z ∈ U2′ . Then Z ∩ U1 and
Z ∩ U2 are nonempty open subsets of the irreducible space Z and hence Z ∩ U1 ∩ U2
is nonempty. Thus (U1 ∩ U2 )′ = U1′ ∩ U2′ . The rule U 7→ U ′ is also compatible with
arbitrary unions (details omitted). Thus it is clear that the collection of U ′ form a
topology on X ′ and that we have a bijection as stated in the lemma.
Next we show that X ′ is sober. Let T ⊂ X ′ be an irreducible closed subset. Let
U ⊂ X be the open such that X ′ \ T = U ′ . Then Z = X \ U is irreducible because
of the properties of the bijection of the lemma. We claim that Z ∈ T is the unique
generic point. Namely, any open of the form V ′ ⊂ X ′ which does not contain Z
must come from an open V ⊂ X which misses Z, i.e., is contained in U .
Finally, we check the universal property. Let f : X → Y be a continuous map to
a sober topological space. Then we let f ′ : X ′ → Y be the map which sends the
 TOPOLOGY 14

irreducible closed Z ⊂ X to the unique generic point of f (Z). It follows immediately

that f ′ ◦ c = f as maps of sets, and the properties of c imply that f ′ is continuous.
We omit the verification that the continuous map f ′ is unique. We also omit the
proof of the statements on Kolmogorov spaces. □
0GM3 Lemma 8.17. Let X be a connected topological space with a finite number of
irreducible components X1 , . . . , Xn . If n > 1 there is an 1 ≤ j ≤ n such that
X ′ = i̸=j Xi is connected.
S

Proof. This is a graph theory problem. Let Γ be the graph with vertices V =
{1, . . . , n} and an edge between i and j if and only if Xi ∩ Xj is nonempty. Con-
nectedness of X means that Γ is connected. Our problem is to find 1 ≤ j ≤ n such
that Γ \ {j} is still connected. You can do this by choosing j, j ′ ∈ E with maximal
distance and then j works (choose a leaf!). Details omitted. □

9. Noetherian topological spaces

0050
0051 Definition 9.1. A topological space is called Noetherian if the descending chain
condition holds for closed subsets of X. A topological space is called locally Noe-
therian if every point has a neighbourhood which is Noetherian.
0052 Lemma 9.2. Let X be a Noetherian topological space.
(1) Any subset of X with the induced topology is Noetherian.
(2) The space X has finitely many irreducible components.
(3) Each irreducible component of X contains a nonempty open of X.
Proof. Let T ⊂ X be a subset of X. Let T1 ⊃ T2 ⊃ . . . be a descending chain of
closed subsets of T . Write Ti = T ∩Zi with Zi ⊂ X closed. Consider the descending
chain of closed subsets Z1 ⊃ Z1 ∩Z2 ⊃ Z1 ∩Z2 ∩Z3 . . . This stabilizes by assumption
and hence the original sequence of Ti stabilizes. Thus T is Noetherian.
Let A be the set of closed subsets of X which do not have finitely many irreducible
components. Assume that A is not empty to arrive at a contradiction. The set A
is partially ordered by inclusion: α ≤ α′ ⇔ Zα ⊂ Zα′ . By the descending chain
condition we may find a smallest element of A, say Z. As Z is not a finite union of
S Z
irreducible components, it is not irreducible. Hence we can write = Z ′ ∪ Z ′′Sand
′ ′ ′′
both are strictly smaller closed subsets. By construction Z =S Zi and S Z = Zj′′
′ ′′
are finite unions of their irreducible components. Hence Z = Zi ∪ Zj is a finite
union of irreducible closed subsets. After removing redundant members of this
expression, this will be the decomposition of Z into its irreducible components
(Lemma 8.4), a contradiction.
Let Z ⊂ X be an irreducible component of X. Let Z1 , . . . , Zn be the other irre-
ducible components of X. Consider U = Z \ (Z1 ∪ . . . ∪ Zn ). This is not empty
since otherwise the irreducible space Z would be contained in one of the other Zi .
Because X = Z ∪ Z1 ∪ . . . Zn (see Lemma 8.3), also U = X \ (Z1 ∪ . . . ∪ Zn ) and
hence open in X. Thus Z contains a nonempty open of X. □
04Z8 Lemma 9.3. Let f : X → Y be a continuous map of topological spaces.
(1) If X is Noetherian, then f (X) is Noetherian.
(2) If X is locally Noetherian and f open, then f (X) is locally Noetherian.
 TOPOLOGY 15

Proof. In case (1), suppose that Z1 ⊃ Z2 ⊃ Z3 ⊃ . . . is a descending chain of

closed subsets of f (X) (as usual with the induced topology as a subset of Y ). Then
f −1 (Z1 ) ⊃ f −1 (Z2 ) ⊃ f −1 (Z3 ) ⊃ . . . is a descending chain of closed subsets of X.
Hence this chain stabilizes. Since f (f −1 (Zi )) = Zi we conclude that Z1 ⊃ Z2 ⊃
Z3 ⊃ . . . stabilizes also. In case (2), let y ∈ f (X). Choose x ∈ X with f (x) = y. By
assumption there exists a neighbourhood E ⊂ X of x which is Noetherian. Then
f (E) ⊂ f (X) is a neighbourhood which is Noetherian by part (1). □
0053 Lemma 9.4. Let X be a topological space. Let Xi ⊂ X, i = 1, . . . , n be a finite
collection
S of subsets. If each Xi is Noetherian (with the induced topology), then
i=1,...,n Xi is Noetherian (with the induced topology).

Proof. Let {Fm }m∈N a decreasing sequence of closed subsets of X ′ = i=1,...,n Xi

S

with the induced topology. Then we can find a decreasing sequence {Gm }m∈N of
closed subsets of X verifying Fm = Gm ∩ X ′ for all m (small detail omitted). As
Xi is noetherian and {Gm ∩ Xi }m∈N a decreasing sequence of closed subsets of Xi ,
there exists mi ∈ N such that for all m ≥ mi we have Gm ∩ Xi = Gmi ∩ Xi . Let
m0 = maxi=1,...,n mi . Then clearly
Fm = Gm ∩ X ′ = Gm ∩ (X1 ∪ . . . ∪ Xn ) = (Gm ∩ X1 ) ∪ . . . (Gm ∩ Xn )
stabilizes for m ≥ m0 and the proof is complete. □
02HZ Example 9.5. Any nonempty, Kolmogorov Noetherian topological space has a
closed point (combine Lemmas 12.8 and 12.13). Let X = {1, 2, 3, . . .}. Define a
topology on X with opens ∅, {1, 2, . . . , n}, n ≥ 1 and X. Thus X is a locally
Noetherian topological space, without any closed points. This space cannot be the
underlying topological space of a locally Noetherian scheme, see Properties, Lemma
5.9.
04MF Lemma 9.6. Let X be a locally Noetherian topological space. Then X is locally
connected.
Proof. Let x ∈ X. Let E be a neighbourhood of x. We have to find a connected
neighbourhood of x contained in E. By assumption there exists a neighbourhood
E ′ of x which is Noetherian. Then E ∩ E ′ is Noetherian, see Lemma 9.2. Let
E ∩ E ′ = Y1 ∪ . . . ∪ YSn be the decomposition into irreducible components, see
Lemma 9.2. Let E ′′ = x∈Yi Yi . This is a connected subset of E ∩ E ′ containing x.
It contains the open E ∩ E ′ \ ( x̸∈Yi Yi ) of E ∩ E ′ and hence it is a neighbourhood
S
of x in X. This proves the lemma. □

10. Krull dimension

0054
0055 Definition 10.1. Let X be a topological space.
(1) A chain of irreducible closed subsets of X is a sequence Z0 ⊂ Z1 ⊂ . . . ⊂
Zn ⊂ X with Zi closed irreducible and Zi ̸= Zi+1 for i = 0, . . . , n − 1.
(2) The length of a chain Z0 ⊂ Z1 ⊂ . . . ⊂ Zn ⊂ X of irreducible closed subsets
of X is the integer n.
(3) The dimension or more precisely the Krull dimension dim(X) of X is the
element of {−∞, 0, 1, 2, 3, . . . , ∞} defined by the formula:
dim(X) = sup{lengths of chains of irreducible closed subsets}
 TOPOLOGY 16

Thus dim(X) = −∞ if and only if X is the empty space.

(4) Let x ∈ X. The Krull dimension of X at x is defined as
dimx (X) = min{dim(U ), x ∈ U ⊂ X open}
the minimum of dim(U ) where U runs over the open neighbourhoods of x
in X.
Note that if U ′ ⊂ U ⊂ X are open then dim(U ′ ) ≤ dim(U ). Hence if dimx (X) =
d then x has a fundamental system of open neighbourhoods U with dim(U ) =
dimx (X).
0B7I Lemma 10.2. Let X be a topological space. Then dim(X) = sup dimx (X) where
the supremum runs over the points x of X.
Proof. It is clear that dim(X) ≥ dimx (X) for all x ∈ X (see discussion following
Definition 10.1). Thus an inequality in one direction. For the converse, let n ≥ 0
and suppose that dim(X) ≥ n. Then we can find a chain of irreducible closed
subsets Z0 ⊂ Z1 ⊂ . . . ⊂ Zn ⊂ X. Pick x ∈ Z0 . For every open neighbourhood U
of x we get a chain of irreducible closed subsets
Z0 ∩ U ⊂ Z1 ∩ U ⊂ . . . ⊂ Zn ∩ U
in U . Namely, the sets U ∩ Zi are irreducible closed in U and the inclusions are
strict (details omitted; hint: the closure of U ∩ Zi is Zi ). In this way we see that
dimx (X) ≥ n which proves the other inequality. □

0056 Example 10.3. The Krull dimension of the usual Euclidean space Rn is 0.
0057 Example 10.4. Let X = {s, η} with open sets given by {∅, {η}, {s, η}}. In
this case a maximal chain of irreducible closed subsets is {s} ⊂ {s, η}. Hence
dim(X) = 1. It is easy to generalize this example to get a (n+1)-element topological
space of Krull dimension n.
0058 Definition 10.5. Let X be a topological space. We say that X is equidimensional
if every irreducible component of X has the same dimension.

11. Codimension and catenary spaces

02I0 We only define the codimension of irreducible closed subsets.
02I3 Definition 11.1. Let X be a topological space. Let Y ⊂ X be an irreducible
closed subset. The codimension of Y in X is the supremum of the lengths e of
chains
Y = Y0 ⊂ Y1 ⊂ . . . ⊂ Ye ⊂ X
of irreducible closed subsets in X starting with Y . We will denote this codim(Y, X).
The codimension is an element of {0, 1, 2, . . .} ∪ {∞}. If codim(Y, X) < ∞, then
every chain can be extended to a maximal chain (but these do not all have to have
the same length).
02I4 Lemma 11.2. Let X be a topological space. Let Y ⊂ X be an irreducible closed
subset. Let U ⊂ X be an open subset such that Y ∩ U is nonempty. Then
codim(Y, X) = codim(Y ∩ U, U )
 TOPOLOGY 17

Proof. The rule T 7→ T defines a bijective inclusion preserving map between the
closed irreducible subsets of U and the closed irreducible subsets of X which meet
U . Using this the lemma easily follows. Details omitted. □
02I5 Example 11.3. Let X = [0, 1] be the unit interval with the following topology:
The sets [0, 1], (1 − 1/n, 1] for n ∈ N, and ∅ are open. So the closed sets are ∅, {0},
[0, 1 − 1/n] for n > 1 and [0, 1]. This is clearly a Noetherian topological space. But
the irreducible closed subset Y = {0} has infinite codimension codim(Y, X) = ∞.
To see this we just remark that all the closed sets [0, 1 − 1/n] are irreducible.
02I1 Definition 11.4. Let X be a topological space. We say X is catenary if for every
pair of irreducible closed subsets T ⊂ T ′ we have codim(T, T ′ ) < ∞ and every
maximal chain of irreducible closed subsets
T = T0 ⊂ T1 ⊂ . . . ⊂ Te = T ′
has the same length (equal to the codimension).
02I2 Lemma 11.5. Let X be a topological space. The following are equivalent:
(1) X is catenary,
(2) X has an open covering by catenary spaces.
Moreover, in this case any locally closed subspace of X is catenary.
Proof. Suppose that X is catenary and that U ⊂ X is an open subset. The rule
T 7→ T defines a bijective inclusion preserving map between the closed irreducible
subsets of U and the closed irreducible subsets of X which meet U . Using this the
lemma easily follows. Details omitted. □
02I6 Lemma 11.6. Let X be a topological space. The following are equivalent:
(1) X is catenary, and
(2) for every pair of irreducible closed subsets Y ⊂ Y ′ we have codim(Y, Y ′ ) <
∞ and for every triple Y ⊂ Y ′ ⊂ Y ′′ of irreducible closed subsets we have
codim(Y, Y ′′ ) = codim(Y, Y ′ ) + codim(Y ′ , Y ′′ ).
Proof. Let suppose that X is catenary. According to Definition 11.4, for every pair
of irreducible closed subsets Y ⊂ Y ′ we have codim(Y, Y ′ ) < ∞. Let Y ⊂ Y ′ ⊂ Y ′′
be a triple of irreducible closed subsets of X. Let
Y = Y0 ⊂ Y1 ⊂ ... ⊂ Ye1 = Y ′
be a maximal chain of irreducible closed subsets between Y and Y ′ where e1 =
codim(Y, Y ′ ). Let also
Y ′ = Ye1 ⊂ Ye1 +1 ⊂ ... ⊂ Ye1 +e2 = Y ′′
be a maximal chain of irreducible closed subsets between Y ′ and Y ′′ where e2 =
codim(Y ′ , Y ′′ ). As the two chains are maximal, the concatenation
Y = Y0 ⊂ Y1 ⊂ ... ⊂ Ye1 = Y ′ = Ye1 ⊂ Ye1 +1 ⊂ ... ⊂ Ye1 +e2 = Y ′′
is maximal too (between Y and Y ′′ ) and its length equals to e1 + e2 . As X is
catenary, each maximal chain has the same length equals to the codimension. Thus
the point (2) that codim(Y, Y ′′ ) = e1 +e2 = codim(Y, Y ′ )+codim(Y ′ , Y ′′ ) is verified.
For the reciprocal, we show by induction that : if Y = Y1 ⊂ ... ⊂ Yn = Y ′ , then
codim(Y, Y ′ ) = codim(Y1 , Y2 ) + ... + codim(Yn−1 , Yn ). Therefore, it forces maximal
chains to have the same length. □
 TOPOLOGY 18

12. Quasi-compact spaces and maps

0059 The phrase “compact” will be reserved for Hausdorff topological spaces. And many
spaces occurring in algebraic geometry are not Hausdorff.
005A Definition 12.1. Quasi-compactness.
(1) We say that a topological space X is quasi-compact if every open covering
of X has a finite subcover.
(2) We say that a continuous map f : X → Y is quasi-compact if the inverse
image f −1 (V ) of every quasi-compact open V ⊂ Y is quasi-compact.
(3) We say a subset Z ⊂ X is retrocompact if the inclusion map Z → X is
quasi-compact.
In many texts on topology a space is called compact if it is quasi-compact and
Hausdorff; and in other texts the Hausdorff condition is omitted. To avoid confu-
sion in algebraic geometry we use the term quasi-compact. The notion of quasi-
compactness of a map is very different from the notion of a “proper map”, since
there we require (besides closedness and separatedness) the inverse image of any
quasi-compact subset of the target to be quasi-compact, whereas in the definition
above we only consider quasi-compact open sets.
005B Lemma 12.2. A composition of quasi-compact maps is quasi-compact.
Proof. This is immediate from the definition. □
005C Lemma 12.3. A closed subset of a quasi-compact topological space is quasi-
compact.
S
Proof. Let E ⊂ X be a closed subset of the quasi-compact space X. Let E = Vj
be an openS covering. Choose Uj ⊂ X open such that Vj = E ∩ Uj . Then X =
(X \ E) ∪ Uj is an open covering of X. Hence X = (X \ E) ∪ Uj1 ∪ . . . ∪ Ujn for
some n and indices ji . Thus E = Vj1 ∪ . . . ∪ Vjn as desired. □
08YB Lemma 12.4. Let X be a Hausdorff topological space.
(1) If E ⊂ X is quasi-compact, then it is closed.
(2) If E1 , E2 ⊂ X are disjoint quasi-compact subsets then there exists opens
Ei ⊂ Ui with U1 ∩ U2 = ∅.
Proof. Proof of (1). Let x ∈ X, x ̸∈ E. For every e ∈ S E we can find disjoint
opens Ve and Ue with e ∈ Ve and x ∈ Ue . Since E ⊂ Ve we can find finitely
many e1 , . . . , en such that E ⊂ Ve1 ∪ . . . ∪ Ven . Then U = Ue1 ∩ . . . ∩ Uen is an open
neighbourhood of x which avoids Ve1 ∪ . . . ∪ Ven . In particular it avoids E. Thus
E is closed.
Proof of (2). In the proof of (1) we have seen that given x ∈ E1 we can find an
open neighbourhood x ∈ Ux and an open E2 ⊂ Vx such that Ux ∩ Vx = ∅. Because
E1 is quasi-compact we can find a finite number xi ∈ E1 such that E1 ⊂ U =
Ux1 ∪ . . . ∪ Uxn . We take V = Vx1 ∩ . . . ∩ Vxn to finish the proof. □
08YC Lemma 12.5. Let X be a quasi-compact Hausdorff space. Let E ⊂ X. The
following are equivalent: (a) E is closed in X, (b) E is quasi-compact.
Proof. The implication (a) ⇒ (b) is Lemma 12.3. The implication (b) ⇒ (a) is
Lemma 12.4. □
 TOPOLOGY 19

The following is really a reformulation of the quasi-compact property.

005D Lemma 12.6. Let X be a quasi-compact topological space. If {Zα }α∈A is a col-
lection of closedTsubsets such that the intersection of each finite subcollection is
nonempty, then α∈A Zα is nonempty.
T S
Proof. We suppose that α∈A Zα = ∅. So S we have α∈A (X \ Zα ) = X by
complementation. As the subsets Zα are closed, α∈A (X\Zα ) is an open covering of
S quasi-compact space X. Thus there exists a finite subset J ⊂ A such
the T that X =
α∈J (X \ Z α ). The complementary is then empty, which meansTthat α∈J Zα = ∅.
It proves there exists a finite subcollection of {Zα }α∈J verifying α∈J Zα = ∅, which
concludes by contraposition. □
04Z9 Lemma 12.7. Let f : X → Y be a continuous map of topological spaces.
(1) If X is quasi-compact, then f (X) is quasi-compact.
(2) If f is quasi-compact, then f (X) is retrocompact.
S S −1
Proof. If f (X) = Vi is an open covering, then X = f (Vi ) is an open
covering. Hence if X is quasi-compact then X = f −1 (Vi1 ) ∪ . . . ∪ f −1 (Vin ) for
some i1 , . . . , in ∈ I and hence f (X) = Vi1 ∪ . . . ∪ Vin . This proves (1). Assume f
is quasi-compact, and let V ⊂ Y be quasi-compact open. Then f −1 (V ) is quasi-
compact, hence by (1) we see that f (f −1 (V )) = f (X) ∩ V is quasi-compact. Hence
f (X) is retrocompact. □
005E Lemma 12.8. Let X be a topological space. Assume that
(1) X is nonempty,
(2) X is quasi-compact, and
(3) X is Kolmogorov.
Then X has a closed point.
Proof. Consider the set
T = {Z ⊂ X | Z = {x} for some x ∈ X}
of all closures of singletons in X. It is nonempty since X is nonempty. Make T
into a partially ordered set using the relation of inclusion. Suppose
T Zα , α ∈ A is
a totally ordered subset of T . By Lemma 12.6 we see that α∈A Zα ̸= ∅. Hence
T
there exists some x ∈ α∈A Zα and we see that Z = {x} ∈ T is a lower bound
for the family. By Zorn’s lemma there exists a minimal element Z ∈ T . As X is
Kolmogorov we conclude that Z = {x} for some x and x ∈ X is a closed point. □
08ZM Lemma 12.9. Let X be a quasi-compact Kolmogorov space. Then the set X0 of
closed points of X is quasi-compact.
S
Proof. Let X0 = Ui,0 be an open covering.
S Write Ui,0 = X0 ∩ Ui for some open
Ui ⊂ X. Consider the complement Z of Ui . This is a closed subset of X, hence
quasi-compact (Lemma 12.3) and Kolmogorov. By Lemma 12.8 S if Z is nonempty it
would haveSa closed point which contradicts the fact that X0 ⊂ Ui . Hence Z = ∅
and X = Ui . Since X is quasi-compact this covering has a finite subcover and
we conclude. □
005F Lemma 12.10. Let X be a topological space. Assume
(1) X is quasi-compact,
 TOPOLOGY 20

(2) X has a basis for the topology consisting of quasi-compact opens, and
(3) the intersection of two quasi-compact opens is quasi-compact.
For any x ∈ X the connected component of X containing x is the intersection of
all open and closed subsets of X containing x.
T
Proof. Let T be the connected component containing x. Let S = α∈A Zα be the
intersection of all open and closed subsets Zα of X containing x. Note that S is
closed in X. Note that any finite intersection of Zα ’s is a Zα . Because T is connected
and x ∈ T we have T ⊂ S. It suffices to show that S is connected. If not, then there
exists a disjoint union decomposition S = B ⨿ C with B and C open and closed in
S. In particular, B and C are closed in X, and so quasi-compact by Lemma 12.3
and assumption (1). By assumption (2) there exist quasi-compact opens U, V ⊂ X
T B = S ∩ U and C = S ∩ V (details omitted). Then U ∩ V ∩ S = ∅. Hence
with
α U ∩ V ∩ Zα = ∅. By assumption (3) the intersection U ∩ V is quasi-compact.
By Lemma 12.6 for some α′ ∈ A we have U ∩ V ∩ Zα′ = ∅. Since X \ (U ∪ V ) is
disjoint from S and closed in X hence quasi-compact, we can use the same lemma
to see that Zα′′ ⊂ U ∪ V for some α′′ ∈ A. Then Zα = Zα′ ∩ Zα′′ is contained in
U ∪ V and disjoint from U ∩ V . Hence Zα = U ∩ Zα ⨿ V ∩ Zα is a decomposition
into two open pieces, hence U ∩ Zα and V ∩ Zα are open and closed in X. Thus, if
x ∈ B say, then we see that S ⊂ U ∩ Zα and we conclude that C = ∅. □
08ZN Lemma 12.11. Let X be a topological space. Assume X is quasi-compact and
Hausdorff. For any x ∈ X the connected component of X containing x is the
intersection of all open and closed subsets of X containing x.
T
Proof. Let T be the connected component containing x. Let S = α∈A Zα be
the intersection of all open and closed subsets Zα of X containing x. Note that
S is closed in X. Note that any finite intersection of Zα ’s is a Zα . Because T is
connected and x ∈ T we have T ⊂ S. It suffices to show that S is connected. If not,
then there exists a disjoint union decomposition S = B ⨿ C with B and C open
and closed in S. In particular, B and C are closed in X, and so quasi-compact
by Lemma 12.3. By Lemma 12.4 there exist disjoint opens U, V ⊂ X with B ⊂ U
and C ⊂ V . Then X \ U ∪ V is closed in X hence quasi-compact (Lemma 12.3).
It follows that (X \ U ∪ V ) ∩ Zα = ∅ for some α by Lemma 12.6. In other words,
Zα ⊂ U ∪ V . Thus Zα = Zα ∩ V ⨿ Zα ∩ U is a decomposition into two open pieces,
hence U ∩ Zα and V ∩ Zα are open and closed in X. Thus, if x ∈ B say, then we
see that S ⊂ U ∩ Zα and we conclude that C = ∅. □
04PL Lemma 12.12. Let X be a topological space. Assume
(1) X is quasi-compact,
(2) X has a basis for the topology consisting of quasi-compact opens, and
(3) the intersection of two quasi-compact opens is quasi-compact.
For a subset T ⊂ X the following are equivalent:
(a) T is an intersection of open and closed subsets of X, and
(b) T is closed in X and is a union of connected components of X.
Proof. It is clear that (a) implies (b). Assume (b). Let x ∈ X, x ̸∈ T . Let
x ∈ C ⊂ X be T the connected component of X containing x. By Lemma 12.10 we
see that C = Vα is the intersection of all open and closed subsets Vα of X which
contain C. In particular, any pairwise intersection Vα ∩ Vβ occurs as a Vα . As T is
 TOPOLOGY 21

T
a union of connected components of X we see that C ∩ T = ∅. Hence T ∩ Vα = ∅.
Since T is quasi-compact as a closed subset of a quasi-compact space (see Lemma
12.3) we deduce that T ∩ Vα = ∅ for some α, see Lemma 12.6. For this α we see
that Uα = X \ Vα is an open and closed subset of X which contains T and not x.
The lemma follows. □
04ZA Lemma 12.13. Let X be a Noetherian topological space.
(1) The space X is quasi-compact.
(2) Any subset of X is retrocompact.
S
Proof. Suppose X = Ui is an open covering of X indexed by the set I which
does not have a refinement by a finite open covering. Choose i1 , i2 , . . . elements of
I inductively in the following way: Choose in+1 such that Uin+1 is not contained in
Ui1 ∪ . . . ∪ Uin . Thus we see that X ⊃ (X \ Ui1 ) ⊃ (X \ Ui1 ∪ Ui2 ) ⊃ . . . is a strictly
decreasing infinite sequence of closed subsets. This contradicts the fact that X is
Noetherian. This proves the first assertion. The second assertion is now clear since
every subset of X is Noetherian by Lemma 9.2. □
04ZB Lemma 12.14. A quasi-compact locally Noetherian space is Noetherian.
Proof. The conditions imply immediately that X has a finite covering by Noether-
ian subsets, and hence is Noetherian by Lemma 9.4. □
08ZP Lemma 12.15 (Alexander subbase theorem). Let X be a topological space. Let B
be a subbase for X. If every covering of X by elements of B has a finite refinement,
then X is quasi-compact.
Proof. Assume there is an open covering of X which does not have a finiteSrefine-
ment. Using Zorn’s lemma we can choose a maximal open covering X = i∈I Ui
which does not have a finite refinement (details omitted). In other words, ifS U ⊂X
is any open which does not occur as one of the Ui , then the covering X = U ∪ i∈I Ui
′
does have S a finite refinement. Let I ⊂ I be the set of indices such that Ui ∈ B.
Then i∈I ′ Ui ̸= X, since otherwise we would S get a finite refinement covering X
by our assumption on B. Pick x ∈ X, x ̸∈ i∈I ′ Ui . Pick i ∈ I with x ∈ Ui . Pick
V1 , . . . , Vn ∈ B such that x ∈ V1 ∩. . .∩Vn ⊂ Ui . This is possible as B is a subbasis for
X. Note that Vj does not occur as a Ui . By maximality of the chosen covering we
see that for each j there exist ij,1 , . . . , ij,nj ∈ I such that X = Vj ∪Uij,1 ∪. . .∪Uij,nj .
S
Since V1 ∩ . . . ∩ Vn ⊂ Ui we conclude that X = Ui ∪ Uij,l a contradiction. □

13. Locally quasi-compact spaces

08ZQ Recall that a neighbourhood of a point need not be open.
0068 Definition 13.1. A topological space X is called locally quasi-compact2 if every
point has a fundamental system of quasi-compact neighbourhoods.
The term locally compact space in the literature often refers to a space as in the
following lemma.

2This may not be standard notation. Alternative notions used in the literature are: (1) Every
point has some quasi-compact neighbourhood, and (2) Every point has a closed quasi-compact
neighbourhood. A scheme has the property that every point has a fundamental system of open
quasi-compact neighbourhoods.
 TOPOLOGY 22

08ZR Lemma 13.2. A Hausdorff space is locally quasi-compact if and only if every point
has a quasi-compact neighbourhood.
Proof. Let X be a Hausdorff space. Let x ∈ X and let x ∈ E ⊂ X be a quasi-
compact neighbourhood. Then E is closed by Lemma 12.4. Suppose that x ∈ U ⊂
X is an open neighbourhood of x. Then Z = E \ U is a closed subset of E not
containing x. Hence we can find a pair of disjoint open subsets W, V ⊂ E of E
such that x ∈ V and Z ⊂ W , see Lemma 12.4. It follows that V ⊂ E is a closed
neighbourhood of x contained in E ∩ U . Also V is quasi-compact as a closed subset
of E (Lemma 12.3). In this way we obtain a fundamental system of quasi-compact
neighbourhoods of x. □
0CQN Lemma 13.3 (Baire category theorem). Let X be a locally quasi-compact
T Haus-
dorff space. Let Un ⊂ X, n ≥ 1 be dense open subsets. Then n≥1 Un is dense in
X.
T
Proof. After replacing Un by i=1,...,n Ui we may assume that U1 ⊃ U2 ⊃ . . ..
T
Let x ∈ X. We will show that x is in the closure of n≥1 Un . Thus let E be a
T
neighbourhood of x. To show that E ∩ n≥1 Un is nonempty we may replace E by
a smaller neighbourhood. After replacing E by a smaller neighbourhood, we may
assume that E is quasi-compact.
Set x0 = x and E0 = E. Below, we will inductively choose a point xi ∈ Ei−1 ∩ Ui
and a quasi-compact neighbourhood Ei of xi with Ei ⊂ Ei−1 ∩ Ui . Because X
⊂ X are closed (Lemma 12.4). Since the ETi are also
is Hausdorff, the subsets Ei T
nonempty we conclude that i≥1 Ei is nonempty (Lemma 12.6). Since i≥1 Ei ⊂
T
E ∩ n≥1 Un this proves the lemma.
The base case i = 0 we have done above. Induction step. Since Ei−1 is a neigh-
bourhood of xi−1 we can find an open xi−1 ∈ W ⊂ Ei−1 . Since Ui is dense in X we
see that W ∩ Ui is nonempty. Pick any xi ∈ W ∩ Ui . By definition of locally quasi-
compact spaces we can find a quasi-compact neighbourhood Ei of xi contained in
W ∩ Ui . Then Ei ⊂ Ei−1 ∩ Ui as desired. □
S
09UV Lemma 13.4. Let X be a Hausdorff and quasi-compact space. S Let X = i∈I Ui
be an open covering. Then there exists an open covering X = i∈I Vi such that
Vi ⊂ Ui for all i.
Proof. Let x ∈ X. Choose an i(x) ∈ I such that x ∈ Ui(x) . Since X \ Ui(x)
and {x} are disjoint closed subsets of X, by Lemmas 12.3 and 12.4 there exists
an open neighbourhood Ux of x whose closure is disjoint from X \ Ui(x) . Thus
Ux ⊂ Ui(x) . Since X is quasi-compact, there S is a finite list of points x1 , . . . , xm such
that X = Ux1 ∪ . . . ∪ Uxm . Setting Vi = i=i(xj ) Uxj the proof is finished. □
S
09UW Lemma 13.5. Let X be a Hausdorff and quasi-compact space. Let X = i∈I Ui
be an open covering. Suppose given an integer pS≥ 0 and for every (p + 1)-tuple
i0 , . . . , ip of I an open covering Ui0 ∩ . . . ∩ Uip = Wi0 ...ip ,k . Then there exists an
S
open covering X = j∈J Vj and a map α : J → I such that Vj ⊂ Uα(j) and such
that each Vj0 ∩ . . . ∩ Vjp is contained in Wα(j0 )...α(jp ),k for some k.
Proof. Since X is quasi-compact, there is a reduction to the case where I is finite
(details omitted). We prove the result for I finite by induction on p. The base
 TOPOLOGY 23

case p = 0 is immediate
S by taking a covering as in Lemma 13.4 refining the open
covering X = Wi0 ,k .
Induction step. Assume theSlemma proven for p − 1. For all p-tuples i′0 , . . . , i′p−1
of I let Ui′0 ∩ . . . ∩ Ui′p−1 = Wi′0 ...i′p−1 ,k be a common refinement of the coverings
Ui0 ∩ . . . ∩ Uip = Wi0 ...ip ,k for those (p + 1)-tuples such that {i′0 , . . . , i′p−1 } =
S
{i0 , . . . , ip } (equality of sets). (There are finitely many of S
these as I is finite.) By
induction there exists a solution for these opens, say X = Vj and α : J → I. At
S
this point the covering X = j∈J Vj and α satisfy Vj ⊂ Uα(j) and each Vj0 ∩. . .∩Vjp
is contained in Wα(j0 )...α(jp ),k for some k if there is a repetition in α(j0 ), . . . , α(jp ).
Of course, we may and do assume that J is finite.
Fix i0 , . . . , ip ∈ I pairwise distinct. Consider (p + 1)-tuples j0 , . . . , jp ∈ J with
i0 = α(j0 ), . . . , ip = α(jp ) such that Vj0 ∩. . .∩Vjp is not contained in Wα(j0 )...α(jp ),k
for any k. Let N be the number of such (p+1)-tuples. We will show how to decrease
N . Since [
Vj0 ∩ . . . ∩ Vjp ⊂ Ui0 ∩ . . . ∩ Uip = Wi0 ...ip ,k
S
we find a finite set K = {k1 , . . . , kt } such that the LHS is contained in k∈K Wi0 ...ip ,k .
Then we consider the open covering
[
Vj0 = (Vj0 \ (Vj1 ∩ . . . ∩ Vjp )) ∪ ( Vj0 ∩ Wi0 ...ip ,k )
k∈K

The first open on the RHS intersects Vj1 ∩ . . . ∩ Vjp in the empty set and the other
opens Vj0 ,k of the RHS satisfy Vj0 ,k ∩ Vj1 . . . ∩ Vjp ⊂ Wα(j0 )...α(jp ),k . Set J ′ = J ⨿ K.
For j ∈ J set Vj′ = Vj if j ̸= j0 and set Vj′0 = Vj0 \ (Vj1 ∩ . . . ∩ Vjp ). For k ∈ K
set Vk′ = Vj0 ,k . Finally, the map α′ : J ′ → I is given by α on J and maps every
element of K to i0 . A simple check shows that N has decreased by one under this
replacement. Repeating this procedure N times we arrive at the situation where
N = 0.
To finish the proof we argue by induction on the number M of (p + 1)-tuples
i0 , . . . , ip ∈ I with pairwise distinct entries for which there exists a (p + 1)-tuple
j0 , . . . , jp ∈ J with i0 = α(j0 ), . . . , ip = α(jp ) such that Vj0 ∩ . . . ∩ Vjp is not
contained in Wα(j0 )...α(jp ),k for any k. To do this, we claim that the operation
performed in the previous paragraph does not increase M . This follows formally
from the fact that the map α′ : J ′ → I factors through a map β : J ′ → J such that
Vj′′ ⊂ Vβ(j ′ ) . □

09UX Lemma 13.6. Let X be a Hausdorff and locally quasi-compact space. Let Z ⊂ X
be a quasi-compact (hence closed) subset. Suppose given an integer p ≥ 0, a set I,
for every i ∈ I an open Ui ⊂ X, and for every (p + 1)-tuple i0 , . . . , ip of I an open
Wi0 ...ip ⊂ Ui0 ∩ . . . ∩ Uip such that
S
(1) Z ⊂ Ui , and
(2) for every i0 , . . . , ip we have Wi0 ...ip ∩ Z = Ui0 ∩ . . . ∩ Uip ∩ Z.
S
Then there exist opens Vi of X such that we have Z ⊂ Vi , for all i we have
Vi ⊂ Ui , and we have Vi0 ∩ . . . ∩ Vip ⊂ Wi0 ...ip for all (p + 1)-tuples i0 , . . . , ip .
Proof. Since Z is quasi-compact, there is a reduction to the case where I is finite
(details omitted). Because X is locally quasi-compact and Z is quasi-compact, we
can find a neighbourhood Z ⊂ E which is quasi-compact, i.e., E is quasi-compact
 TOPOLOGY 24

and contains an open neighbourhood of Z in X. If we prove the result after replacing

X by E, then the result follows. Hence we may assume X is quasi-compact.
We prove the result in case I is finite and X is quasi-compact by
S induction on p.
The base case is p = 0. In this case S we have X = (X \ Z) ∪ Wi . By Lemma
13.4 we can find a covering X = V ∪ Vi by opens Vi ⊂ Wi and V ⊂ X \ Z with
Vi ⊂ Wi for all i. Then we see that we obtain a solution of the problem posed by
the lemma.
Induction step. Assume the lemma proven for p − 1. Set Wj0 ...jp−1 equal to the
intersection of all Wi0 ...ip with {j0 , . . . , jp−1 } = {i0 , . . . , ip } (equality of sets). By
induction there exists a solution for these opens, say Vi ⊂ Ui . It follows from our
choice of Wj0 ...jp−1 that we have Vi0 ∩ . . . ∩ Vip ⊂ Wi0 ...ip for all (p + 1)-tuples
i0 , . . . , ip where ia = ib for some 0 ≤ a < b ≤ p. Thus we only need to modify our
choice of Vi if Vi0 ∩ . . . ∩ Vip ̸⊂ Wi0 ...ip for some (p + 1)-tuple i0 , . . . , ip with pairwise
distinct elements. In this case we have
T = Vi0 ∩ . . . ∩ Vip \ Wi0 ...ip ⊂ Vi0 ∩ . . . ∩ Vip \ Wi0 ...ip
is a closed subset of X contained in Ui0 ∩ . . . ∩ Uip not meeting Z. Hence we can
replace Vi0 by Vi0 \ T to “fix” the problem. After repeating this finitely many times
for each of the problem tuples, the lemma is proven. □
0CY5 Lemma 13.7. Let X be a topological space. Let Z ⊂ X be a quasi-compact subset
such that any two points of Z have disjoint open neighbourhoods in X. Suppose
given an integer p ≥ 0, a set I, for every i ∈ I an open Ui ⊂ X, and for every
(p + 1)-tuple i0 , . . . , ip of I an open Wi0 ...ip ⊂ Ui0 ∩ . . . ∩ Uip such that
S
(1) Z ⊂ Ui , and
(2) for every i0 , . . . , ip we have Wi0 ...ip ∩ Z = Ui0 ∩ . . . ∩ Uip ∩ Z.
Then there exist opens Vi of X such that
S
(1) Z ⊂ Vi ,
(2) Vi ⊂ Ui for all i,
(3) Vi ∩ Z ⊂ Ui for all i, and
(4) Vi0 ∩ . . . ∩ Vip ⊂ Wi0 ...ip for all (p + 1)-tuples i0 , . . . , ip .
Proof. Since Z is quasi-compact, there is a reduction to the case where I is finite
(details omitted). We prove the result in case I is finite by induction on p.
The base case is p = 0. For z ∈ Z ∩ Ui and z ′ ∈ Z \ Ui there exist disjoint opens
z ∈ Vz,z′ and z ′ ∈ Wz,z′ of X. Since Z \ Ui is quasi-compact (Lemma 12.3), we can
choose a finite number z1′ , . . . , zr′ such that Z \ Ui ⊂ Wz,z1′ ∪ . . . ∪ Wz,zr′ . Then we
see that Vz = Vz,z1′ ∩ . . . ∩ Vz,zr′ ∩ Ui is an open neighbourhood of z contained in
Ui with the property that Vz ∩ Z ⊂ Ui . Since z and i where arbitrary and since Z
is quasi-compact we can find a finite list z1 , i1 , . . . , zt , it and opens Vzj ⊂ Uij with
S S
Vzj ∩ Z ⊂ Uij and Z ⊂ Vzj . Then we can set Vi = Wi ∩ ( j:i=ij Vzj ) to solve the
problem for p = 0.
Induction step. Assume the lemma proven for p − 1. Set Wj0 ...jp−1 equal to the
intersection of all Wi0 ...ip with {j0 , . . . , jp−1 } = {i0 , . . . , ip } (equality of sets). By
induction there exists a solution for these opens, say Vi ⊂ Ui . It follows from our
choice of Wj0 ...jp−1 that we have Vi0 ∩ . . . ∩ Vip ⊂ Wi0 ...ip for all (p + 1)-tuples
i0 , . . . , ip where ia = ib for some 0 ≤ a < b ≤ p. Thus we only need to modify our
 TOPOLOGY 25

choice of Vi if Vi0 ∩ . . . ∩ Vip ̸⊂ Wi0 ...ip for some (p + 1)-tuple i0 , . . . , ip with pairwise
distinct elements. In this case we have
T = Vi0 ∩ . . . ∩ Vip \ Wi0 ...ip ⊂ Vi0 ∩ . . . ∩ Vip \ Wi0 ...ip
is a closed subset of X not meeting Z by our property (3) of the opens Vi . Hence
we can replace Vi0 by Vi0 \ T to “fix” the problem. After repeating this finitely
many times for each of the problem tuples, the lemma is proven. □

14. Limits of spaces

08ZS The category of topological spaces has products. Namely, Q if I is a set and for
i ∈ I we are given a topological space Xi then we endow i∈IQ Xi with the product
topology. As a basis for the topology we use sets of the form Ui where Ui ⊂ Xi
is open and Ui = Xi for almost all i.
The category of topological spaces has equalizers. Namely, if a, b : X → Y are
morphisms of topological spaces, then the equalizer of a and b is the subset {x ∈
X | a(x) = b(x)} ⊂ X endowed with the induced topology.
08ZT Lemma 14.1. The category of topological spaces has limits and the forgetful func-
tor to sets commutes with them.
Proof. This follows from the discussion above and Categories, Lemma 14.11. It
follows from the description above that the forgetful functor commutes with lim-
its. Another way to see this is to use Categories, Lemma 24.5 and use that the
forgetful functor has a left adjoint, namely the functor which assigns to a set the
corresponding discrete topological space. □

0A2P Lemma 14.2. Let I be a cofiltered category. Let i 7→ Xi be a diagram of topological

spaces over I. Let X = lim Xi be the limit with projection maps fi : X → Xi .
(1) Any open of X is of the form j∈J fj−1 (Uj ) for some subset J ⊂ I and
S
opens Uj ⊂ Xj .
(2) Any quasi-compact open of X is of the form fi−1 (Ui ) for some i and some
Ui ⊂ Xi open.
Q
Proof. The construction of the limit given aboveQshows that X ⊂ X Qi with the
induced topology. A basis for the topology of Xi are the opens Ui where
Ui ⊂ Xi is open and Ui = Xi for almost all i. Say i1 , . . . , in ∈ Ob(I) are the
objects such that Uij ̸= Xij . Then
Y
X∩ Ui = fi−1
1
(Ui1 ) ∩ . . . ∩ fi−1
n
(Uin )
For a general limit of topological spaces these form a basis for the topology on X.
However, if I is cofiltered as in the statement of the lemma, then we can pick a
j ∈ Ob(I) and morphisms j → il , l = 1, . . . , n. Let
Uj = (Xj → Xi1 )−1 (Ui1 ) ∩ . . . ∩ (Xj → Xin )−1 (Uin )
Then it is clear that X ∩ Ui = fj−1 (Uj ). Thus for any open W of X there is a set
Q
S −1
A and a map α : A → Ob(I) and opens Ua ⊂ Xα(a) such that W = fα(a) (Ua ).
S S −1
Set J = Im(α) and for j ∈ J set Uj = α(a)=j Ua to see that W = j∈J fj (Uj ).
This proves (1).
 TOPOLOGY 26

To see (2) suppose that j∈J fj−1 (Uj ) is quasi-compact. Then it is equal to
S

fj−1
1
(Uj1 ) ∪ . . . ∪ fj−1
m
(Ujm ) for some j1 , . . . , jm ∈ J. Since I is cofiltered, we can
pick a i ∈ Ob(I) and morphisms i → jl , l = 1, . . . , m. Let
Ui = (Xi → Xj1 )−1 (Uj1 ) ∪ . . . ∪ (Xi → Xjm )−1 (Ujm )
Then our open equals fi−1 (Ui ) as desired. □

0A2Q Lemma 14.3. Let I be a cofiltered category. Let i 7→ Xi be a diagram of topological

spaces over I. Let X be a topological space such that
(1) X = lim Xi as a set (denote fi the projection maps),
(2) the sets fi−1 (Ui ) for i ∈ Ob(I) and Ui ⊂ Xi open form a basis for the
topology of X.
Then X is the limit of the Xi as a topological space.
Proof. Follows from the description of the limit topology in Lemma 14.2. □

08ZU Theorem 14.4 (Tychonov). A product of quasi-compact spaces is quasi-compact.

Proof. Let Q I be a set and for i ∈ I let Xi be a quasi-compact topological
Q space.
Set X = Xi . Let B be the set of subsets of X of the form Ui × i′ ∈I,i′ ̸=i Xi′
where Ui ⊂ Xi is open. By construction this family is a subbasis for theS topology
on X. By Lemma 12.15 it suffices to show that any covering X = j∈J Bj by
`
elements Bj of B has aQ finite refinement. We can decompose SJ = Ji so that if
j ∈ Ji , then Bj = Uj × i′ ̸=i Xi′ with Uj ⊂ Xi open. If Xi = j∈Ji Uj , then there
S
is a finite refinement and we conclude that X = j∈J Bj has a finite refinement. If
thisSis not the case, then for every i we can choose an point xi ∈ Xi which is not
in U . But then the point x = (xi )i∈I is an element of X not contained in
S j∈Ji j
B
j∈J j a contradiction.
, □

The following lemma does not hold if one drops the assumption that the spaces Xi
are Hausdorff, see Examples, Section 4.
08ZV Lemma 14.5. Let I be a category and let i 7→ Xi be a diagram over I in the
category of topological spaces. If each Xi is quasi-compact and Hausdorff, then
lim Xi is quasi-compact.
Q
Proof. Recall that lim Xi is a subspace of Xi . By Theorem 14.4 this product Q is
quasi-compact. Hence it suffices to show that lim Xi is a closed subspace of Xi
(Lemma 12.3). If φ : j → k is a morphism of I, then let Γφ ⊂ Xj × Xk denote the
graph of the corresponding continuous map Xj → Xk . By Lemma 3.2 this graph
is closed. It is clear that lim Xi is the intersection of the closed subsets
Y Y
Γφ × Xl ⊂ Xi
l̸=j,k

Thus the result follows. □

The following lemma generalizes Categories, Lemma 21.7 and partially generalizes
Lemma 12.6.
0A2R Lemma 14.6. Let I be a cofiltered category and let i 7→ Xi be a diagram over I
in the category of topological spaces. If each Xi is quasi-compact, Hausdorff, and
nonempty, then lim Xi is nonempty.
 TOPOLOGY 27

Proof. In the proof of Lemma 14.5 we have seen that X = lim Xi is the intersection
of the closed subsets Y
Zφ = Γφ × Xl
l̸=j,k
Q
inside the quasi-compact space Xi where φ : j → k is a morphism of I and
Γφ ⊂ Xj × Xk is the graph of the corresponding morphism Xj → Xk . Hence by
Lemma 12.6 it suffices to show any finite intersection of these subsets is nonempty.
Assume φt : jt → kt , t = 1, . . . , n is a finite collection of morphisms of I. Since I
is cofiltered, we can pick an object j and a morphism ψt : j → jt for each t. For
each pair t, t′ such that either (a) jt = jt′ , or (b) jt = kt′ , or (c) kt = kt′ we obtain
two morphisms j → l with l = jt in case (a), (b) or l = kt in case (c). Because I is
cofiltered and since there are finitely many pairs (t, t′ ) we may choose a map j ′ → j
which equalizes these two morphisms for all such pairs (t, t′ ). Pick an element
x ∈ Xj ′ and for each t let xjt , resp. xkt be the image of x under the morphism
Xj ′ → Xj → Xjt , resp. Xj ′ → Xj → Xjt → Xkt . For any index l ∈ Ob(I) which
is not equal to jt or kt for some t we pick an arbitrary element xl ∈ Xl (using the
axiom of choice). Then (xi )i∈Ob(I) is in the intersection
Zφ1 ∩ . . . ∩ Zφn
by construction and the proof is complete. □

15. Constructible sets

04ZC
005G Definition 15.1. Let X be a topological space. Let E ⊂ X be a subset of X.
(1) We say E is constructible3 in X if E is a finite union of subsets of the form
U ∩ V c where U, V ⊂ X are open and retrocompact in X.
S E is locally constructible in X if there exists an open covering
(2) We say
X = Vi such that each E ∩ Vi is constructible in Vi .
005H Lemma 15.2. The collection of constructible sets is closed under finite intersec-
tions, finite unions and complements.
Proof. Note that if U1 , U2 are open and retrocompact in X then so is U1 ∪ U2
because the union of two quasi-compact subsets of X is quasi-compact. It is also
true that U1 ∩ U2 is retrocompact. Namely, suppose U ⊂ X is quasi-compact
open, then U2 ∩ U is quasi-compact because U2 is retrocompact in X, and then we
conclude U1 ∩ (U2 ∩ U ) is quasi-compact because U1 is retrocompact in X. From
this it is formal to show that the complement of a constructible set is constructible,
that finite unions of constructibles are constructible, and that finite intersections
of constructibles are constructible. □

005I Lemma 15.3. Let f : X → Y be a continuous map of topological spaces. If the

inverse image of every retrocompact open subset of Y is retrocompact in X, then
inverse images of constructible sets are constructible.
Proof. This is true because f −1 (U ∩ V c ) = f −1 (U ) ∩ f −1 (V )c , combined with the
definition of constructible sets. □
3In the second edition of EGA I [GD71] this was called a “globally constructible” set and a
the terminology “constructible” was used for what we call a locally constructible set.
 TOPOLOGY 28

005J Lemma 15.4. Let U ⊂ X be open. For a constructible set E ⊂ X the intersection
E ∩ U is constructible in U .
Proof. Suppose that V ⊂ X is retrocompact open in X. It suffices to show that
V ∩ U is retrocompact in U by Lemma 15.3. To show this let W ⊂ U be open and
quasi-compact. Then W is open and quasi-compact in X. Hence V ∩W = V ∩U ∩W
is quasi-compact as V is retrocompact in X. □
09YD Lemma 15.5. Let U ⊂ X be a retrocompact open. Let E ⊂ U . If E is constructible
in U , then E is constructible in X.
Proof. Suppose that V, W ⊂ U are retrocompact open in U . Then V, W are
retrocompact open in X (Lemma 12.2). Hence V ∩ (U \ W ) = V ∩ (X \ W ) is
constructible in X. We conclude since every constructible subset of U is a finite
union of subsets of the form V ∩ (U \ W ). □
053W Lemma 15.6. Let X be a topological space. Let E ⊂ X be a subset. Let X =
V1 ∪ . . . ∪ Vm be a finite covering by retrocompact opens. Then E is constructible in
X if and only if E ∩ Vj is constructible in Vj for each j = 1, . . . , m.
Proof. If E is constructible in X, then by Lemma 15.4 we see that E ∩ Vj is
constructible in Vj for all j. Conversely,
S suppose that E ∩ Vj is constructible in Vj
for each j = 1, . . . , m. Then E = E ∩ Vj is a finite union of constructible sets by
Lemma 15.5 and hence constructible. □
09YE Lemma 15.7. Let X be a topological space. Let Z ⊂ X be a closed subset such
that X \ Z is quasi-compact. Then for a constructible set E ⊂ X the intersection
E ∩ Z is constructible in Z.
Proof. Suppose that V ⊂ X is retrocompact open in X. It suffices to show that
V ∩ Z is retrocompact in Z by Lemma 15.3. To show this let W ⊂ Z be open
and quasi-compact. The subset W ′ = W ∪ (X \ Z) is quasi-compact, open, and
W = Z ∩W ′ . Hence V ∩Z ∩W = V ∩Z ∩W ′ is a closed subset of the quasi-compact
open V ∩ W ′ as V is retrocompact in X. Thus V ∩ Z ∩ W is quasi-compact by
Lemma 12.3. □
09YF Lemma 15.8. Let X be a topological space. Let T ⊂ X be a subset. Suppose
(1) T is retrocompact in X,
(2) quasi-compact opens form a basis for the topology on X.
Then for a constructible set E ⊂ X the intersection E ∩ T is constructible in T .
Proof. Suppose that V ⊂ X is retrocompact open in X. It suffices to show that
V ∩ T is retrocompact in T by Lemma 15.3. To show this let W ⊂ T be open
and quasi-compact. By assumption (2) we can find a quasi-compact open W ′ ⊂ X
such that W = T ∩ W ′ (details omitted). Hence V ∩ T ∩ W = V ∩ T ∩ W ′ is the
intersection of T with the quasi-compact open V ∩ W ′ as V is retrocompact in X.
Thus V ∩ T ∩ W is quasi-compact. □
09YG Lemma 15.9. Let Z ⊂ X be a closed subset whose complement is retrocompact
open. Let E ⊂ Z. If E is constructible in Z, then E is constructible in X.
Proof. Suppose that V ⊂ Z is retrocompact open in Z. Consider the open subset
Ṽ = V ∪ (X \ Z) of X. Let W ⊂ X be quasi-compact open. Then
W ∩ Ṽ = (V ∩ W ) ∪ ((X \ Z) ∩ W ) .
 TOPOLOGY 29

The first part is quasi-compact as V ∩ W = V ∩ (Z ∩ W ) and (Z ∩ W ) is quasi-

compact open in Z (Lemma 12.3) and V is retrocompact in Z. The second part
is quasi-compact as (X \ Z) is retrocompact in X. In this way we see that Ṽ is
retrocompact in X. Thus if V1 , V2 ⊂ Z are retrocompact open, then
V1 ∩ (Z \ V2 ) = Ṽ1 ∩ (X \ Ṽ2 )
is constructible in X. We conclude since every constructible subset of Z is a finite
union of subsets of the form V1 ∩ (Z \ V2 ). □
09YH Lemma 15.10. Let X be a topological space. Every constructible subset of X is
retrocompact.
Proof. Let E = i=1,...,n Ui ∩ Vic with Ui , Vi retrocompact open in X. Let W ⊂ X
S

be quasi-compact open. Then E ∩ W = i=1,...,n Ui ∩ Vic ∩ W . Thus it suffices

S
to show that U ∩ V c ∩ W is quasi-compact if U, V are retrocompact open and W
is quasi-compact open. This is true because U ∩ V c ∩ W is a closed subset of the
quasi-compact U ∩ W so Lemma 12.3 applies. □
Question: Does the following lemma also hold if we assume X is a quasi-compact
topological space? Compare with Lemma 15.7.
09YI Lemma 15.11. Let X be a topological space. Assume X has a basis consisting
of quasi-compact opens. For E, E ′ constructible in X, the intersection E ∩ E ′ is
constructible in E.
Proof. Combine Lemmas 15.8 and 15.10. □
09YJ Lemma 15.12. Let X be a topological space. Assume X has a basis consisting of
quasi-compact opens. Let E be constructible in X and F ⊂ E constructible in E.
Then F is constructible in X.
Proof. Observe that any retrocompact subset T of X has a basis for the induced
topology consisting of quasi-compact opens. In particular this holds for any con-
structible subset (Lemma 15.10). Write E = E1 ∪ . . . ∪ En with Ei = Ui ∩ Vic where
Ui , Vi ⊂ X are retrocompact open. Note that Ei = E ∩ Ei is constructible in E
by Lemma 15.11. Hence F ∩ Ei is constructible in Ei by Lemma 15.11. Thus it
suffices to prove the lemma in case E = U ∩ V c where U, V ⊂ X are retrocompact
open. In this case the inclusion E ⊂ X is a composition
E =U ∩Vc →U →X
Then we can apply Lemma 15.9 to the first inclusion and Lemma 15.5 to the
second. □
0F2K Lemma 15.13. Let X be a quasi-compact topological space having a basis con-
sisting of quasi-compact opens such that the intersection of any two quasi-compact
opens is quasi-compact. Let T ⊂ X be a locally closed subset such that T is quasi-
compact and T c is retrocompact in X. Then T is constructible in X.
Proof. Note that T is quasi-compact and open in T . Using our basis of quasi-
compact opens we can write T = U ∩ T where U is quasi-compact open in X. Then
V = U \ T = U ∩ T c is retrocompact in U as T c is retrocompact in X. Hence V
is quasi-compact. Since the intersection of any two quasi-compact opens is quasi-
compact any quasi-compact open of X is retrocompact. Thus T = U ∩ V c with U
and V = U \ T retrocompact opens of X. A fortiori, T is constructible in X. □
 TOPOLOGY 30

09YK Lemma 15.14. Let X be a topological space which has a basis for the topology
consisting of quasi-compact opens. Let E ⊂ X be a subset. Let X = E1 ∪ . . . ∪ Em
be a finite covering by constructible subsets. Then E is constructible in X if and
only if E ∩ Ej is constructible in Ej for each j = 1, . . . , m.
Proof. Combine Lemmas 15.11 and 15.12. □
005K Lemma 15.15. Let X be a topological space. Suppose that Z ⊂ X is irreducible.
Let E ⊂ X be a finite union of locally closed subsets (e.g. E is constructible). The
following are equivalent
(1) The intersection E ∩ Z contains an open dense subset of Z.
(2) The intersection E ∩ Z is dense in Z.
If Z has a generic point ξ, then this is also equivalent to
(3) We have ξ ∈ E.
Proof. The implication (1) ⇒ (2) is clear. Assume (2). Note that E ∩ Z is a finite
union of locally closed subsets Zi of Z. Since Z is irreducible, one of the Zi must
be dense in Z. Then this Zi is dense open in Z as it is open in its closure. Hence
(1) holds.
Suppose that ξ ∈ Z is a generic point. If the equivalent conditions (1) and (2) hold,
then ξ ∈ E. Conversely, if ξ ∈ E then ξ ∈ E ∩ Z and hence E ∩ Z is dense in Z. □

16. Constructible sets and Noetherian spaces

053X
005L Lemma 16.1. Let X be a Noetherian topological space. The constructible sets in
X are precisely the finite unions of locally closed subsets of X.
Proof. This follows immediately from Lemma 12.13. □
053Y Lemma 16.2. Let f : X → Y be a continuous map of Noetherian topological
spaces. If E ⊂ Y is constructible in Y , then f −1 (E) is constructible in X.
Proof. Follows immediately from Lemma 16.1 and the definition of a continuous
map. □
053Z Lemma 16.3. Let X be a Noetherian topological space. Let E ⊂ X be a subset.
The following are equivalent:
(1) E is constructible in X, and
(2) for every irreducible closed Z ⊂ X the intersection E ∩ Z either contains a
nonempty open of Z or is not dense in Z.
Proof. Assume E is constructible and Z ⊂ X irreducible closed. Then E ∩ Z is
constructible in Z by Lemma 16.2. Hence E ∩ Z is a finite union of nonempty
locally closed subsets Ti of Z. Clearly if none of the Ti is open in Z, then E ∩ Z is
not dense in Z. In this way we see that (1) implies (2).
Conversely, assume (2) holds. Consider the set S of closed subsets Y of X such that
E ∩ Y is not constructible in Y . If S ≠ ∅, then it has a smallest element Y as X
is Noetherian. Let Y = Y1 ∪ . . . ∪ Yr be the decomposition of Y into its irreducible
components, see Lemma 9.2. If r > 1, then each Yi ∩ E is constructible in Yi and
hence a finite union of locally closed subsets of Yi . Thus E ∩ Y is a finite union of
locally closed subsets of Y too and we conclude that E ∩ Y is constructible in Y by
 TOPOLOGY 31

Lemma 16.1. This is a contradiction and so r = 1. If r = 1, then Y is irreducible,

and by assumption (2) we see that E ∩ Y either (a) contains an open V of Y or
(b) is not dense in Y . In case (a) we see, by minimality of Y , that E ∩ (Y \ V ) is a
finite union of locally closed subsets of Y \ V . Thus E ∩ Y is a finite union of locally
closed subsets of Y and is constructible by Lemma 16.1. This is a contradiction and
so we must be in case (b). In case (b) we see that E ∩ Y = E ∩ Y ′ for some proper
closed subset Y ′ ⊂ Y . By minimality of Y we see that E ∩ Y ′ is a finite union of
locally closed subsets of Y ′ and we see that E ∩ Y ′ = E ∩ Y is a finite union of
locally closed subsets of Y and is constructible by Lemma 16.1. This contradiction
finishes the proof of the lemma. □
0540 Lemma 16.4. Let X be a Noetherian topological space. Let x ∈ X. Let E ⊂ X
be constructible in X. The following are equivalent:
(1) E is a neighbourhood of x, and
(2) for every irreducible closed subset Y of X which contains x the intersection
E ∩ Y is dense in Y .
Proof. It is clear that (1) implies (2). Assume (2). Consider the set S of closed
subsets Y of X containing x such that E ∩ Y is not a neighbourhood of x in Y . If
S= ̸ ∅, then it has a minimal element Y as X is Noetherian. Suppose Y = Y1 ∪ Y2
with two smaller nonempty closed subsets Y1 , Y2 . If x ∈ Yi for i = 1, 2, then Yi ∩ E
is a neighbourhood of x in Yi and we conclude Y ∩ E is a neighbourhood of x in Y
which is a contradiction. If x ∈ Y1 but x ̸∈ Y2 (say), then Y1 ∩E is a neighbourhood
of x in Y1 and hence also in Y , which is a contradiction as well. We conclude that Y
is irreducible closed. By assumption (2) we see that E ∩Y is dense in Y . Thus E ∩Y
contains an open V of Y , see Lemma 16.3. If x ∈ V then E ∩ Y is a neighbourhood
of x in Y which is a contradiction. If x ̸∈ V , then Y ′ = Y \ V is a proper closed
subset of Y containing x. By minimality of Y we see that E ∩ Y ′ contains an open
neighbourhood V ′ ⊂ Y ′ of x in Y ′ . But then V ′ ∪ V is an open neighbourhood of
x in Y contained in E, a contradiction. This contradiction finishes the proof of the
lemma. □
0541 Lemma 16.5. Let X be a Noetherian topological space. Let E ⊂ X be a subset.
The following are equivalent:
(1) E is open in X, and
(2) for every irreducible closed subset Y of X the intersection E ∩ Y is either
empty or contains a nonempty open of Y .
Proof. This follows formally from Lemmas 16.3 and 16.4. □

17. Characterizing proper maps

005M We include a section discussing the notion of a proper map in usual topology. We
define a continuous map of topological spaces to be proper if it is universally closed
and separated. Although this matches well with the definition of a proper morphism
in algebraic geometry, this is different from the definition in Bourbaki. With our
definition of a proper map of topological spaces, the proper base change theorem
(Cohomology, Theorem 18.2) holds without any further assumptions. Furthermore,
given a morphism f : X → Y of finite type schemes over C one has: f is proper
as a morphism of schemes if and only if the continuous map f : X(C) → Y (C) on
C-points with the classical topology is proper. This is explained in [Gro71, Exp.
 TOPOLOGY 32

XII, Prop. 3.2(v)] which also has a footnote pointing out that they take properness
in topology to be Bourbaki’s notion with separatedness added on.
We find it useful to have names for three distinct concepts: separated, universally
closed, and both of those together (i.e., properness). For a continuous map f : X →
Y of locally compact Hausdorff spaces the word “proper” has long been used for the
notion “f −1 (compact) = compact” and this is equivalent to universal closedness for
such nice spaces. In fact, we will see the preimage condition formulated for clarity
using the word “quasi-compact” is equivalent to universal closedness in general, if
one includes the assumption of the map being closed. See also [Lan93, Exercises 22-
26 in Chapter II] but beware that Lang uses “proper” as a synonym for “universally
closed”, like Bourbaki does.
005N Lemma 17.1 (Tube lemma). Let X and Y be topological spaces. Let A ⊂ X and
B ⊂ Y be quasi-compact subsets. Let A × B ⊂ W ⊂ X × Y with W open in X × Y .
Then there exists opens A ⊂ U ⊂ X and B ⊂ V ⊂ Y such that U × V ⊂ W .
Proof. For every a ∈ A and b ∈ B there exist opens U(a,b) of X and V(a,b) of Y
such that (a, b) ∈ U(a,b) × V(a,b) ⊂ W . Fix b and we see there exist a finite number
a1 , . . . , an such that A ⊂ U(a1 ,b) ∪ . . . ∪ U(an ,b) . Hence
A × {b} ⊂ (U(a1 ,b) ∪ . . . ∪ U(an ,b) ) × (V(a1 ,b) ∩ . . . ∩ V(an ,b) ) ⊂ W.
Thus for every b ∈ B there exists opens Ub ⊂ X and Vb ⊂ Y such that A × {b} ⊂
Ub × Vb ⊂ W . As above there exist a finite number b1 , . . . , bm such that B ⊂
Vb1 ∪. . .∪Vbm . Then we win because A×B ⊂ (Ub1 ∩. . .∩Ubm )×(Vb1 ∪. . .∪Vbm ). □
The notation in the following definition may be slightly different from what you are
used to.
005O Definition 17.2. Let f : X → Y be a continuous map between topological spaces.
(1) We say that the map f is closed if the image of every closed subset is closed.
(2) We say that the map f is Bourbaki-proper4 if the map Z × X → Z × Y is
closed for any topological space Z.
(3) We say that the map f is quasi-proper if the inverse image f −1 (V ) of every
quasi-compact subset V ⊂ Y is quasi-compact.
(4) We say that f is universally closed if the map f ′ : Z ×Y X → Z is closed
for any continuous map g : Z → Y .
(5) We say that f is proper if f is separated and universally closed.
The following lemma is useful later.
005P Lemma 17.3. A topological space X is quasi-compact if and only if the projection Combination of
map Z × X → Z is closed for any topological space Z. [Bou71, I, p. 75,
Lemme 1] and
S remark below.) If X is not quasi-compact, there exists an open
Proof. (See also
[Bou71, I, p. 76,
covering X = i∈I Ui such that no finite number of Ui cover X. Let Z be the
Corrolaire 1].
subset of the power set P(I) of I consisting of I and all nonempty finite subsets of
I. Define a topology on Z with as a basis for the topology the following sets:
(1) All subsets of Z \ {I}.
(2) For every finite subset K of I the set UK := {J ⊂ I | J ∈ Z, K ⊂ J}).
4This is the terminology used in [Bou71]. Sometimes this property may be called “universally
closed” in the literature.
 TOPOLOGY 33

It is left to the reader to verify this is the basis for a topology. Consider the subset
of Z × X defined by the formula
\
M = {(J, x) | J ∈ Z, x ∈ Uic )}
i∈J
If (J, x) ̸∈ M , then x ∈ Ui for some i ∈ J. Hence U{i} × Ui ⊂ Z × X is an open
subset containing (J, x) and not intersecting M . Hence M is closed. The projection
of M to Z is Z − {I} which is not closed. Hence Z × X → Z is not closed.
Assume X is quasi-compact. Let Z be a topological space. Let M ⊂ Z × X be
closed. Let z ∈ Z be a point which is not in pr1 (M ). By the Tube Lemma 17.1
there exists an open U ⊂ Z such that U × X is contained in the complement of M .
Hence pr1 (M ) is closed. □
005Q Remark 17.4. Lemma 17.3 is a combination of [Bou71, I, p. 75, Lemme 1] and
[Bou71, I, p. 76, Corollaire 1].
005R Theorem 17.5. Let f : X → Y be a continuous map between topological spaces. In [Bou71, I, p. 75,
The following conditions are equivalent: Theorem 1] you can
(1) The map f is quasi-proper and closed. find: (2) ⇔ (4). In
(2) The map f is Bourbaki-proper. [Bou71, I, p. 77,
(3) The map f is universally closed. Proposition 6] you
(4) The map f is closed and f −1 (y) is quasi-compact for any y ∈ Y . can find: (2) ⇒ (1).

Proof. (See also the remark below.) If the map f satisfies (1), it automatically
satisfies (4) because any single point is quasi-compact.
Assume map f satisfies (4). We will prove it is universally closed, i.e., (3) holds.
Let g : Z → Y be a continuous map of topological spaces and consider the diagram
Z ×Y X /X
g′
f′ f
 
Z
g
/Y
During the proof we will use that Z ×Y X → Z × X is a homeomorphism onto its
image, i.e., that we may identify Z×Y X with the corresponding subset of Z×X with
the induced topology. The image of f ′ : Z ×Y X → Z is Im(f ′ ) = {z : g(z) ∈ f (X)}.
Because f (X) is closed, we see that Im(f ′ ) is a closed subspace of Z. Consider a
closed subset P ⊂ Z ×Y X. Let z ∈ Z, z ̸∈ f ′ (P ). If z ̸∈ Im(f ′ ), then Z \ Im(f ′ )
is an open neighbourhood which avoids f ′ (P ). If z is in Im(f ′ ) then (f ′ )−1 {z} =
{z} × f −1 {g(z)} and f −1 {g(z)} is quasi-compact by assumption. Because P is a
closed subset of Z ×Y X, we have a closed P ′ of Z × X such that P = P ′ ∩ Z ×Y X.
Since (f ′ )−1 {z} is a subset of P c = P ′c ∪ (Z ×Y X)c , and since (f ′ )−1 {z} is disjoint
from (Z ×Y X)c we see that (f ′ )−1 {z} is contained in P ′c . We may apply the Tube
Lemma 17.1 to (f ′ )−1 {z} = {z} × f −1 {g(z)} ⊂ (P ′ )c ⊂ Z × X. This gives V × U
containing (f ′ )−1 {z} where U and V are open sets in X and Z respectively and
V × U has empty intersection with P ′ . Then the set V ∩ g −1 (Y − f (U c )) is open
in Z since f is closed, contains z, and has empty intersection with the image of P .
Thus f ′ (P ) is closed. In other words, the map f is universally closed.
The implication (3) ⇒ (2) is trivial. Namely, given any topological space Z consider
the projection morphism g : Z × Y → Y . Then it is easy to see that f ′ is the map
Z × X → Z × Y , in other words that (Z × Y ) ×Y X = Z × X. (This identification
 TOPOLOGY 34

is a purely categorical property having nothing to do with topological spaces per

se.)
Assume f satisfies (2). We will prove it satisfies (1). Note that f is closed as
f can be identified with the map {pt} × X → {pt} × Y which is assumed closed.
Choose any quasi-compact subset K ⊂ Y . Let Z be any topological space. Because
Z × X → Z × Y is closed we see the map Z × f −1 (K) → Z × K is closed (if T is
closed in Z × f −1 (K), write T = Z × f −1 (K) ∩ T ′ for some closed T ′ ⊂ Z × X).
Because K is quasi-compact, K × Z → Z is closed by Lemma 17.3. Hence the
composition Z × f −1 (K) → Z × K → Z is closed and therefore f −1 (K) must be
quasi-compact by Lemma 17.3 again. □

005S Remark 17.6. Here are some references to the literature. In [Bou71, I, p. 75,
Theorem 1] you can find: (2) ⇔ (4). In [Bou71, I, p. 77, Proposition 6] you can
find: (2) ⇒ (1). Of course, trivially we have (1) ⇒ (4). Thus (1), (2) and (4) are
equivalent. The equivalence of (3) and (4) is [Lan93, Chapter II, Exercise 25].
08YD Lemma 17.7. Let f : X → Y be a continuous map of topological spaces. If X is
quasi-compact and Y is Hausdorff, then f is universally closed.
Proof. Since every point of Y is closed, we see from Lemma 12.3 that the closed
subset f −1 (y) of X is quasi-compact for all y ∈ Y . Thus, by Theorem 17.5 it
suffices to show that f is closed. If E ⊂ X is closed, then it is quasi-compact
(Lemma 12.3), hence f (E) ⊂ Y is quasi-compact (Lemma 12.7), hence f (E) is
closed in Y (Lemma 12.4). □

08YE Lemma 17.8. Let f : X → Y be a continuous map of topological spaces. If f is

bijective, X is quasi-compact, and Y is Hausdorff, then f is a homeomorphism.
Proof. It suffices to prove f is closed, because this implies that f −1 is continuous.
If T ⊂ X is closed, then T is quasi-compact by Lemma 12.3, hence f (T ) is quasi-
compact by Lemma 12.7, hence f (T ) is closed by Lemma 12.4. □

18. Jacobson spaces

005T
005U Definition 18.1. Let X be a topological space. Let X0 be the set of closed points
of X. We say that X is Jacobson if every closed subset Z ⊂ X is the closure of
Z ∩ X0 .
Note that a topological space X is Jacobson if and only if every nonempty locally
closed subset of X has a point closed in X.
Let X be a Jacobson space and let X0 be the set of closed points of X with the
induced topology. Clearly, the definition implies that the morphism X0 → X
induces a bijection between the closed subsets of X0 and the closed subsets of X.
Thus many properties of X are inherited by X0 . For example, the Krull dimensions
of X and X0 are the same.
005V Lemma 18.2. Let X be a topological space. Let X0 be the set of closed points of
X. Suppose that for every point x ∈ X the intersection X0 ∩ {x} is dense in {x}.
Then X is Jacobson.
 TOPOLOGY 35

Proof. Let Z be closed subset of X and U be and open subset of X such that
U ∩ Z is nonempty. Then for x ∈ U ∩ Z we have that {x} ∩ U is a nonempty subset
of Z ∩ U , and by hypothesis it contains a point closed in X as required. □
02I7 Lemma 18.3. Let X be a Kolmogorov topological space with a basis of quasi-
compact open sets. If X is not Jacobson, then there exists a non-closed point x ∈ X
such that {x} is locally closed.
Proof. As X is not Jacobson there exists a closed set Z and an open set U in X
such that Z ∩ U is nonempty and does not contain points closed in X. As X has
a basis of quasi-compact open sets we may replace U by an open quasi-compact
neighborhood of a point in Z ∩ U and so we may assume that U is quasi-compact
open. By Lemma 12.8, there exists a point x ∈ Z ∩ U closed in Z ∩ U , and so {x}
is locally closed but not closed in X. □
S
005W Lemma 18.4. Let X be a topological space. Let X = Ui be an open covering.
Then S X is Jacobson if and only if each Ui is Jacobson. Moreover, in this case
X0 = Ui,0 .
Proof. Let X be a topological space. Let X0 be the set of closed points of X. Let
Ui,0 be the set of closed points of Ui . Then X0 ∩ Ui ⊂ Ui,0 but equality may not
hold in general.
First, assume that each Ui is Jacobson. We claim that in this case X0 ∩ Ui = Ui,0 .
Namely, suppose that x ∈ Ui,0 , i.e., x is closed in Ui . Let {x} be the closure in X.
Consider {x} ∩ Uj . If x ̸∈ Uj , then {x} ∩ Uj = ∅. If x ∈ Uj , then Ui ∩ Uj ⊂ Uj is
an open subset of Uj containing x. Let T ′ = Uj \ Ui ∩ Uj and T = {x} ⨿ T ′ . Then
T , T ′ are closed subsets of Uj and T contains x. As Uj is Jacobson we see that the
closed points of Uj are dense in T . Because T = {x} ⨿ T ′ this can only be the case
if x is closed in Uj . Hence {x} ∩ Uj = {x}. We conclude that {x} = {x} as desired.
Let Z ⊂ X be a closed subset (still assuming each Ui is Jacobson). Since now we
know that X0 ∩ Z ∩ Ui = Ui,0 ∩ Z are dense in Z ∩ Ui it follows immediately that
X0 ∩ Z is dense in Z.
Conversely, assume that X is Jacobson. Let Z ⊂ Ui be closed. Then X0 ∩Z is dense
in Z. Hence also X0 ∩ Z is dense in Z, because Z \ Z is closed. As X0 ∩ Ui ⊂ Ui,0
we see that Ui,0 ∩ Z is dense in Z. Thus Ui is Jacobson as desired. □
005X Lemma 18.5. Let X be Jacobson. The following types of subsets T ⊂ X are
Jacobson:
(1) Open subspaces.
(2) Closed subspaces.
(3) Locally closed subspaces.
(4) Unions of locally closed subspaces.
(5) Constructible sets.
(6) Any subset T ⊂ X which locally on X is a union of locally closed subsets.
In each of these cases closed points of T are closed in X.
Proof. Let X0 be the set of closed points of X. For any subset T ⊂ X we let (∗)
denote the property:
(*) Every nonempty locally closed subset of T has a point closed in X.
 TOPOLOGY 36

Note that always X0 ∩ T ⊂ T0 . Hence property (∗) implies that T is Jacobson. In

addition it clearly implies that every closed point of T is closed in X.
S
Suppose that T = i Ti with Ti locally closed in X. Take A ⊂ T a locally closed
nonempty subset in T , then there exists a Ti such that A ∩ Ti is nonempty, it is
locally closed in Ti and so in X. As X is Jacobson A has a point closed in X. □
07JU Lemma 18.6. A finite Jacobson space is discrete. A Jacobson space with finitely
many closed points is discrete.
Proof. If X is finite then the set X0 ⊂ X of closed points is finite. Assume X0
is is finite and X is S Jacobson. Then X0 = X by the Jacobson property. Now
X0 = {x1 , . . . , xn } = i=1,...,n {xi } is a finite union of closed sets, hence closed, so
X = X0 = X0 . Every point is closed, and by finiteness, every point is open. □
005Z Lemma 18.7. Suppose X is a Jacobson topological space. Let X0 be the set of
closed points of X. There is a bijective, inclusion preserving correspondence
{finite unions loc. closed subsets of X} ↔ {finite unions loc. closed subsets of X0 }
given by E 7→ E ∩ X0 . This correspondence preserves the subsets of locally closed,
of open and of closed subsets.
Proof. We just prove that the correspondence E 7→ E ∩ X0 is injective. Indeed if
E ̸= E ′ then without loss of generality E \ E ′ is nonempty, and it is a finite union of
locally closed sets (details omitted). As X is Jacobson, we see that (E \ E ′ ) ∩ X0 =
E ∩ X0 \ E ′ ∩ X0 is not empty. □
005Y Lemma 18.8. Suppose X is a Jacobson topological space. Let X0 be the set of
closed points of X. There is a bijective, inclusion preserving correspondence
{constructible subsets of X} ↔ {constructible subsets of X0 }
given by E 7→ E ∩ X0 . This correspondence preserves the subset of retrocompact
open subsets, as well as complements of these.
Proof. From Lemma 18.7 above, we just have to see that if U is open in X then
U ∩ X0 is retrocompact in X0 if and only if U is retrocompact in X. This follows
if we prove that for U open in X then U ∩ X0 is quasi-compact if and only if U
is quasi-compact. From Lemma 18.5 it follows that we may replace X by U and
assume that U = X. Finally notice that any collection of opens U of X cover X S if
and only if they cover X0 , using the Jacobson property of X in the closed X \ U
to find a point in X0 if it were nonempty. □

19. Specialization
0060
0061 Definition 19.1. Let X be a topological space.
(1) If x, x′ ∈ X then we say x is a specialization of x′ , or x′ is a generalization
of x if x ∈ {x′ }. Notation: x′ ⇝ x.
(2) A subset T ⊂ X is stable under specialization if for all x′ ∈ T and every
specialization x′ ⇝ x we have x ∈ T .
(3) A subset T ⊂ X is stable under generalization if for all x ∈ T and every
generalization x′ ⇝ x we have x′ ∈ T .
0062 Lemma 19.2. Let X be a topological space.
 TOPOLOGY 37

(1) Any closed subset of X is stable under specialization.

(2) Any open subset of X is stable under generalization.
(3) A subset T ⊂ X is stable under specialization if and only if the complement
T c is stable under generalization.

Proof. Let F be a closed subset of X, if y ∈ F then {y} ⊂ F , so {y} ⊂ F = F as

F is closed. Thus for all specialization x of y, we have x ∈ F .
Let x, y ∈ X such that x ∈ {y} and let T be a subset of X. Saying that T is stable
under specialization means that y ∈ T implies x ∈ T and reciprocally saying that
T is stable under generalization means that x ∈ T implies y ∈ T . Therefore (3) is
proven using contraposition.
The second property follows from (1) and (3) by considering the complement. □

0EES Lemma 19.3. Let T ⊂ X be a subset of a topological space X. The following are
equivalent
(1) T is stable under specialization, and
(2) T is a (directed) union of closed subsets of X.
Proof. Suppose thatST is stable under specialization, then for all y ∈ T we have
{y} ⊂ T . Thus T = y∈T {y} which is an union of closed subsets of X. Recipro-
S
cally, suppose that T = i∈I Fi where Fi are closed subsets of X. If y ∈ T then
there exists i ∈ I such that y ∈ Fi . As Fi is closed, we have {y} ⊂ Fi ⊂ T , which
proves that T is stable under specialization. □

0063 Definition 19.4. Let f : X → Y be a continuous map of topological spaces.

(1) We say that specializations lift along f or that f is specializing if given
y ′ ⇝ y in Y and any x′ ∈ X with f (x′ ) = y ′ there exists a specialization
x′ ⇝ x of x′ in X such that f (x) = y.
(2) We say that generalizations lift along f or that f is generalizing if given
y ′ ⇝ y in Y and any x ∈ X with f (x) = y there exists a generalization
x′ ⇝ x of x in X such that f (x′ ) = y ′ .
0064 Lemma 19.5. Suppose f : X → Y and g : Y → Z are continuous maps of
topological spaces. If specializations lift along both f and g then specializations lift
along g ◦ f . Similarly for “generalizations lift along”.
Proof. Let z ′ ⇝ z be a specialization in Z and let x′ ∈ X such as g ◦ f (x′ ) = z ′ .
Then because specializations lift along g, there exists a specialization f (x′ ) ⇝ y
of f (x′ ) in Y such that g(y) = z. Likewise, because specializations lift along f ,
there exists a specialization x′ ⇝ x of x′ in X such that f (x) = y. It provides a
specialization x′ ⇝ x of x′ in X such that g◦f (x) = z. In other words, specialization
lift along g ◦ f . □

0065 Lemma 19.6. Let f : X → Y be a continuous map of topological spaces.

(1) If specializations lift along f , and if T ⊂ X is stable under specialization,
then f (T ) ⊂ Y is stable under specialization.
(2) If generalizations lift along f , and if T ⊂ X is stable under generalization,
then f (T ) ⊂ Y is stable under generalization.
 TOPOLOGY 38

Proof. Let y ′ ⇝ y be a specialization in Y where y ′ ∈ f (T ) and let x′ ∈ T such

that f (x′ ) = y ′ . Because specialization lift along f , there exists a specialization
x′ ⇝ x of x′ in X such that f (x) = y. But T is stable under specialization so x ∈ T
and then y ∈ f (T ). Therefore f (T ) is stable under specialization.
The proof of (2) is identical, using that generalizations lift along f . □

0066 Lemma 19.7. Let f : X → Y be a continuous map of topological spaces.

(1) If f is closed then specializations lift along f .
(2) If f is open, X is a Noetherian topological space, each irreducible closed
subset of X has a generic point, and Y is Kolmogorov then generalizations
lift along f .
Proof. Assume f is closed. Let y ′ ⇝ y in Y and any x′ ∈ X with f (x′ ) = y ′
be given. Consider the closed subset T = {x′ } of X. Then f (T ) ⊂ Y is a closed
subset, and y ′ ∈ f (T ). Hence also y ∈ f (T ). Hence y = f (x) with x ∈ T , i.e.,
x′ ⇝ x.
Assume f is open, X Noetherian, every irreducible closed subset of X has a generic
point, and Y is Kolmogorov. Let y ′ ⇝ y in Y and any x ∈ X with f (x) = y be
given. Consider T = f −1 ({y ′ }) ⊂ X. Take an open neighbourhood x ∈ U ⊂ X of
x. Then f (U ) ⊂ Y is open and y ∈ f (U ). Hence also y ′ ∈ f (U ). In other words,
T ∩U ̸= ∅. This proves that x ∈ T . Since X is Noetherian, T is Noetherian (Lemma
9.2). Hence it has a decomposition T = T1 ∪ . . . ∪ Tn into irreducible components.
Then correspondingly T = T1 ∪ . . . ∪ Tn . By the above x ∈ Ti for some i. By
assumption there exists a generic point x′ ∈ Ti , and we see that x′ ⇝ x. As x′ ∈ T
we see that f (x′ ) ∈ {y ′ }. Note that f (Ti ) = f ({x′ }) ⊂ {f (x′ )}. If f (x′ ) ̸= y ′ , then
since Y is Kolmogorov f (x′ ) is not a generic point of the irreducible closed subset
{y ′ } and the inclusion {f (x′ )} ⊂ {y ′ } is strict, i.e., y ′ ̸∈ f (Ti ). This contradicts the
fact that f (Ti ) = {y ′ }. Hence f (x′ ) = y ′ and we win. □

06NA Lemma 19.8. Suppose that s, t : R → U and π : U → X are continuous maps of

topological spaces such that
(1) π is open,
(2) U is sober,
(3) s, t have finite fibres,
(4) generalizations lift along s, t,
(5) (t, s)(R) ⊂ U × U is an equivalence relation on U and X is the quotient of
U by this equivalence relation (as a set).
Then X is Kolmogorov.
Proof. Properties (3) and (5) imply that a point x corresponds to an finite equiv-
alence class {u1 , . . . , un } ⊂ U of the equivalence relation. Suppose that x′ ∈ X is
a second point corresponding to the equivalence class {u′1 , . . . , u′m } ⊂ U . Suppose
that ui ⇝ u′j for some i, j. Then for any r′ ∈ R with s(r′ ) = u′j by (4) we can find
r ⇝ r′ with s(r) = ui . Hence t(r) ⇝ t(r′ ). Since {u′1 , . . . , u′m } = t(s−1 ({u′j })) we
conclude that every element of {u′1 , . . . , u′m } is the specialization of an element of
{u1 , . . . , un }. Thus {u1 } ∪ . . . ∪ {un } is a union of equivalence classes, hence of the
form π −1 (Z) for some subset Z ⊂ X. By (1) we see that Z is closed in X and in
fact Z = {x} because π({ui }) ⊂ {x} for each i. In other words, x ⇝ x′ if and only
 TOPOLOGY 39

if some lift of x in U specializes to some lift of x′ in U , if and only if every lift of x′

in U is a specialization of some lift of x in U .
Suppose that both x ⇝ x′ and x′ ⇝ x. Say x corresponds to {u1 , . . . , un } and
x′ corresponds to {u′1 , . . . , u′m } as above. Then, by the results of the preceding
paragraph, we can find a sequence
. . . ⇝ u′j3 ⇝ ui3 ⇝ u′j2 ⇝ ui2 ⇝ u′j1 ⇝ ui1
which must repeat, hence by (2) we conclude that {u1 , . . . , un } = {u′1 , . . . , u′m },
i.e., x = x′ . Thus X is Kolmogorov. □
02JF Lemma 19.9. Let f : X → Y be a morphism of topological spaces. Suppose
that Y is a sober topological space, and f is surjective. If either specializations or
generalizations lift along f , then dim(X) ≥ dim(Y ).
Proof. Assume specializations lift along f . Let Z0 ⊂ Z1 ⊂ . . . Ze ⊂ Y be a chain of
irreducible closed subsets of X. Let ξe ∈ X be a point mapping to the generic point
of Ze . By assumption there exists a specialization ξe ⇝ ξe−1 in X such that ξe−1
maps to the generic point of Ze−1 . Continuing in this manner we find a sequence
of specializations
ξe ⇝ ξe−1 ⇝ . . . ⇝ ξ0
with ξi mapping to the generic point of Zi . This clearly implies the sequence of
irreducible closed subsets
{ξ0 } ⊂ {ξ1 } ⊂ . . . {ξe }
is a chain of length e in X. The case when generalizations lift along f is similar. □
0542 Lemma 19.10. Let X be a Noetherian sober topological space. Let E ⊂ X be a
subset of X.
(1) If E is constructible and stable under specialization, then E is closed.
(2) If E is constructible and stable under generalization, then E is open.
Proof. Let E be constructible and stable under generalization. Let Y ⊂ X be an
irreducible closed subset with generic point ξ ∈ Y . If E ∩ Y is nonempty, then
it contains ξ (by stability under generalization) and hence is dense in Y , hence it
contains a nonempty open of Y , see Lemma 16.3. Thus E is open by Lemma 16.5.
This proves (2). To prove (1) apply (2) to the complement of E in X. □

20. Dimension functions

02I8 It scarcely makes sense to consider dimension functions unless the space considered
is sober (Definition 8.6). Thus the definition below can be improved by considering
the sober topological space associated to X. Since the underlying topological space
of a scheme is sober we do not bother with this improvement.
02I9 Definition 20.1. Let X be a topological space.
(1) Let x, y ∈ X, x ̸= y. Suppose x ⇝ y, that is y is a specialization of x. We
say y is an immediate specialization of x if there is no z ∈ X \ {x, y} with
x ⇝ z and z ⇝ y.
(2) A map δ : X → Z is called a dimension function5 if
(a) whenever x ⇝ y and x ̸= y we have δ(x) > δ(y), and
5This is likely nonstandard notation. This notion is usually introduced only for (locally)
Noetherian schemes, in which case condition (a) is implied by (b).
 TOPOLOGY 40

(b) for every immediate specialization x ⇝ y in X we have δ(x) = δ(y)+1.

It is clear that if δ is a dimension function, then so is δ + t for any t ∈ Z. Here is a
fun lemma.
02IA Lemma 20.2. Let X be a topological space. If X is sober and has a dimension
function, then X is catenary. Moreover, for any x ⇝ y we have
 
δ(x) − δ(y) = codim {y}, {x} .

Proof. Suppose Y ⊂ Y ′ ⊂ X are irreducible closed subsets. Let ξ ∈ Y , ξ ′ ∈

Y ′ be their generic points. Then we see immediately from the definitions that
codim(Y, Y ′ ) ≤ δ(ξ) − δ(ξ ′ ) < ∞. In fact the first inequality is an equality. Namely,
suppose
Y = Y0 ⊂ Y1 ⊂ . . . ⊂ Ye = Y ′
is any maximal chain of irreducible closed subsets. Let ξi ∈ Yi denote the generic
point. Then we see that ξi ⇝ ξi+1 is an immediate specialization. Hence we see that
e = δ(ξ) − δ(ξ ′ ) as desired. This also proves the last statement of the lemma. □

02IB Lemma 20.3. Let X be a topological space. Let δ, δ ′ be two dimension functions
on X. If X is locally Noetherian and sober then δ − δ ′ is locally constant on X.
Proof. Let x ∈ X be a point. We will show that δ − δ ′ is constant in a neigh-
bourhood of x. We may replace X by an open neighbourhood of x in X which is
Noetherian. Hence we may assume X is Noetherian and sober. Let Z1 , . . . , Zr be
the irreducible components of X passing through x. (There are finitely many as X
is Noetherian, see Lemma 9.2.) Let ξi ∈ Zi be the generic point. Note Z1 ∪ . . . ∪ Zr
is a neighbourhood of x in X (not necessarily closed). We claim that δ − δ ′ is
constant on Z1 ∪ . . . ∪ Zr . Namely, if y ∈ Zi , then
δ(x) − δ(y) = δ(x) − δ(ξi ) + δ(ξi ) − δ(y) = −codim({x}, Zi ) + codim({y}, Zi )
by Lemma 20.2. Similarly for δ ′ . Whence the result. □

02IC Lemma 20.4. Let X be locally Noetherian, sober and catenary. Then any point
has an open neighbourhood U ⊂ X which has a dimension function.
Proof. We will use repeatedly that an open subspace of a catenary space is cate-
nary, see Lemma 11.5 and that a Noetherian topological space has finitely many
irreducible components, see Lemma 9.2. In the proof of Lemma 20.3 we saw how
to construct such a function. Namely, we first replace X by a Noetherian open
neighbourhood of x. Next, we let Z1 , . . . , Zr ⊂ X be the irreducible components of
X. Let [
Zi ∩ Zj = Zijk
be the decomposition into irreducible components. We replace X by
[ [ 
X\ Zi ∪ Zijk
x̸∈Zi x̸∈Zijk

so that we may assume x ∈ Zi for all i and x ∈ Zijk for all i, j, k. For y ∈ X choose
any i such that y ∈ Zi and set
δ(y) = −codim({x}, Zi ) + codim({y}, Zi ).
 TOPOLOGY 41

We claim this is a dimension function. First we show that it is well defined, i.e.,
independent of the choice of i. Namely, suppose that y ∈ Zijk for some i, j, k. Then
we have (using Lemma 11.6)
δ(y) = −codim({x}, Zi ) + codim({y}, Zi )
= −codim({x}, Zijk ) − codim(Zijk , Zi ) + codim({y}, Zijk ) + codim(Zijk , Zi )
= −codim({x}, Zijk ) + codim({y}, Zijk )
which is symmetric in i and j. We omit the proof that it is a dimension function. □
02ID Remark 20.5. Combining Lemmas 20.3 and 20.4 we see that on a catenary,
locally Noetherian, sober topological space the obstruction to having a dimension
function is an element of H 1 (X, Z).

21. Nowhere dense sets

03HM
03HN Definition 21.1. Let X be a topological space.
(1) Given a subset T ⊂ X the interior of T is the largest open subset of X
contained in T .
(2) A subset T ⊂ X is called nowhere dense if the closure of T has empty
interior.
03HO Lemma 21.2. Let X be a topological space. The union of a finite number of
nowhere dense sets is a nowhere dense set.
Proof. It suffices to prove the lemma for two nowhere dense sets as the result in
general will follow by induction. Let A, B ⊂ X be nowhere dense subsets. We have
A ∪ B = A ∪ B. Hence, if U ⊂ A ∪ B is an open subset of X, then U \ U ∩ B
is an open subset of U and of X and contained in A and hence empty. Similarly
U \ U ∩ A is empty. Thus U = ∅ as desired. □
03J0 Lemma 21.3. Let X be a topological space. Let U ⊂ X be an open. Let T ⊂ U
be a subset. If T is nowhere dense in U , then T is nowhere dense in X.
Proof. Assume T is nowhere dense in U . Suppose that x ∈ X is an interior point
of the closure T of T in X. Say x ∈ V ⊂ T with V ⊂ X open in X. Note that
T ∩ U is the closure of T in U . Hence the interior of T ∩ U being empty implies
V ∩ U = ∅. Thus x cannot be in the closure of U , a fortiori cannot be in the closure
of T , a contradiction. □
S
03HP Lemma 21.4. Let X be a topological space. Let X = Ui be an open covering.
Let T ⊂ X be a subset. If T ∩ Ui is nowhere dense in Ui for all i, then T is nowhere
dense in X.
Proof. Denote T i the closure of T ∩ Ui in Ui . We have T ∩ Ui = T i . Taking the
interior commutes with intersection with opens, thus
(interior of T ) ∩ Ui = interior of (T ∩ Ui ) = interior in Ui of T i
By assumption the last of these is empty. Hence T is nowhere dense in X. □
03HQ Lemma 21.5. Let f : X → Y be a continuous map of topological spaces. Let
T ⊂ X be a subset. If f is a homeomorphism of X onto a closed subset of Y and
T is nowhere dense in X, then also f (T ) is nowhere dense in Y .
 TOPOLOGY 42

Proof. Because f (X) is closed in Y and f is a homeomorphism of X onto f (X),

we see that the closure of f (T ) in Y equals f (T ). Hence if V ⊂ Y is open and
contained in the closure of f (T ), then U = f −1 (V ) is open and contained in T .
Hence U = ∅, which in turn shows that V = ∅ as desired. □

03HR Lemma 21.6. Let f : X → Y be a continuous map of topological spaces. Let

T ⊂ Y be a subset. If f is open and T is a closed nowhere dense subset of Y , then
also f −1 (T ) is a closed nowhere dense subset of X. If f is surjective and open,
then T is closed nowhere dense if and only if f −1 (T ) is closed nowhere dense.
Proof. Omitted. (Hint: In the first case the interior of f −1 (T ) maps into the
interior of T , and in the second case the interior of f −1 (T ) maps onto the interior
of T .) □

22. Profinite spaces

08ZW Here is the definition.
08ZX Definition 22.1. A topological space is profinite if it is homeomorphic to a limit
of a diagram of finite discrete spaces.
This is not the most convenient characterization of a profinite space.
08ZY Lemma 22.2. Let X be a topological space. The following are equivalent
(1) X is a profinite space, and
(2) X is Hausdorff, quasi-compact, and totally disconnected.
If this is true, then X is a cofiltered limit of finite discrete spaces.
Proof. Assume (1). Choose a diagram i 7→ Xi of finite discrete spaces such that
X = lim Xi . As each Xi is Hausdorff and quasi-compact we find that X is quasi-
compact by Lemma 14.5. If x, x′ ∈ X are distinct points, then x and x′ map to
distinct points in some Xi . Hence x and x′ have disjoint open neighbourhoods, i.e.,
X is Hausdorff. In exactly the same way we see that X is totally disconnected.
`
Assume (2). Let I be the set of finite disjoint union decompositions X = i∈I Ui
with Ui nonempty open (and closed) for all i ∈ I. For each I ∈ I there is a
continuous map X → I sending a point of Ui to i. We define a partial ordering:
I ≤ I ′ for I, I ′ ∈ I if and only if the covering corresponding to I ′ refines the
covering corresponding to I. In this case we obtain a canonical map I ′ → I. In
other words we obtain an inverse system of finite discrete spaces over I. The maps
X → I fit together and we obtain a continuous map
X −→ limI∈I I
We claim this map is a homeomorphism, which finishes the proof. (The final as-
sertion follows too as the partially ordered set I is directed: given two disjoint
union decompositions of X we can find a third refining both.) Namely, the map is
injective as X is totally disconnected and hence {x} is the intersection of all open
and closed subsets of X containing x (Lemma 12.11) and the map is surjective by
Lemma 12.6. By Lemma 17.8 the map is a homeomorphism. □

0ET8 Lemma 22.3. A limit of profinite spaces is profinite.

 TOPOLOGY 43

Proof. Let i 7→ Xi be a diagram of profinite spaces over the index category I. Let
us use the characterization of profinite spaces in Lemma 22.2. In particular each Xi
is Hausdorff, quasi-compact, and totally disconnected. By Lemma 14.1 the limit
X = lim Xi exists. By Lemma 14.5 the limit X is quasi-compact. Let x, x′ ∈ X
be distinct points. Then there exists an i such that x and x′ have distinct images
xi and x′i in Xi under the projection X → Xi . Then xi and x′i have disjoint open
neighbourhoods in Xi . Taking the inverse images of these opens we conclude that
X is Hausdorff. Similarly, xi and x′i are in distinct connected components of Xi
whence necessarily x and x′ must be in distinct connected components of X. Hence
X is totally disconnected. This finishes the proof. □
08ZZ Lemma 22.4. Let X be a profinite
` space. Every open covering of X has a refine-
ment by a finite covering X = Ui with Ui open and closed.
Proof. Write X = lim Xi as a limit of an inverse system of finite discrete spaces
over a directed set I (Lemma 22.2). Denote fi : X → Xi the projection. For every
point x = (xi ) ∈ X a fundamental system of open neighbourhoods is the collection
fi−1 ({xi }). Thus, as X is quasi-compact, we may assume we have an open covering
X = fi−1
1
({xi1 }) ∪ . . . ∪ fi−1
n
({xin })
Choose i ∈ I with i ≥ ij for j = 1, . . . , n (this is possible as I is a directed set).
Then we see that the covering
a
X= fi−1 ({t})
t∈Xi
refines the given covering and is of the desired form. □
0900 Lemma 22.5. Let X be a topological space. If X is quasi-compact and every con-
nected component of X is the intersection of the open and closed subsets containing
it, then π0 (X) is a profinite space.
Proof. We will use Lemma 22.2 to prove this. Since π0 (X) is the image of a
quasi-compact space it is quasi-compact (Lemma 12.7). It is totally disconnected
by construction (Lemma
T 7.9). Let C, D ⊂ X be distinct connected components
of X. Write C = Uα as the intersection of the open and closed T subsets of X
containing C. Any finite intersection of Uα ’s is another. Since Uα ∩ D = ∅ we
conclude that Uα ∩ D = ∅ for some α (use Lemmas 7.3, 12.3 and 12.6) Since Uα is
open and closed, it is the union of the connected components it contains, i.e., Uα
is the inverse image of some open and closed subset Vα ⊂ π0 (X). This proves that
the points corresponding to C and D are contained in disjoint open subsets, i.e.,
π0 (X) is Hausdorff. □

23. Spectral spaces

08YF The material in this section is taken from [Hoc69] and [Hoc67]. In his thesis
Hochster proves (among other things) that the spectral spaces are exactly the topo-
logical spaces that occur as the spectrum of a ring.
08YG Definition 23.1. A topological space X is called spectral if it is sober, quasi-
compact, the intersection of two quasi-compact opens is quasi-compact, and the
collection of quasi-compact opens forms a basis for the topology. A continuous
map f : X → Y of spectral spaces is called spectral if the inverse image of a
quasi-compact open is quasi-compact.
 TOPOLOGY 44

In other words a continuous map of spectral spaces is spectral if and only if it is

quasi-compact (Definition 12.1).
Let X be a spectral space. The constructible topology on X is the topology which
has as a subbase of opens the sets U and U c where U is a quasi-compact open of
X. Note that since X is spectral an open U ⊂ X is retrocompact if and only if
U is quasi-compact. Hence the constructible topology can also be characterized as
the coarsest topology such that every constructible subset of X is both open and
closed (see Section 15 for definitions and properties of constructible sets). It follows
that a subset of X is open, resp. closed in the constructible topology if and only
if it is a union, resp. intersection of constructible subsets. Since the collection of
quasi-compact opens is a basis for the topology on X we see that the constructible
topology is stronger than the given topology on X.
0901 Lemma 23.2. Let X be a spectral space. The constructible topology is Hausdorff,
totally disconnected, and quasi-compact.
Proof. Let x, y ∈ X with x ̸= y. Since X is sober, there is an open subset U
containing exactly one of the two points x, y. Say x ∈ U . We may replace U by a
quasi-compact open neighbourhood of x contained in U . Then U and U c are open
and closed in the constructible topology. Hence X is Hausdorff in the constructible
topology because x ∈ U and y ∈ U c are disjoint opens in the constructible topology.
The existence of U also implies x and y are in distinct connected components in
the constructible topology, whence X is totally disconnected in the constructible
topology.
Let B be the collection of subsets B ⊂ X with B either quasi-compact open or
closed with quasi-compactScomplement. If B ∈ B then B c ∈ B. It suffices to
show every covering X = i∈I Bi with Bi ∈ B has a finite refinement, see Lemma
12.15. Taking complements we see that we have to show that any family {Bi }i∈I
of elements of B such that Bi1 ∩ . . . ∩ Bin ̸= ∅ for all n and all i1 , . . . , in ∈ I has a
common point of intersection. We may and do assume Bi ̸= Bi′ for i ̸= i′ .
To get a contradiction assume {Bi }i∈I is a family of elements of B having the finite
intersection property but empty intersection. An application of Zorn’s lemma shows
that we may assume our family is maximal T (details omitted). Let I ′ ⊂ I be those
indices such that Bi is closed and set Z = i∈I ′ Bi . This is a closed subset of X
which is nonempty by Lemma 12.6. If Z is reducible, then we can write Z = Z ′ ∪Z ′′
as a union of two closed subsets, neither equal to Z. This means in particular that
we can find a quasi-compact open U ′ ⊂ X meeting Z ′ but not Z ′′ . Similarly, we
can find a quasi-compact open U ′′ ⊂ X meeting Z ′′ but not Z ′ . Set B ′ = X \ U ′
and B ′′ = X \ U ′′ . Note that Z ′′ ⊂ B ′ and Z ′ ⊂ B ′′ . If there exist a finite
number of indices i1 , . . . , in ∈ I such that B ′ ∩ Bi1 ∩ . . . ∩ Bin = ∅ as well as a finite
number of indices j1 , . . . , jm ∈ I such that B ′′ ∩Bj1 ∩. . .∩Bjm = ∅ then we find that
Z∩Bi1 ∩. . .∩Bin ∩Bj1 ∩. . .∩Bjm = ∅. However, the set Bi1 ∩. . .∩Bin ∩Bj1 ∩. . .∩Bjm
is quasi-compact hence we would find a finite number of indices i′1 , . . . , i′l ∈ I ′ with
Bi1 ∩ . . . ∩ Bin ∩ Bj1 ∩ . . . ∩ Bjm ∩ Bi′1 ∩ . . . ∩ Bi′l = ∅, a contradiction. Thus we
see that we may add either B ′ or B ′′ to the given family contradicting maximality.
We conclude that Z is irreducible. However, this leads to a contradiction as well,
as now every nonempty (by the same argument as above) open Z ∩ Bi for i ∈ I \ I ′
contains the unique generic point of Z. This contradiction proves the lemma. □
 TOPOLOGY 45

0A2S Lemma 23.3. Let f : X → Y be a spectral map of spectral spaces. Then

(1) f is continuous in the constructible topology,
(2) the fibres of f are quasi-compact, and
(3) the image is closed in the constructible topology.
Proof. Let X ′ and Y ′ denote X and Y endowed with the constructible topology
which are quasi-compact Hausdorff spaces by Lemma 23.2. Part (1) says X ′ → Y ′
is continuous and follows immediately from the definitions. Part (3) follows as
f (X ′ ) is a quasi-compact subset of the Hausdorff space Y ′ , see Lemma 12.4. We
have a commutative diagram
X′ /X

 
Y′ /Y
of continuous maps of topological spaces. Since Y ′ is Hausdorff we see that the
fibres Xy′ are closed in X ′ . As X ′ is quasi-compact we see that Xy′ is quasi-compact
(Lemma 12.3). As Xy′ → Xy is a surjective continuous map we conclude that Xy
is quasi-compact (Lemma 12.7). □

0G1J Lemma 23.4. Let X and Y be spectral spaces. Let f : X → Y be a continuous

map. Then f is spectral if and only if f is continuous in the constructible topology.
Proof. The only if part of this is Lemma 23.3. Assume f is continuous in the
constructible topology. Let V ⊂ Y be quasi-compact open. Then V is open and
closed in the constructible topology. Hence f −1 (V ) is open and closed in the con-
structible topology. Hence f −1 (V ) is quasi-compact in the constructible topology
as X is quasi-compact in the constructible topology by Lemma 23.2. Since the iden-
tity f −1 (V ) → f −1 (V ) is surjective and continuous from the constructible topology
to the usual topology, we conclude that f −1 (V ) is quasi-compact in the topology
of X by Lemma 12.7. This finishes the proof. □

0902 Lemma 23.5. Let X be a spectral space. Let E ⊂ X be closed in the constructible
topology (for example constructible or closed). Then E with the induced topology is
a spectral space.
Proof. Let Z ⊂ E be a closed irreducible subset. Let η be the generic point of the
closure Z of Z in X. To prove that E is sober, we show that η ∈ E. If not, then
since E is closed in the constructible topology, there exists a constructible subset
F ⊂ X such that η ∈ F and F ∩ E = ∅. By Lemma 15.15 this implies F ∩ Z
contains a nonempty open subset of Z. But this is impossible as Z is the closure
of Z and Z ∩ F = ∅.
Since E is closed in the constructible topology, it is quasi-compact in the con-
structible topology (Lemmas 12.3 and 23.2). Hence a fortiori it is quasi-compact
in the topology coming from X. If U ⊂ X is a quasi-compact open, then E ∩ U
is closed in the constructible topology, hence quasi-compact (as seen above). It
follows that the quasi-compact open subsets of E are the intersections E ∩ U with
U quasi-compact open in X. These form a basis for the topology. Finally, given two
U, U ′ ⊂ X quasi-compact opens, the intersection (E ∩ U ) ∩ (E ∩ U ′ ) = E ∩ (U ∩ U ′ )
and U ∩ U ′ is quasi-compact as X is spectral. This finishes the proof. □
 TOPOLOGY 46

0903 Lemma 23.6. Let X be a spectral space. Let E ⊂ X be a subset closed in the
constructible topology (for example constructible).
(1) If x ∈ E, then x is the specialization of a point of E.
(2) If E is stable under specialization, then E is closed.
(3) If E ′ ⊂ X is open in the constructible topology (for example constructible)
and stable under generalization, then E ′ is open.
Proof. Proof of (1). Let x ∈ E. Let {Ui } be the set of quasi-compact open
neighbourhoods of x. A finite intersection of the Ui is another one. The intersection
Ui ∩ E is nonempty for T all i. Since the subsets Ui ∩ E are closed in the constructible
topology we see that (Ui ∩ E) is nonempty by Lemma 23.2 and Lemma 12.6. T
Since {Ui } is a fundamental system of open neighbourhoods of x, we see that Ui
is the set of generalizations of x. Thus x is a specialization of a point of E.
Part (2) is immediate from (1).
Proof of (3). Assume E ′ is as in (3). The complement of E ′ is closed in the
constructible topology (Lemma 15.2) and closed under specialization (Lemma 19.2).
Hence the complement is closed by (2), i.e., E ′ is open. □
0904 Lemma 23.7. Let X be a spectral space. Let x, y ∈ X. Then either there exists a
third point specializing to both x and y, or there exist disjoint open neighbourhoods
containing x and y.
Proof. Let {Ui } be the set of quasi-compact open neighbourhoods of x. A finite
intersection of the Ui is another one. Let {Vj } be the set of quasi-compact open
neighbourhoods of y. A finite intersection of the Vj is another one. If Ui ∩ Vj is
empty for some i, j we are done. If not, then the intersection Ui ∩ Vj is nonempty
Ti ∩ Vj are closed in the constructible topology on X. By
for all i and j. The sets U
Lemma 23.2 we see that (Ui ∩ Vj ) is nonempty (Lemma 12.6). Since X is a sober
space
T and {Ui } is a fundamental system of open neighbourhoods
T of x, we see that
Ui is the set of generalizations
T of x. Similarly, V j is the set of generalizations
of y. Thus any element of (Ui ∩ Vj ) specializes to both x and y. □
0905 Lemma 23.8. Let X be a spectral space. The following are equivalent:
(1) X is profinite,
(2) X is Hausdorff,
(3) X is totally disconnected,
(4) every quasi-compact open is closed,
(5) there are no nontrivial specializations between points,
(6) every point of X is closed,
(7) every point of X is the generic point of an irreducible component of X,
(8) the constructible topology equals the given topology on X, and
(9) add more here.
Proof. Lemma 22.2 shows the implication (1) ⇒ (3). Irreducible components are
closed, so if X is totally disconnected, then every point is closed. So (3) implies
(6). The equivalence of (6) and (5) is immediate, and (6) ⇔ (7) holds because
X is sober. Assume (5). Then all constructible subsets of X are closed (Lemma
23.6), in particular all quasi-compact opens are closed. So (5) implies (4). Since X
is sober, for any two points there is a quasi-compact open containing exactly one
of them, hence (4) implies (2). Parts (4) and (8) are equivalent by the definition
 TOPOLOGY 47

of the constructible topology. It remains to prove (2) implies (1). Suppose X is

Hausdorff. Every quasi-compact open is also closed (Lemma 12.4). This implies X
is totally disconnected. Hence it is profinite, by Lemma 22.2. □
0906 Lemma 23.9. If X is a spectral space, then π0 (X) is a profinite space.
Proof. Combine Lemmas 12.10 and 22.5. □
0907 Lemma 23.10. The product of two spectral spaces is spectral.
Proof. Let X, Y be spectral spaces. Denote p : X × Y → X and q : X × Y → Y
the projections. Let Z ⊂ X × Y be a closed irreducible subset. Then p(Z) ⊂ X
is irreducible and q(Z) ⊂ Y is irreducible. Let x ∈ X be the generic point of
the closure of p(X) and let y ∈ Y be the generic point of the closure of q(Y ). If
(x, y) ̸∈ Z, then there exist opens x ∈ U ⊂ X, y ∈ V ⊂ Y such that Z ∩ U × V = ∅.
Hence Z is contained in (X \ U ) × Y ∪ X × (Y \ V ). Since Z is irreducible, we see
that either Z ⊂ (X \ U ) × Y or Z ⊂ X × (Y \ V ). In the first case p(Z) ⊂ (X \ U )
and in the second case q(Z) ⊂ (Y \ V ). Both cases are absurd as x is in the closure
of p(Z) and y is in the closure of q(Z). Thus we conclude that (x, y) ∈ Z, which
means that (x, y) is the generic point for Z.
A basis of the topology of X × Y are the opens of the form U × V with U ⊂ X
and V ⊂ Y quasi-compact open (here we use that X and Y are spectral). Then
U × V is quasi-compact as the product of quasi-compact spaces is quasi-compact.
Moreover, any quasi-compact open of X × Y is a finite union of such quasi-compact
rectangles U ×V . It follows that the intersection of two such is again quasi-compact
(since X and Y are spectral). This concludes the proof. □
09XU Lemma 23.11. Let f : X → Y be a continuous map of topological spaces. If
(1) X and Y are spectral,
(2) f is spectral and bijective, and
(3) generalizations (resp. specializations) lift along f .
Then f is a homeomorphism.
Proof. Since f is spectral it defines a continuous map between X and Y in the
constructible topology. By Lemmas 23.2 and 17.8 it follows that X → Y is a
homeomorphism in the constructible topology. Let U ⊂ X be quasi-compact open.
Then f (U ) is constructible in Y . Let y ∈ Y specialize to a point in f (U ). By the
last assumption we see that f −1 (y) specializes to a point of U . Hence f −1 (y) ∈ U .
Thus y ∈ f (U ). It follows that f (U ) is open, see Lemma 23.6. Whence f is a
homeomorphism. To prove the lemma in case specializations lift along f one shows
instead that f (Z) is closed if X \ Z is a quasi-compact open of X. □
09XV Lemma 23.12. The inverse limit of a directed inverse system of finite sober
topological spaces is a spectral topological space.
Proof. Let I be a directed set. Let Xi be an inverse system of finite sober spaces
over I. Let X = lim Xi which exists by Lemma 14.1. As a set X = lim Xi . Denote
pi : X → Xi the projection. Because I is directed we may apply Lemma 14.2. A
basis for the topology is given by the opens p−1 i (Ui ) for Ui ⊂ Xi open. Since an
open covering of p−1 i (Ui ) is in particular an open covering in the profinite topology,
we conclude that p−1 i (Ui ) is quasi-compact. Given Ui ⊂ Xi and Uj ⊂ Xj , then
p−1
i (U i ) ∩ p−1
j (Uj ) = p−1
k (Uk ) for some k ≥ i, j and open Uk ⊂ Xk . Finally, if
 TOPOLOGY 48

Z ⊂ X is irreducible and closed, then pi (Z) ⊂ Xi is irreducible and therefore has

a unique generic point ξi (because Xi is a finite sober topological space). Then
ξ = lim ξi is a generic point of Z (it is a point of Z as Z is closed). This finishes
the proof. □
09XW Lemma 23.13. Let W be the topological space with two points, one closed, the
other not. A topological space is spectral if and only if it is homeomorphic to a
subspace of a product of copies of W which is closed in the constructible topology.
Proof. Write W = {0, 1}
Q where 0 is a specialization of 1 but not vice versa. Let
I be a set. The
Q space i∈I W is spectral by Lemma 23.12. Thus we see that
a subspace of i∈I W closed in the constructible topology is a spectral space by
Lemma 23.5.
For the converse, let X be a spectral space. Let U ⊂ X be a quasi-compact open.
Consider the continuous map
fU : X −→ W
which maps every point in U to 1 and every point in X \U to 0. Taking the product
of these maps we obtain a continuous map
Y Y
f= fU : X −→ W
U
By construction the map f : X → Y is spectral. By Lemma 23.3 the image of f is
closed in the constructible topology. If x′ , x ∈ X are distinct, then since X is sober
either x′ is not a specialization of x or conversely. In either case (as the quasi-
compact opens form a basis for the topology of X) there exists a quasi-compact
open U ⊂ X such that fU (x′ ) ̸= fU (x). Thus f is injective. Let Y = f (X) endowed
with the induced topology. Let y ′ ⇝ y be a specialization in Y and say f (x′ ) = y ′
and f (x) = y. Arguing as above we see that x′ ⇝ x, since otherwise there is a U
such that x ∈ U and x′ ̸∈ U , which would imply fU (x′ ) ̸⇝ fU (x). We conclude
that f : X → Y is a homeomorphism by Lemma 23.11. □
09XX Lemma 23.14. A topological space is spectral if and only if it is a directed inverse
limit of finite sober topological spaces.
Proof. One direction is given by Lemma
Q 23.12. For the converse, assume X is
spectral. Then we may assume X ⊂ i∈I W is a subset closed in the constructible
topology where W = {0, 1} as in Lemma 23.13. We can write
Y Y
W = limJ⊂I finite W
i∈I j∈J
Q
as a cofiltered limit. For each J, let XJ ⊂ j∈J W be the image of X. Then
we see that X = lim XJ as sets because X is closed in the product with the
constructible topology (detail omitted). A formal argument (omitted) on limits
shows that X = lim XJ as topological spaces. □
0A2T Lemma 23.15. Let X be a topological space and let c : X → X ′ be the universal
map from X to a sober topological space, see Lemma 8.16.
(1) If X is quasi-compact, so is X ′ .
(2) If X is quasi-compact, has a basis of quasi-compact opens, and the inter-
section of two quasi-compact opens is quasi-compact, then X ′ is spectral.
(3) If X is Noetherian, then X ′ is a Noetherian spectral space.
 TOPOLOGY 49

Proof. Let U ⊂ X be open and let U ′ ⊂ X ′ be the corresponding open, i.e., the
open such that c−1 (U ′ ) = U . Then U is quasi-compact if and only if U ′ is quasi-
compact, as pulling back by c is a bijection between the opens of X and X ′ which
commutes with unions. This in particular proves (1).
Proof of (2). It follows from the above that X ′ has a basis of quasi-compact opens.
Since c−1 also commutes with intersections of pairs of opens, we see that the in-
tersection of two quasi-compact opens X ′ is quasi-compact. Finally, X ′ is quasi-
compact by (1) and sober by construction. Hence X ′ is spectral.
Proof of (3). It is immediate that X ′ is Noetherian as this is defined in terms of
the acc for open subsets which holds for X. We have already seen in (2) that X ′ is
spectral. □

24. Limits of spectral spaces

0A2U Lemma 23.14 tells us that every spectral space is a cofiltered limit of finite sober
spaces. Every finite sober space is a spectral space and every continuous map of
finite sober spaces is a spectral map of spectral spaces. In this section we prove some
lemmas concerning limits of systems of spectral topological spaces along spectral
maps.
0A2V Lemma 24.1. Let I be a category. Let i 7→ Xi be a diagram of spectral spaces
such that for a : j → i in I the corresponding map fa : Xj → Xi is spectral.
(1) Given subsets Zi ⊂ Xi closed in the constructible topology with fa (Zj ) ⊂ Zi
for all a : j → i in I, then lim Zi is quasi-compact.
(2) The space X = lim Xi is quasi-compact.
Proof. The limit Z = lim Zi exists by Lemma 14.1. Denote Xi′ the space Xi
endowed with the constructible topology and Zi′ the corresponding subspace of
Xi′ . Let a : j → i in I be a morphism. As fa is spectral it defines a continuous
map fa : Xj′ → Xi′ . Thus fa |Zj : Zj′ → Zi′ is a continuous map of quasi-compact
Hausdorff spaces (by Lemmas 23.2 and 12.3). Thus Z ′ = lim Zi is quasi-compact
by Lemma 14.5. The maps Zi′ → Zi are continuous, hence Z ′ → Z is continuous
and a bijection on underlying sets. Hence Z is quasi-compact as the image of the
surjective continuous map Z ′ → Z (Lemma 12.7). □
0A2W Lemma 24.2. Let I be a cofiltered category. Let i 7→ Xi be a diagram of spectral
spaces such that for a : j → i in I the corresponding map fa : Xj → Xi is spectral.
(1) Given nonempty subsets Zi ⊂ Xi closed in the constructible topology with
fa (Zj ) ⊂ Zi for all a : j → i in I, then lim Zi is nonempty.
(2) If each Xi is nonempty, then X = lim Xi is nonempty.
Proof. Denote Xi′ the space Xi endowed with the constructible topology and Zi′
the corresponding subspace of Xi′ . Let a : j → i in I be a morphism. As fa is
spectral it defines a continuous map fa : Xj′ → Xi′ . Thus fa |Zj : Zj′ → Zi′ is a
continuous map of quasi-compact Hausdorff spaces (by Lemmas 23.2 and 12.3).
By Lemma 14.6 the space lim Zi′ is nonempty. Since lim Zi′ = lim Zi as sets we
conclude. □
0A2X Lemma 24.3. Let I be a cofiltered category. Let i 7→ Xi be a diagram of spectral
spaces such that for a : j → i in I the corresponding map fa : Xj → Xi is spectral.
Let X = lim Xi with projections pi : X → Xi . Let i ∈ Ob(I) and let E, F ⊂ Xi be
 TOPOLOGY 50

subsets with E closed in the constructible topology and F open in the constructible
topology. Then p−1 −1
i (E) ⊂ pi (F ) if and only if there is a morphism a : j → i in I
−1 −1
such that fa (E) ⊂ fa (F ).
Proof. Observe that
p−1 −1 −1 −1
i (E) \ pi (F ) = lima:j→i fa (E) \ fa (F )

Since fa is a spectral map, it is continuous in the constructible topology hence the

set fa−1 (E) \ fa−1 (F ) is closed in the constructible topology. Hence Lemma 24.2
applies to show that the LHS is nonempty if and only if each of the spaces of the
RHS is nonempty. □

0A2Y Lemma 24.4. Let I be a cofiltered category. Let i 7→ Xi be a diagram of spectral

spaces such that for a : j → i in I the corresponding map fa : Xj → Xi is spectral.
Let X = lim Xi with projections pi : X → Xi . Let E ⊂ X be a constructible
subset. Then there exists an i ∈ Ob(I) and a constructible subset Ei ⊂ Xi such
that p−1
i (Ei ) = E. If E is open, resp. closed, we may choose Ei open, resp. closed.

Proof. Assume E is a quasi-compact open of X. By Lemma 14.2 S we can write

E = p−1i (Ui ) for some i and some open Ui ⊂ Xi . Write Ui = Ui,α as a union
of quasi-compact opens. As E is quasi-compact we can find α1 , . . . , αn such that
E = p−1
i (Ui,α1 ∪ . . . ∪ Ui,αn ). Hence Ei = Ui,α1 ∪ . . . ∪ Ui,αn works.

Assume E is a constructible closed subset. Then E c is quasi-compact open. So

E c = p−1i (Fi ) for some i and quasi-compact open Fi ⊂ Xi by the result of the
previous paragraph. Then E = p−1 c
i (Fi ) as desired.
S
If E is general we can write E = l=1,...,n Ul ∩ Zl with Ul constructible open and Zl
constructible closed. By the result of the previous paragraphs we may write Ul =
p−1 −1
il (Ul,il ) and Zl = pjl (Zl,jl ) with Ul,il ⊂ Xil constructible open and Zl,jl ⊂ Xjl
constructible closed. As I is cofiltered we may choose
S an object k of I and morphism
−1
al : k → il and bl : k → jl . Then taking Ek = l=1,...,n fa−1 l
(Ul,il
) ∩ f b l
(Zl,jl ) we
obtain a constructible subset of Xk whose inverse image in X is E. □

0A2Z Lemma 24.5. Let I be a cofiltered index category. Let i 7→ Xi be a diagram of

spectral spaces such that for a : j → i in I the corresponding map fa : Xj → Xi is
spectral. Then the inverse limit X = lim Xi is a spectral topological space and the
projection maps pi : X → Xi are spectral.
Proof. The limit X = lim Xi exists (Lemma 14.1) and is quasi-compact by Lemma
24.1.
Denote pi : X → Xi the projection. Because I is cofiltered we can apply Lemma
14.2. Hence a basis for the topology on X is given by the opens p−1
i (Ui ) for Ui ⊂ Xi
open. Since a basis for the topology of Xi is given by the quasi-compact open, we
conclude that a basis for the topology on X is given by p−1 i (Ui ) with Ui ⊂ Xi
quasi-compact open. A formal argument shows that
pi−1 (Ui ) = lima:j→i fa−1 (Ui )
as topological spaces. Since each fa is spectral the sets fa−1 (Ui ) are closed in the
constructible topology of Xj and hence p−1i (Ui ) is quasi-compact by Lemma 24.1.
Thus X has a basis for the topology consisting of quasi-compact opens.
 TOPOLOGY 51

Any quasi-compact open U of X is of the form U = p−1 i (Ui ) for some i and some
quasi-compact open Ui ⊂ Xi (see Lemma 24.4). Given Ui ⊂ Xi and Uj ⊂ Xj quasi-
compact open, then p−1 −1 −1
i (Ui ) ∩ pj (Uj ) = pk (Uk ) for some k and quasi-compact
open Uk ⊂ Xk . Namely, choose k and morphisms k → i and k → j and let Uk
be the intersection of the pullbacks of Ui and Uj to Xk . Thus we see that the
intersection of two quasi-compact opens of X is quasi-compact open.

Finally, let Z ⊂ X be irreducible and closed. Then pi (Z) ⊂ Xi is irreducible and

therefore Zi = pi (Z) has a unique generic point ξi (because Xi is a spectral space).
Then fa (ξj ) = ξi for a : j → i in I because fa (Zj ) = Zi . Hence ξ = lim ξi is a point
of X. Claim: ξ ∈ Z. Namely, if not we can find a quasi-compact open containing ξ
disjoint from Z. This would be of the form p−1 i (Ui ) for some i and quasi-compact
open Ui ⊂ Xi . Then ξi ∈ Ui but pi (Z) ∩ Ui = ∅ which contradicts ξi ∈ pi (Z). So
ξ ∈ Z and hence {ξ} ⊂ Z. Conversely, every z ∈ Z is in the closure of ξ. Namely,
given a quasi-compact open neighbourhood U of z we write U = p−1 i (Ui ) for some
i and quasi-compact open Ui ⊂ Xi . We see that pi (z) ∈ Ui hence ξi ∈ Ui hence
ξ ∈ U . Thus ξ is a generic point of Z. We omit the proof that ξ is the unique
generic point of Z (hint: show that a second generic point has to be equal to ξ by
showing that it has to map to ξi in Xi since by spectrality of Xi the irreducible Zi
has a unique generic point). This finishes the proof. □

0A30 Lemma 24.6. Let I be a cofiltered index category. Let i 7→ Xi be a diagram of

spectral spaces such that for a : j → i in I the corresponding map fa : Xj → Xi is
spectral. Set X = lim Xi and denote pi : X → Xi the projection.
(1) Given any quasi-compact open U ⊂ X there exists an i ∈ Ob(I) and a
quasi-compact open Ui ⊂ Xi such that p−1 i (Ui ) = U .
(2) Given Ui ⊂ Xi and Uj ⊂ Xj quasi-compact opens such that p−1 i (Ui ) ⊂
p−1
j (U j ) there exist k ∈ Ob(I) and morphisms a : k → i and b : k → j such
−1 −1
that fa (Ui ) ⊂ fb (Uj ).
(3) If Ui , U1,i , . . . , Un,i ⊂ Xi are quasi-compact opens and p−1 −1
i (Ui ) = pi (U1,i )∪
−1 −1 −1 −1
. . . ∪ pi (Un,i ) then fa (Ui ) = fa (U1,i ) ∪ . . . ∪ fa (Un,i ) for some mor-
phism a : j → i in I.
(4) Same statement as in (3) but for intersections.

Proof. Part (1) is a special case of Lemma 24.4. Part (2) is a special case of Lemma
24.3 as quasi-compact opens are both open and closed in the constructible topology.
Parts (3) and (4) follow formally from (1) and (2) and the fact that taking inverse
images of subsets commutes with taking unions and intersections. □

0A31 Lemma 24.7. Let W be a subset of a spectral space X. The following are equiv-
alent:
(1) W is an intersection of constructible sets and closed under generalizations,
(2) W is quasi-compact and closed under generalizations,
(3) there exists a quasi-compact subset E ⊂ X such that W is the set of points
specializing to E,
(4) W is an intersection of quasi-compact open subsets,
0ANZ (5) there existsTa nonempty set I and quasi-compact opens Ui ⊂ X, i ∈ I such
that W = Ui and for all i, j ∈ I there exists a k ∈ I with Uk ⊂ Ui ∩ Uj .
 TOPOLOGY 52

In this case we have (a) W is a spectral space, (b) W = lim Ui as topological spaces,
and (c) for any open U containing W there exists an i with Ui ⊂ U .
Proof. Let W ⊂ X satisfy (1). Then W is closed in the constructible topology,
hence quasi-compact in the constructible topology (by Lemmas 23.2 and 12.3),
hence quasi-compact in the topology of X (because opens in X are open in the
constructible topology). Thus (2) holds.
It is clear that (2) implies (3) by taking E = W .
Let X be a spectral space and let E ⊂ W be as in (3). Since every point of W
specializes to a point of E we see that an open of W which contains E is equal to
W . Hence since E is quasi-compact, so is W . If x ∈ X, x ̸∈ W , then Z = {x} is
disjoint from W . Since W is quasi-compact we can find a quasi-compact open U
with W ⊂ U and U ∩ Z = ∅. We conclude that (4) holds.
T
If W = j∈J Uj then setting I equal to the set of finite subsets of J and Ui =
Uj1 ∩ . . . ∩ Ujr for i = {j1 , . . . , jr } shows that (4) implies (5). It is immediate that
(5) implies (1).
T
Let I and Ui be as in (5). Since W = Ui we have W = lim Ui by the universal
property of limits. Then W is a spectral space by Lemma 24.5. Let U ⊂ X be an
open neighbourhood of W . Then Ei = Ui ∩ (X \ U ) is a family of constructible
subsets of the spectral space Z = X \ U with empty intersection. Using that the
spectral topology on Z is quasi-compact (Lemma 23.2) we conclude from Lemma
12.6 that Ei = ∅ for some i. □
0AP0 Lemma 24.8. Let X be a spectral space. Let E ⊂ X be a constructible subset. Let
W ⊂ X be the set of points
T of X which specialize to a point of E. Then W \ E is a
spectral space. If W = Ui with Ui as in Lemma 24.7 (5) then W \E = lim(Ui \E).
Proof. Since E is constructible, it is quasi-compact and hence Lemma 24.7 applies
to W . If E is constructible, then E is constructible
T in Ui for all i ∈ I. Hence Ui \ E
is spectral by Lemma 23.5. Since W \ E = (Ui \ E) we have W \ E = lim Ui \ E
by the universal property of limits. Then W \ E is a spectral space by Lemma
24.5. □

25. Stone-Čech compactification

0908 The Stone-Čech compactification of a topological space X is a map X → β(X)
from X to a Hausdorff quasi-compact space β(X) which is universal for such maps.
We prove this exists by a standard argument using the following simple lemma.
0909 Lemma 25.1. Let f : X → Y be a continuous map of topological spaces. Assume
that f (X) is dense in Y and that Y is Hausdorff. Then the cardinality of Y is at
most the cardinality of P (P (X)) where P is the power set operation.
Proof. Let S = f (X) ⊂ Y . Let D be the set of all closed domains of Y , i.e.,
subsets D ⊂ Y which equal the closure of its interior. Note that the closure of an
open subset of Y is a closed domain. For y ∈ Y consider the set
Iy = {T ⊂ S | there exists D ∈ D with T = S ∩ D and y ∈ D}.
Since S is dense in Y for every closed domain D we see that S ∩ D is dense in D.
Hence, if D ∩ S = D′ ∩ S for D, D′ ∈ D, then D = D′ . Thus Iy = Iy′ implies that
 TOPOLOGY 53

y = y ′ because the Hausdorff condition assures us that we can find a closed domain
containing y but not y ′ . The result follows. □
Let X be a topological space. By Lemma 25.1, there is a set I of isomorphism
classes of continuous maps f : X → Y which have dense image and where Y is
Hausdorff and quasi-compact. For i ∈ I choose a representative fi : X → Yi .
Consider the map Y Y
fi : X −→ Yi
i∈I
and denote β(X) the closure of the image. Since each Yi is Hausdorff, so is β(X).
Since each Yi is quasi-compact, so is β(X) (use Theorem 14.4 and Lemma 12.3).
Let us show the canonical map X → β(X) satisfies the universal property with
respect to maps to Hausdorff, quasi-compact spaces. Namely, let f : X → Y be
such a morphism. Let Z ⊂ Y be the closure of f (X). Then X → Z is isomorphic
Q one of the∼maps fi : X → Yi , say fi0 : X → Yi0 . Thus f factors as X → β(X) →
to
Yi → Yi0 = Z → Y as desired.
090A Lemma 25.2. Let X be a Hausdorff, locally quasi-compact space. There exists a
map X → X ∗ which identifies X as an open subspace of a quasi-compact Hausdorff
space X ∗ such that X ∗ \X is a singleton (one point compactification). In particular,
the map X → β(X) identifies X with an open subspace of β(X).
Proof. Set X ∗ = X ⨿ {∞}. We declare a subset V of X ∗ to be open if either
V ⊂ X is open in X, or ∞ ∈ V and U = V ∩ X is an open of X such that X \ U
is quasi-compact. We omit the verification that this defines a topology. It is clear
that X → X ∗ identifies X with an open subspace of X.
Since X is locally quasi-compact, every point x ∈ X has a quasi-compact neighbour-
hood x ∈ E ⊂ X. Then E is closed (Lemma 12.4 part (1)) and V = (X \E)⨿{∞} is
an open neighbourhood of ∞ disjoint from the interior of E. Thus X ∗ is Hausdorff.
Let X ∗ = Vi be an open covering. Then for some i, say i0 , we have ∞
S
S ∈ Vi0 . By
construction Z = X ∗ \ Vi0 is quasi-compact. Hence the covering Z ⊂ i̸=i0 Z ∩ Vi
has a finite refinement which implies that the given covering of X ∗ has a finite
refinement. Thus X ∗ is quasi-compact.
The map X → X ∗ factors as X → β(X) → X ∗ by the universal property of the
Stone-Čech compactification. Let φ : β(X) → X ∗ be this factorization. Then
X → φ−1 (X) is a section to φ−1 (X) → X hence has closed image (Lemma 3.3).
Since the image of X → β(X) is dense we conclude that X = φ−1 (X). □

26. Extremally disconnected spaces

08YH The material in this section is taken from [Gle58] (with a slight modification as
in [Rai59]). In Gleason’s paper it is shown that in the category of quasi-compact
Hausdorff spaces, the “projective objects” are exactly the extremally disconnected
spaces.
08YI Definition 26.1. A topological space X is called extremally disconnected if the
closure of every open subset of X is open.
If X is Hausdorff and extremally disconnected, then X is totally disconnected (this
isn’t true in general). If X is quasi-compact, Hausdorff, and extremally discon-
nected, then X is profinite by Lemma 22.2, but the converse does not hold in
 TOPOLOGY 54

general. For example the p-adic integers Zp = lim Z/pn Z is a profinite space which
is not extremally disconnected. Namely, if U ⊂ Zp is the set of nonzero elements
whose valuation is even, then U is open but its closure is U ∪ {0} which is not open.

08YJ Lemma 26.2. Let f : X → Y be a continuous map of topological spaces. Assume

f is surjective and f (E) ̸= Y for all proper closed subsets E ⊂ X. Then for U ⊂ X
open the subset f (U ) is contained in the closure of Y \ f (X \ U ).

Proof. Pick y ∈ f (U ) and let V ⊂ Y be any open neighbourhood of y. We will

show that V intersects Y \ f (X \ U ). Note that W = U ∩ f −1 (V ) is a nonempty
open subset of X, hence f (X \ W ) ̸= Y . Take y ′ ∈ Y , y ′ ̸∈ f (X \ W ). It is
elementary to show that y ′ ∈ V and y ′ ∈ Y \ f (X \ U ). □

08YK Lemma 26.3. Let X be an extremally disconnected space. If U, V ⊂ X are disjoint

open subsets, then U and V are disjoint too.

Proof. By assumption U is open, hence V ∩ U is open and disjoint from U , hence

empty because U is the intersection of all the closed subsets of X containing U .
This means the open V ∩ U avoids V hence is empty by the same argument. □

08YL Lemma 26.4. Let f : X → Y be a continuous map of Hausdorff quasi-compact

topological spaces. If Y is extremally disconnected, f is surjective, and f (Z) ̸= Y
for every proper closed subset Z of X, then f is a homeomorphism.

Proof. By Lemma 17.8 it suffices to show that f is injective. Suppose that x, x′ ∈

X are distinct points with y = f (x) = f (x′ ). Choose disjoint open neighbourhoods
U, U ′ ⊂ X of x, x′ . Observe that f is closed (Lemma 17.7) hence T = f (X \ U ) and
T ′ = f (X \ U ′ ) are closed in Y . Since X is the union of X \ U and X \ U ′ we see
that Y = T ∪ T ′ . By Lemma 26.2 we see that y is contained in the closure of Y \ T
and the closure of Y \ T ′ . On the other hand, by Lemma 26.3, this intersection is
empty. In this way we obtain the desired contradiction. □

08YM Lemma 26.5. Let f : X → Y be a continuous surjective map of Hausdorff quasi-

compact topological spaces. There exists a quasi-compact subset E ⊂ X such that
f (E) = Y but f (E ′ ) ̸= Y for all proper closed subsets E ′ ⊂ E.

Proof. We will use without further mention that the quasi-compact subsets of
X are exactly the closed subsets (Lemma 12.5). Consider the collection E of all
quasi-compact subsets E ⊂ X with f (E) = Y ordered by inclusion. We will use
Zorn’s lemma to show that E has a minimal element. To do this it suffices T
to show
that given a totally ordered family Eλ of elements of E the intersection Eλ is
an element of E. It is quasi-compact as it is closed. For every T y ∈ Y−1the sets
−1
E λ ∩ f ({y}) are nonempty and closed, hence the intersection Eλ ∩ f ({y}) =
(Eλ ∩ f −1 ({y})) is nonempty by Lemma 12.6. This finishes the proof.
T
□

08YN Proposition 26.6. Let X be a Hausdorff, quasi-compact topological space. The

following are equivalent
(1) X is extremally disconnected,
(2) for any surjective continuous map f : Y → X with Y Hausdorff quasi-
compact there exists a continuous section, and
 TOPOLOGY 55

(3) for any solid commutative diagram

>Y


X /Z
of continuous maps of quasi-compact Hausdorff spaces with Y → Z surjec-
tive, there is a dotted arrow in the category of topological spaces making the
diagram commute.
Proof. It is clear that (3) implies (2). On the other hand, if (2) holds and X → Z
and Y → Z are as in (3), then (2) assures there is a section to the projection
X ×Z Y → X which implies a suitable dotted arrow exists (details omitted). Thus
(3) is equivalent to (2).
Assume X is extremally disconnected and let f : Y → X be as in (2). By Lemma
26.5 there exists a quasi-compact subset E ⊂ Y such that f (E) = X but f (E ′ ) ̸= X
for all proper closed subsets E ′ ⊂ E. By Lemma 26.4 we find that f |E : E → X is
a homeomorphism, the inverse of which gives the desired section.
Assume (2). Let U ⊂ X be open with complement Z. Consider the continuous
surjection f : U ⨿ Z → X. Let σ be a section. Then U = σ −1 (U ) is open. Thus X
is extremally disconnected. □
090B Lemma 26.7. Let f : X → X be a surjective continuous selfmap of a Hausdorff
topological space. If f is not idX , then there exists a proper closed subset E ⊂ X
such that X = E ∪ f (E).
Proof. Pick p ∈ X with f (p) ̸= p. Choose disjoint open neighbourhoods p ∈ U ,
f (p) ∈ V and set E = X \ U ∩ f −1 (V ). Then p ̸∈ E hence E is a proper closed
subset. If x ∈ X, then either x ∈ E, or if not, then x ∈ U ∩ f −1 (V ). Writing
x = f (y) (possible as f is surjective). If y ∈ U ∩ f −1 (V ) then we would have
x = f (y) ∈ V which is a contradiction with x ∈ U . Hence y ∈ E and x ∈ f (E). □
090C Example 26.8. We can use Proposition 26.6 to see that the Stone-Čech com-
pactification β(X) of a discrete space X is extremally disconnected. Namely, let
f : Y → β(X) be a continuous surjection where Y is quasi-compact and Hausdorff.
Then we can lift the map X → β(X) to a continuous (!) map X → Y as X is
discrete. By the universal property of the Stone-Čech compactification we see that
we obtain a factorization X → β(X) → Y . Since β(X) → Y → β(X) equals the
identity on the dense subset X we conclude that we get a section. In particular,
we conclude that the Stone-Čech compactification of a discrete space is totally dis-
connected, whence profinite (see discussion following Definition 26.1 and Lemma
22.2).
Using the supply of extremally disconnected spaces given by Example 26.8 we can
prove that every quasi-compact Hausdorff space has a “projective cover” in the
category of quasi-compact Hausdorff spaces.
090D Lemma 26.9. Let X be a quasi-compact Hausdorff space. There exists a contin-
uous surjection X ′ → X with X ′ quasi-compact, Hausdorff, and extremally discon-
nected. If we require that every proper closed subset of X ′ does not map onto X,
then X ′ is unique up to isomorphism.
 TOPOLOGY 56

Proof. Let Y = X but endowed with the discrete topology. Let X ′ = β(Y ). The
continuous map Y → X factors as Y → X ′ → X. This proves the first statement
of the lemma by Example 26.8.

By Lemma 26.5 we can find a quasi-compact subset E ⊂ X ′ surjecting onto X

such that no proper closed subset of E surjects onto X. Because X ′ is extremally
disconnected there exists a continuous map f : X ′ → E over X (Proposition 26.6).
Composing f with the map E → X ′ gives a continuous selfmap f |E : E → E.
Observe that f |E has to be surjective as otherwise the image would be a proper
closed subset surjecting onto X. Hence f |E has to be idE as otherwise Lemma
26.7 shows that E isn’t minimal. Thus the idE factors through the extremally
disconnected space X ′ . A formal, categorical argument (using the characterization
of Proposition 26.6) shows that E is extremally disconnected.

To prove uniqueness, suppose we have a second X ′′ → X minimal cover. By the

lifting property proven in Proposition 26.6 we can find a continuous map g : X ′ →
X ′′ over X. Observe that g is a closed map (Lemma 17.7). Hence g(X ′ ) ⊂ X ′′ is
a closed subset surjecting onto X and we conclude g(X ′ ) = X ′′ by minimality of
X ′′ . On the other hand, if E ⊂ X ′ is a proper closed subset, then g(E) ̸= X ′′ as
E does not map onto X by minimality of X ′ . By Lemma 26.4 we see that g is an
isomorphism. □

090E Remark 26.10. Let X be a quasi-compact Hausdorff space. Let κ be an infinite

cardinal bigger or equal than the cardinality of X. Then the cardinality of the
minimal quasi-compact, Hausdorff, extremally disconnected cover X ′ → X (Lemma
κ
26.9) is at most 22 . Namely, choose a subset S ⊂ X ′ mapping bijectively to X.
κ
By minimality of X ′ the set S is dense in X ′ . Thus |X ′ | ≤ 22 by Lemma 25.1.

27. Miscellany
0067 The following lemma applies to the underlying topological space associated to a
quasi-separated scheme.

0069 Lemma 27.1. Let X be a topological space which

(1) has a basis of the topology consisting of quasi-compact opens, and
(2) has the property that the intersection of any two quasi-compact opens is
quasi-compact.
Then
(1) X is locally quasi-compact,
(2) a quasi-compact open U ⊂ X is retrocompact,
(3) any quasi-compact
S open U ⊂ X has a cofinal system of open coverings
U : U = j∈J Uj with J finite and all Uj and Uj ∩ Uj ′ quasi-compact,
(4) add more here.

Proof. Omitted. □

06RM Definition 27.2. Let X be a topological space. We say x ∈ X is an isolated point

of X if {x} is open in X.
 TOPOLOGY 57

28. Partitions and stratifications

09XY Stratifications can be defined in many different ways. We welcome comments on
the choice of definitions in this section.
09XZ Definition
` 28.1. Let X be a topological space. A partition of X is a decomposition
X = Xi into locally closed subsets Xi . The Xi are called the parts of the
partition. Given two partitions of X we say one refines the other if the parts of one
are unions of parts of the other.
Any topological space X has a partition into connected components. If X has
finitely manyTirreducibleS components Z1 , . . . , Zr , then there is a partition with
parts XI = i∈I Zi \ ( i̸∈I Zi ) whose indices are subsets I ⊂ {1, . . . , r} which
refines the partition into connected components.
09Y0 ` Let X be a topological space. A good stratification of X is a
Definition 28.2.
partition X = Xi such that for all i, j ∈ I we have
Xi ∩ Xj ̸= ∅ ⇒ Xi ⊂ Xj .
`
Given a good stratification X = i∈I Xi we obtain a partial ordering on I by
setting i ≤ j if and only if Xi ⊂ Xj . Then we see that
[
Xj = Xi
i≤j

However, what often happens in algebraic geometry is that one just has that the
left hand side is a subset of the right hand side in the last displayed formula. This
leads to the following definition.
09Y1 Definition 28.3.` Let X be a topological space. A stratification of X is given by
a partition X = i∈I Xi and a partial ordering on I such that for each j ∈ I we
have [
Xj ⊂ Xi
i≤j
The parts Xi are called the strata of the stratification.
We often impose additional conditions on the stratification. For example, stratifi-
cations are particularly nice if they are locally finite, which means that every point
has a neighbourhood which meets only finitely many strata. More generally we
introduce the following definition.
0BDS Definition 28.4. Let X be a topological space. Let I be a set and for i ∈ I let
Ei ⊂ X be a subset. We say the collection {Ei }i∈I is locally finite if for all x ∈ X
there exists an open neighbourhood U of x such that {i ∈ I|Ei ∩ U ̸= ∅} is finite.
`
09Y2 Remark 28.5. Given a locally finite stratificationS X = Xi of a topological
space X, we obtain a family of closed subsets Zi = j≤i Xj of X indexed by I such
that [
Zi ∩ Zj = Zk
k≤i,j
Conversely,Sgiven closed subsets Zi ⊂ X indexed by a partially ordered set I such
that X = Zi , such that every point has a neighbourhood meeting only finitely
many Zi , and such that the displayed formula
S holds, then we obtain a locally finite
stratification of X by setting Xi = Zi \ j<i Zj .
 TOPOLOGY 58

`
09Y3 Lemma 28.6. Let X be a topological space. Let X = Xi be a finite partition of
X. Then there exists a finite stratification of X refining it.
Proof. Let Ti = Xi and ∆i = Ti \ Xi . Let S be the set of all intersections of Ti
and ∆i . (For example T1 ∩ T2 ∩ ∆4 is an element of S.) Then S = {Zs } is a finite
′
collection of closed subsets of X such that Zs ∩ Zs′ ∈ S
S for all s, s ∈ S. Define a
partial ordering on S by inclusion. Then set Ys = Zs \ s′ <s Zs′ to get the desired
stratification. □
09Y4 Lemma 28.7. Let X be a topological space. Suppose X = T1 ∪ . . . ∪ Tn is written
`
as a union of constructible subsets. There exists a finite stratification X = Xi
with each Xi constructible such that each Tk is a union of strata.
Proof. By definition of constructible subsets, we can write each Ti as a finite union
of U ∩ V c with U, V ⊂ X retrocompact open. Hence we may assume that Ti =
Ui ∩Vic with Ui , Vi ⊂ X retrocompact open. Let S be the finite set of closed subsets
of X consisting of ∅, X, Uic , Vic and finite intersections of these. If Z ∈ S, then Z is
constructible in X (Lemma 15.2). Moreover, Z ∩ Z ′ ∈ S for all ′
S Z, Z ∈ S. Define a
partial ordering on S by inclusion. For Z ∈ S set XZ = Z \ Z ′ <Z, Z ′ ∈S Z ′ to get
`
a stratification X = Z∈S XZ satisfying the properties stated in the lemma. □
09Y5 Lemma 28.8. Let X be a Noetherian topological space. Any finite partition of X
can be refined by a finite good stratification.
`
Proof. Let X = S Xi be a finite partition of X. Let Z be an irreducible component
of X. Since X = Xi with finite index set, there is an i such that Z ⊂ Xi . Since Xi
is locally closed this implies that Z ∩Xi contains an open of Z. Thus Z ∩Xi contains
an open U of X (Lemma 9.2). Write Xi = U ⨿ Xi1 ⨿ Xi2 with Xi1 = (Xi \ U ) ∩ U
c c
and Xi2 = (Xi \ U ) ∩ U . For i′ ̸= i we set Xi1′ = Xi′ ∩ U and Xi2′ = Xi′ ∩ U . Then
a
X \U = Xlk
is a partition such that U \ U = Xl1 . Note that X \ U is closed and strictly
S
smaller than X. By Noetherian ` induction we can refine ` this partition by a finite
good stratification X \ U = α∈A Tα . Then X = U ⨿ α∈A Tα is a finite good
stratification of X refining the partition we started with. □

29. Colimits of spaces

0B1W The category of topological spaces has coproducts. Namely, if I`is a set and for
i ∈ I we are given a topological space Xi then we endow the set i∈I Xi with the
coproduct topology. As a basis for this topology we use sets of the form Ui where
Ui ⊂ Xi is open.
The category of topological spaces has coequalizers. Namely, if a, b : X → Y are
morphisms of topological spaces, then the coequalizer of a and b is the coequalizer
Y / ∼ in the category of sets endowed with the quotient topology (Section 6).
0B1X Lemma 29.1. The category of topological spaces has colimits and the forgetful
functor to sets commutes with them.
Proof. This follows from the discussion above and Categories, Lemma 14.12. An-
other proof of existence of colimits is sketched in Categories, Remark 25.2. It follows
from the above that the forgetful functor commutes with colimits. Another way to
 TOPOLOGY 59

see this is to use Categories, Lemma 24.5 and use that the forgetful functor has a
right adjoint, namely the functor which assigns to a set the corresponding chaotic
(or indiscrete) topological space. □

30. Topological groups, rings, modules

0B1Y This is just a short section with definitions and elementary properties.
0B1Z Definition 30.1. A topological group is a group G endowed with a topology such
that multiplication G × G → G, (x, y) 7→ xy and inverse G → G, x 7→ x−1 are
continuous. A homomorphism of topological groups is a homomorphism of groups
which is continuous.
If G is a topological group and H ⊂ G is a subgroup, then H with the induced
topology is a topological group. If G is a topological group and G → H is a
surjection of groups, then H endowed with the quotient topology is a topological
group.
0BMC Example 30.2. Let E be a set. We can endow the set of self maps Map(E, E)
with the compact open topology, i.e., the topology such that given f : E → E a
fundamental system of neighbourhoods of f is given by the sets US (f ) = {f ′ : E →
E | f ′ |S = f |S } where S ⊂ E is finite. With this topology the action
Map(E, E) × E −→ E, (f, e) 7−→ f (e)
is continuous when E is given the discrete topology. If X is a topological space and
X × E → E is a continuous map, then the map X → Map(E, E) is continuous.
In other words, the compact open topology is the coarsest topology such that the
“action” map displayed above is continuous. The composition
Map(E, E) × Map(E, E) → Map(E, E)
is continuous as well (as is easily verified using the description of neighbourhoods
above). Finally, if Aut(E) ⊂ Map(E, E) is the subset of invertible maps, then the
inverse i : Aut(E) → Aut(E), f 7→ f −1 is continuous too. Namely, say S ⊂ E is
finite, then i−1 (US (f −1 )) = Uf −1 (S) (f ). Hence Aut(E) is a topological group as in
Definition 30.1.
0B20 Lemma 30.3. The category of topological groups has limits and limits commute
with the forgetful functors to (a) the category of topological spaces and (b) the cat-
egory of groups.
Proof. It is enough to prove the existence and commutation for products and
Let Gi , i ∈ I be a collection of topological
equalizers, see Categories, Lemma 14.11. Q
groups. TakeQ the usual product G = Gi with the product topology. Since
G × G = (Gi × Gi ) as a topological space (because products commutes with
products in any category), we see that multiplication on G is continuous. Similarly
for the inverse map. Let a, b : G → H be two homomorphisms of topological
groups. Then as the equalizer we can simply take the equalizer of a and b as maps
of topological spaces, which is the same thing as the equalizer as maps of groups
endowed with the induced topology. □
0BR1 Lemma 30.4. Let G be a topological group. The following are equivalent
(1) G as a topological space is profinite,
(2) G is a limit of a diagram of finite discrete topological groups,
 TOPOLOGY 60

(3) G is a cofiltered limit of finite discrete topological groups.

Proof. We have the corresponding result for topological spaces, see Lemma 22.2.
Combined with Lemma 30.3 we see that it suffices to prove that (1) implies (3).
We first prove that every neighbourhood E of the neutral element e contains an
open subgroup. Namely, since G is the cofiltered limit of finite discrete topological
spaces (Lemma 22.2), we can choose a continuous map f : G → T to a finite discrete
space T such that f −1 (f ({e})) ⊂ E. Consider
H = {g ∈ G | f (gg ′ ) = f (g ′ ) for all g ′ ∈ G}
This is a subgroup of G and contained in E. Thus it suffices to show that H is
open. Pick t ∈ T and set W = f −1 ({t}). Observe that W ⊂ G is open and closed,
in particular quasi-compact. For each w ∈ W there exist open neighbourhoods
e ∈ Uw ⊂ G and w ∈ Uw′ ⊂ W suchSthat Uw Uw′ ⊂ W . By quasi-compactness we
can find w1 , . . . , wn such that W = Uw′ i . Then Ut = Uw1 ∩ . . . ∩ Uwn is an open
neighbourhood
T of e such that f (gw) = t for all w ∈ W . Since T is finite we see
that t∈T Ut ⊂ H is an open neighbourhood of e. Since H ⊂ G is a subgroup it
follows that H is open.
Suppose that H ⊂ G is an open subgroup. Since G is quasi-compact T we see that
the index of H in G is finite. Say G = Hg1 ∪. . .∪Hgn . Then N = i=1,...,n gi Hgi−1
is an open normal subgroup contained in H. Since N also has finite index we see
that G → G/N is a surjection to a finite discrete topological group.
Consider the map
G −→ limN ⊂G open and normal G/N
We claim that this map is an isomorphism of topological groups. This finishes the
proof of the lemma as the limit on the right is cofiltered (the intersection of two open
normal subgroups is open and normal). The map is continuous as each G → G/N
is continuous. The map is injective as G is Hausdorff and every neighbourhood of
e contains an N by the arguments above. The map is surjective by Lemma 12.6.
By Lemma 17.8 the map is a homeomorphism. □
0BR2 Definition 30.5. A topological group is called a profinite group if it satisfies the
equivalent conditions of Lemma 30.4.
If G1 → G2 → G3 → . . . is a system of topological groups then the colimit G =
colim Gn as a topological group (Lemma 30.6) is in general different from the colimit
as a topological space (Lemma 29.1) even though these have the same underlying
set. See Examples, Section 78.
0B21 Lemma 30.6. The category of topological groups has colimits and colimits com-
mute with the forgetful functor to the category of groups.
Proof. We will use the argument of Categories, Remark 25.2 to prove existence
of colimits. Namely, suppose that I → Top, i 7→ Gi is a functor into the category
TopGroup of topological groups. Then we can consider
F : TopGroup −→ Sets, H 7−→ limI MorTopGroup (Gi , H)
This functor commutes with limits. Moreover, given any topological group H and
′
` (φi : Gi → H) of F (H), there is a subgroup H ⊂ H of cardinality
an element
at most | Gi | (coproduct in the category of groups, i.e., the free product on the
 TOPOLOGY 61

Gi ) such that the morphisms φi map into H ′ . Namely, we can take the induced
topology on the subgroup generated by the images of the φi . Thus it is clear that
the hypotheses of Categories, Lemma 25.1 are satisfied and we find a topological
group G representing the functor F , which precisely means that G is the colimit of
the diagram i 7→ Gi .
To see the statement on commutation with the forgetful functor to groups we will
use Categories, Lemma 24.5. Indeed, the forgetful functor has a right adjoint,
namely the functor which assigns to a group the corresponding chaotic (or indis-
crete) topological group. □
0B22 Definition 30.7. A topological ring is a ring R endowed with a topology such that
addition R × R → R, (x, y) 7→ x + y and multiplication R × R → R, (x, y) 7→ xy
are continuous. A homomorphism of topological rings is a homomorphism of rings
which is continuous.
In the Stacks project rings are commutative with 1. If R is a topological ring, then
(R, +) is a topological group since x 7→ −x is continuous. If R is a topological ring
and R′ ⊂ R is a subring, then R′ with the induced topology is a topological ring. If
R is a topological ring and R → R′ is a surjection of rings, then R′ endowed with
the quotient topology is a topological ring.
0B23 Lemma 30.8. The category of topological rings has limits and limits commute with
the forgetful functors to (a) the category of topological spaces and (b) the category
of rings.
Proof. It is enough to prove the existence and commutation for products and
equalizers, see Categories, Lemma 14.11.
Q Let Ri , i ∈ I be a collection of topological
rings.
Q Take the usual product R = R i with the product topology. Since R × R =
(Ri × Ri ) as a topological space (because products commutes with products in
any category), we see that addition and multiplication on R are continuous. Let
a, b : R → R′ be two homomorphisms of topological rings. Then as the equalizer we
can simply take the equalizer of a and b as maps of topological spaces, which is the
same thing as the equalizer as maps of rings endowed with the induced topology. □
0B24 Lemma 30.9. The category of topological rings has colimits and colimits commute
with the forgetful functor to the category of rings.
Proof. The exact same argument as used in the proof of Lemma 30.6 shows exis-
tence of colimits. To see the statement on commutation with the forgetful functor
to rings we will use Categories, Lemma 24.5. Indeed, the forgetful functor has a
right adjoint, namely the functor which assigns to a ring the corresponding chaotic
(or indiscrete) topological ring. □
0B25 Definition 30.10. Let R be a topological ring. A topological module is an R-
module M endowed with a topology such that addition M × M → M and scalar
multiplication R×M → M are continuous. A homomorphism of topological modules
is a homomorphism of modules which is continuous.
If R is a topological ring and M is a topological module, then (M, +) is a topological
group since x 7→ −x is continuous. If R is a topological ring, M is a topological
module and M ′ ⊂ M is a submodule, then M ′ with the induced topology is a
topological module. If R is a topological ring, M is a topological module, and
 TOPOLOGY 62

M → M ′ is a surjection of modules, then M ′ endowed with the quotient topology

is a topological module.
0B26 Lemma 30.11. Let R be a topological ring. The category of topological modules
over R has limits and limits commute with the forgetful functors to (a) the category
of topological spaces and (b) the category of R-modules.
Proof. It is enough to prove the existence and commutation for products and
i ∈ I be a collection of topological
equalizers, see Categories, Lemma 14.11. Let Mi , Q
modules over R. Q Take the usual product M = Mi with the product topology.
Since M × M = (Mi × Mi ) as a topological space (because products commutes
with products in any category), we see that addition on M is continuous. Similarly
for multiplication R × M → M . Let a, b : M → M ′ be two homomorphisms of
topological modules over R. Then as the equalizer we can simply take the equalizer
of a and b as maps of topological spaces, which is the same thing as the equalizer
as maps of modules endowed with the induced topology. □
0B27 Lemma 30.12. Let R be a topological ring. The category of topological modules
over R has colimits and colimits commute with the forgetful functor to the category
of modules over R.
Proof. The exact same argument as used in the proof of Lemma 30.6 shows exis-
tence of colimits. To see the statement on commutation with the forgetful functor
to R-modules we will use Categories, Lemma 24.5. Indeed, the forgetful functor has
a right adjoint, namely the functor which assigns to a module the corresponding
chaotic (or indiscrete) topological module. □

31. Other chapters

Preliminaries (21) Cohomology on Sites

(22) Differential Graded Algebra
(1) Introduction (23) Divided Power Algebra
(2) Conventions (24) Differential Graded Sheaves
(3) Set Theory (25) Hypercoverings
(4) Categories
Schemes
(5) Topology
(6) Sheaves on Spaces (26) Schemes
(7) Sites and Sheaves (27) Constructions of Schemes
(8) Stacks (28) Properties of Schemes
(9) Fields (29) Morphisms of Schemes
(10) Commutative Algebra (30) Cohomology of Schemes
(11) Brauer Groups (31) Divisors
(12) Homological Algebra (32) Limits of Schemes
(13) Derived Categories (33) Varieties
(14) Simplicial Methods (34) Topologies on Schemes
(15) More on Algebra (35) Descent
(16) Smoothing Ring Maps (36) Derived Categories of Schemes
(17) Sheaves of Modules (37) More on Morphisms
(18) Modules on Sites (38) More on Flatness
(19) Injectives (39) Groupoid Schemes
(20) Cohomology of Sheaves (40) More on Groupoid Schemes
 TOPOLOGY 63

(41) Étale Morphisms of Schemes Topics in Geometry

Topics in Scheme Theory (82) Chow Groups of Spaces
(42) Chow Homology (83) Quotients of Groupoids
(43) Intersection Theory (84) More on Cohomology of Spaces
(44) Picard Schemes of Curves (85) Simplicial Spaces
(45) Weil Cohomology Theories (86) Duality for Spaces
(46) Adequate Modules (87) Formal Algebraic Spaces
(47) Dualizing Complexes (88) Algebraization of Formal Spaces
(48) Duality for Schemes (89) Resolution of Surfaces Revisited
(49) Discriminants and Differents Deformation Theory
(50) de Rham Cohomology (90) Formal Deformation Theory
(51) Local Cohomology (91) Deformation Theory
(52) Algebraic and Formal Geometry (92) The Cotangent Complex
(53) Algebraic Curves (93) Deformation Problems
(54) Resolution of Surfaces
Algebraic Stacks
(55) Semistable Reduction
(56) Functors and Morphisms (94) Algebraic Stacks
(57) Derived Categories of Varieties (95) Examples of Stacks
(58) Fundamental Groups of Schemes (96) Sheaves on Algebraic Stacks
(59) Étale Cohomology (97) Criteria for Representability
(60) Crystalline Cohomology (98) Artin’s Axioms
(61) Pro-étale Cohomology (99) Quot and Hilbert Spaces
(62) Relative Cycles (100) Properties of Algebraic Stacks
(101) Morphisms of Algebraic Stacks
(63) More Étale Cohomology
(102) Limits of Algebraic Stacks
(64) The Trace Formula
(103) Cohomology of Algebraic Stacks
Algebraic Spaces (104) Derived Categories of Stacks
(65) Algebraic Spaces (105) Introducing Algebraic Stacks
(66) Properties of Algebraic Spaces (106) More on Morphisms of Stacks
(67) Morphisms of Algebraic Spaces (107) The Geometry of Stacks
(68) Decent Algebraic Spaces Topics in Moduli Theory
(69) Cohomology of Algebraic Spaces
(70) Limits of Algebraic Spaces (108) Moduli Stacks
(71) Divisors on Algebraic Spaces (109) Moduli of Curves
(72) Algebraic Spaces over Fields Miscellany
(73) Topologies on Algebraic Spaces (110) Examples
(74) Descent and Algebraic Spaces (111) Exercises
(75) Derived Categories of Spaces (112) Guide to Literature
(76) More on Morphisms of Spaces (113) Desirables
(77) Flatness on Algebraic Spaces (114) Coding Style
(78) Groupoids in Algebraic Spaces (115) Obsolete
(79) More on Groupoids in Spaces (116) GNU Free Documentation Li-
(80) Bootstrap cense
(81) Pushouts of Algebraic Spaces (117) Auto Generated Index

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