# 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.

# 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 .

# Cardinality of fraction field

This is a ridiculous lemma that I came up with.

Lemma. Let be a (commutative) ring, and let be its total ring of fractions. Then and have the same cardinality.

Proof. If is finite, my previous post shows that . If is infinite, then is a subquotient of , hence . But injects into , so .

Corollary. If is a domain, then .

Proof. This is a special case of the lemma.