# Not every open immersion is an open immersion

An immersion (or locally closed immersion) of schemes is a morphism that can be factored as , where is a closed immersion and is an open immersion. If it is moreover an open morphism, it need not be an open immersion:

Example. Let be a nonreduced scheme, and let be the reduction. This is a closed immersion, whose underlying set is the entire space. Thus, it is a homeomorphism, hence an open morphism. It is not an open immersion, for that would force it to be an isomorphism.

Remark. However, every closed immersion is a closed immersion; see Tag 01IQ.

# Higher pushforwards along finite morphisms

This post is about one of my favourite answers I have given on MathOverflow, although it seems to have gone by mostly unnoticed. In the post, Qixiao asks (essentially) the following:

Question. If is a finite morphism of schemes, is the pushforward exact?

Note that this is true on the subcategory of quasicoherent sheaves because affine morphisms have no quasicoherent higher pushforwards. Also, in the étale topology the pushforward along a finite morphism is exact on the category of all abelian sheaves; see e.g. Tag 03QP.

However, we show that the answer to the question above is negative.

Example. Let be the spectrum of a DVR , let be a finite extension of domains such that has exactly two primes above , and let . For example, and , or and if you prefer a more geometric example.

By my previous post, the global sections functor is exact. If the same were true for , then the global sections functor on would be exact as well. Thus, it suffices to prove that this is not the case, i.e. to produce a surjection of sheaves on such that the map on global sections is not surjective.

The topological space of consists of closed points and a generic point . Let and ; then is open and is closed. Hence, for any sheaf on , we have a short exact sequence (see e.g. Tag 02UT)

where and are the inclusions. Let be the constant sheaf ; then the same goes for and . Then the map

is given by the diagonal map , since is connected by has two connected components. This is visibly not surjective.

# Cohomology of a local scheme

This is a continuation of my previous post on local schemes. Here is a ridiculous lemma.

Lemma. Let be a local scheme, and let be any abelian sheaf on . Then for all .

Proof. It suffices to show that the global sections functor is exact. Let be a surjection of abelian sheaves on , and let be a global section. Then can be lifted to a section of in an open neighbourhood of . But the only open neighbourhood of is . Thus, can be lifted to a section of .

What’s going on is that the functors and are naturally isomorphic, due to the absence of open neighbourhoods of .

Remark. It seems believable that there are suitable site-theoretic versions of this lemma as well. For example, a strictly Henselian local ring has no higher cohomology in the étale topology. The argument is essentially the same: every open neighbourhood of the closed point has a section; see e.g. the proof of Tag 03QO.

# 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 .

# Furstenberg’s proof of the infinitude of the primes

Another very classical proof that’s just too beautiful to ignore. This time, the theorem is millennia old, but it’s really the proof that I’m interested in.

Lemma. There are infinitely many primes.

Proof. Define the sets

for with . Note that if ; say it contains some , then

Hence, the intersection of two sets of this form is again of this form (or empty). Thus, they form the basis for a topology on . Notice that the sets

are also open.

Now suppose that there were finitely many primes. Then the intersection

is a finite intersection of opens, hence open. On the other hand, it equals the set of integers divisible by no prime number, which is . But all basic open sets are infinite, so this set can never be open.

Remark. This proof is in essence the usual proof, where we consider and conclude that some prime has to divide it and this prime can be none of the . Indeed, when we showed that the intersection of basic opens is open, we went to a set whose period is (which for gives period ). Since 1 is in , so should be.

That said, it doesn’t seem possible to adopt this proof to reprove other theorems like ‘there are infinitely many primes congruent to 3 modulo 4’. The problem is that the above proof seems to rely on a global analysis of the set : it has to be an infinite set. So the method is too crude to prove theorems with congruence restrictions.

# Separation properties for topological groups

Although this is quite a classical result, I really like it.

Lemma. Let be a topological group. Then is if and only if is Hausdorff.

Proof. One implication is clear. Conversely, suppose is . Then the identity element is closed. The map

is continuous. Hence, the inverse image of the identity is closed. But this is the diagonal, hence is Hausdorff.

Exercise. Prove that Hausdorff is in fact equivalent to .