# Local schemes

Consider the following definition. It seems to be standard, although I have not found a place where it is actually spelled out in this way.

Definition. A pointed scheme is local if is contained in every nonempty closed subset of .

Example. If is a local ring, then is a local scheme. Indeed, is contained in every nonempty closed subset , because every strict ideal is contained in .

We prove that this is actually the only example.

Lemma. Let be a local scheme. Then is affine, and is a local ring whose maximal ideal corresponds to the point .

Proof. Let be an affine open neighbourhood of . Then the complement is a closed set not containing , hence . Thus, is affine. Let . Let be a maximal ideal of ; then . Since this contains , we must have , i.e. corresponds to the (necessarily unique) maximal ideal .

# Classification of compact objects in Top

In my previous post, I showed that compact objects in the category of topological spaces have to be finite. Today we improve this to a full characterisation.

Lemma. Let be a topological space. Then is a compact object in if and only if is finite discrete.

This result dates back to Gabriel and Ulmer [GU71, 6.4], as was pointed out to me by Jiří Rosický in reply to my MO question and answer of this account (of which this post is essentially a retelling). Our proof is different from the one given in [GU71], instead using a variant of an argument given in the n-Lab.

Before giving the proof, we construct an auxiliary space against which we will be testing compactness. It is essentially the colimit constructed in the n-Lab, except that we swapped the roles of and (the reason for this will become clear in the proof).

Definition. For all , let be the topological space , where the nonempty open sets are given by for . They form a topology since

Define the map by

This is continuous since equals if and if . Let be the colimit of this diagram.

Since the elements map to the same element in , we conclude that is the two-point space , where the map is the second coordinate projection. Moreover, the colimit topology on is the indiscrete topology. Indeed, neither nor are open.

Proof of Lemma. If is compact, then my previous post shows that is finite. Let be any subset, and let be the indicator function . It is continuous because has the indiscrete topology. Since is a compact object, has to factor through some . Let be the first coordinate projection, i.e.

Let be a number such that for all ; this exists because is finite. Then , which shows that is open. Since was arbitrary, we conclude that is discrete.

Conversely, every finite discrete space is a compact object. Indeed, any map out of is continuous, and finite sets are compact in .

[GU71] Gabriel, Peter and Ulmer, Friedrich, Lokal präsentierbare Kategorien. Lecture Notes in Mathematics 221. Springer-Verlag, Berlin-New York, 1971. DOI: 10.1007/BFb0059396.

# Compact objects in the category of topological spaces

We compare two competing notions of compactness for topological spaces. Besides the usual notion, there is the following:

Definition. Let be a cocomplete category. Then an object is compact if commutes with filtered colimits.

Exercise. An -module is compact if and only if it is finitely generated.

We want to study compact objects in the category of topological spaces. One would hope that this corresponds to compact topological spaces. However, this is very far off:

Lemma. Let be a compact object. Then is finite.

Proof. Let be the set with the indiscrete topology, i.e. . It is the union of all its finite subsets, and this gives it the colimit topology because a subset is open if and only if its intersection with each finite subset is. Indeed, if were neither nor , then there exist with and . But then is not open, because inherits the indiscrete topology from .

Therefore, if is a compact object, then the identity map factors through one of these finite subsets, hence is finite.

However, the converse is not true. In fact the indiscrete space on a two element set is not a compact object, as is explained here.

Corollary. Let be a compact object. Then is a compact topological space.

Proof. It is finite by the lemma above. Every finite topological space is compact.

Originally, this post relied on the universal open covering of my previous post to show that a compact object in is compact; however the above proof shows something much stronger.

# A fun example of a representable functor

This post is about representable functors:

Definition. Let be a functor. Then is representable if it is isomorphic to for some . In this case, we say that represents .

Exercise. If such exists, then it is unique up to unique isomorphism.

Really one should encode the isomorphism as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation is uniquely determined by the element of corresponding to the identity of .

When is a natural isomorphism, the corresponding element is called the universal object of . It has the property that for every and any , there exists a unique morphism such that .

Example. The forgetful functor is represented by . Indeed, the natural map

is an isomorphism. The universal element is .

Example. Similarly, the forgetful functor is represented by . The universal element is .

A fun exercise (for the rest of your life!) is to see whether functors you encounter in your work are representable. See for example this post about some more geometric examples.

The main example for today is the following:

Lemma. The functor that associates to a topological space its topology is representable.

Proof. Consider the topological space with topology . Then there is a natural map

Conversely, given an open set , we can associate the characteristic function . This gives an inverse of the map above.

The space we constructed is called the Sierpiński space. The universal open set is .

Remark. The space represents the data of open sets for : for any continuous map , we have , where . If denotes the complementary open, then the form a cover of if and only if . This corresponds to the statement that lands in .

Thus, the open cover is the universal open cover, i.e. for every open covering there exists a unique continuous map such that .