Not every open immersion is an open immersion

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

Example. Let X be a nonreduced scheme, and let X_{\text{red}} \to X 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. \qedsymbol

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 f \colon X \to Y is a finite morphism of schemes, is the pushforward f_* \colon \Sh(X) \to \Sh(Y) 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 Y be the spectrum of a DVR (R,\mathfrak m), let R \to S be a finite extension of domains such that S has exactly two primes \mathfrak p, \mathfrak q above \mathfrak m, and let X = \Spec S. For example, R = \Z_{(5)} and S = \Z_{(5)}[i], or R = k[x]_{(x)} and S = k[x]_{(x)}[\sqrt{x+1}] if you prefer a more geometric example.

By my previous post, the global sections functor \Gamma \colon \Sh(Y) \to \Ab is exact. If the same were true for f_* \colon \Sh(X) \to \Sh(Y), then the global sections functor on X would be exact as well. Thus, it suffices to prove that this is not the case, i.e. to produce a surjection \mathscr F \to \mathscr G of sheaves on X such that the map on global sections is not surjective.

The topological space of X consists of closed points x,y and a generic point \eta. Let U = \{\eta\} and Z = U^{\operatorname{c}} = \{x,y\}; then U is open and Z is closed. Hence, for any sheaf \mathscr F on X, we have a short exact sequence (see e.g. Tag 02UT)

    \[0 \to j_! (\mathscr F|_U) \to \mathscr F \to i_* (\mathscr F|_Z) \to 0,\]

where j \colon U \to X and i \colon Z \to X are the inclusions. Let \mathscr F be the constant sheaf \Z; then the same goes for \mathscr F|_U and \mathscr F|_Z. Then the map

    \[H^0(X,\mathscr F) \to H^0(X,i_*(\mathscr F|_Z)) = H^0(Z,\mathscr F|_Z)\]

is given by the diagonal map \Z \to \Z \oplus \Z, since X is connected by Z has two connected components. This is visibly not surjective. \qedsymbol

Cohomology of a local scheme

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

Lemma. Let (X,x) be a local scheme, and let \mathscr F be any abelian sheaf on X. Then H^i(X,\mathscr F) = 0 for all i > 0.

Proof. It suffices to show that the global sections functor \Gamma \colon \Sh(X) \to \Ab is exact. Let \mathscr F \to \mathscr G be a surjection of abelian sheaves on X, and let s \in \mathscr G(X) be a global section. Then s can be lifted to a section of \mathscr F in an open neighbourhood U of x. But the only open neighbourhood of x is X. Thus, s can be lifted to a section of \mathscr F(X). \qedsymbol

What’s going on is that the functors \mathscr F \mapsto \Gamma(X,\mathscr F) and \mathscr F \mapsto \mathscr F_x are naturally isomorphic, due to the absence of open neighbourhoods of x.

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 (X,x) is local if x is contained in every nonempty closed subset of X.

Example. If (A,\mathfrak m) is a local ring, then (\Spec A,\mathfrak m) is a local scheme. Indeed, \mathfrak m is contained in every nonempty closed subset V(I) \subseteq X, because every strict ideal I \subsetneq A is contained in \mathfrak m.

We prove that this is actually the only example.

Lemma. Let (X,x) be a local scheme. Then X is affine, and A = \Gamma(X,\mathcal O_X) is a local ring whose maximal ideal corresponds to the point x \in X = \Spec A.

Proof. Let U be an affine open neighbourhood of x. Then the complement V is a closed set not containing x, hence V = \varnothing. Thus, X = U is affine. Let A = \Gamma(X,\mathcal O_X). Let \mathfrak m be a maximal ideal of A; then V(\mathfrak m) = \{\mathfrak m\}. Since this contains x, we must have x = \mathfrak m, i.e. x corresponds to the (necessarily unique) maximal ideal \mathfrak m \subseteq A. \qedsymbol

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 X be a topological space. Then X is a compact object in \operatorname{\underline{Top}} if and only if X 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 0 and 1 (the reason for this will become clear in the proof).

Definition. For all n \in \mathbb N, let X_n be the topological space \mathbb N_{\geq n} \times \{0,1\}, where the nonempty open sets are given by U_{n,m} = \mathbb N_{\geq m} \times \{0\} \cup \mathbb N_{\geq n} \times \{1\} for m \geq n. They form a topology since

    \begin{align*} U_{n,m_1} \cap U_{n,m_2} &= U_{n, \max(m_1,m_2)}, \\ \bigcup_i U_{n,m_i} &= U_{n,\min\{m_i\}}. \end{align*}

Define the map f_n \colon X_n \to X_{n+1} by

    \[(x,\varepsilon) \mapsto \left\{\begin{array}{ll} (x,\varepsilon), & x > n, \\ (n+1,\varepsilon), & x = n. \end{array}\right.\]

This is continuous since f_n^{-1}(U_{n+1,m}) equals U_{n,m} if m > n+1 and U_{n,n} if m = n+1. Let X_\infty be the colimit of this diagram.

Since the elements (x,\varepsilon), (y,\varepsilon) \in X_n map to the same element in X_{\max(x,y)}, we conclude that X_\infty is the two-point space \{0,1\}, where the map X_n \to X_\infty = \{0,1\} is the second coordinate projection. Moreover, the colimit topology on \{0,1\} is the indiscrete topology. Indeed, neither \mathbb N_{\geq n} \times \{0\} \subseteq X_n nor \mathbb N_{\geq n} \times \{1\} \subseteq X_n are open.

Proof of Lemma. If X is compact, then my previous post shows that X is finite. Let U \subseteq X be any subset, and let f \colon X \to X_\infty = \{0,1\} be the indicator function \mathbb I_U. It is continuous because X_\infty has the indiscrete topology. Since X is a compact object, f has to factor through some g \colon X \to X_n. Let h \colon X \to X_n \to \N_{\geq n} be the first coordinate projection, i.e.

    \[g(x) = \left\{\begin{array}{ll}(h(x),1), & x \in U, \\ (h(x),0), & x \not\in U. \end{array}\right.\]

Let m \in \N_{\geq n} be a number such that m > h(x) for all x \not\in U; this exists because X is finite. Then g^{-1}(U_{n,m}) = U, which shows that U is open. Since U was arbitrary, we conclude that X is discrete.

Conversely, every finite discrete space X is a compact object. Indeed, any map out of X is continuous, and finite sets are compact in \operatorname{\underline{Set}}. \qedsymbol

[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 \mathscr C be a cocomplete category. Then an object X \in \ob \mathscr C is compact if \Hom(X,-) commutes with filtered colimits.

Exercise. An R-module M 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 X \in \Top be a compact object. Then X is finite.

Proof. Let Y be the set X with the indiscrete topology, i.e. \mathcal T_Y = \{\varnothing, Y\}. It is the union of all its finite subsets, and this gives it the colimit topology because a subset U \subseteq Y is open if and only if its intersection with each finite subset is. Indeed, if U were neither \varnothing nor Y, then there exist y_1, y_2 \in Y with y_1 \in U and y_2 \not\in U. But then U \cap \{y_1,y_2\} is not open, because \{y_1,y_2\} inherits the indiscrete topology from Y.

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

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 X \in \Top be a compact object. Then X is a compact topological space.

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

Originally, this post relied on the universal open covering of my previous post to show that a compact object in \Top 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 F \colon \mathscr C \to \Set be a functor. Then F is representable if it is isomorphic to \Hom(A,-) for some A \in \ob \mathscr C. In this case, we say that A represents F.

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

Really one should encode the isomorphism \Hom(A,-) \stackrel\sim\to F as well, but this is often dropped from the notation. By the Yoneda lemma, every natural transformation \Hom(A,-) \to F is uniquely determined by the element of F(A) corresponding to the identity of A.

When \Hom(A,-) \to F is a natural isomorphism, the corresponding element a \in F(A) is called the universal object of F. It has the property that for every B \in \mathscr C and any b \in F(B), there exists a unique morphism f \colon A \to B such that f(a) = b.

Example. The forgetful functor \Ab \to \Set is represented by \Z. Indeed, the natural map

    \begin{align*} \Hom(\Z,M) &\to M\\ f &\mapsto f(1) \end{align*}

is an isomorphism. The universal element is 1 \in \Z.

Example. Similarly, the forgetful functor \Ring \to \Set is represented by \Z[x]. The universal element is x.

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 \Top\op \to \Set that associates to a topological space (X,\mathcal T_X) its topology \mathcal T_X is representable.

Proof. Consider the topological space Y = \{0,1\} with topology \{\varnothing, \{1\},\{0,1\}\}. Then there is a natural map

    \begin{align*} \Hom(X,Y) &\to \mathcal T_X\\ f &\mapsto f^{-1}(\{1\}). \end{align*}

Conversely, given an open set U, we can associate the characteristic function \mathbb I_U. This gives an inverse of the map above. \qedsymbol

The space Y we constructed is called the Sierpiński space. The universal open set is \{1\}.

Remark. The space Y^I represents the data of open sets U_i for i \in I: for any continuous map f \colon X \to Y^I, we have U_i = f^{-1}(Y_i), where Y_i = \pi_i^{-1}(\{1\}) \subseteq Y^I. If Z_i denotes the complementary open, then the U_i form a cover of X if and only if \bigcap_{i \in I} Z_i = \varnothing. This corresponds to the statement that f lands in Y^I\setminus\{(0,0,\ldots)\}.

Thus, the open cover Y^I\setminus\{0\} = \bigcup_{i \in I} Y_i is the universal open cover, i.e. for every open covering X = \bigcup U_i there exists a unique continuous map f \colon X \to Y^I\setminus\{0\} such that U_i = f^{-1}(Y_i).

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

    \[ U_{a,n} = a + n\Z = \{x \in \Z\ |\ x \equiv a \pmod{n}\}, \]

for a, n \in \Z with n \neq 0. Note that if U_{a,n} \cap U_{b,m} \neq \varnothing; say it contains some c \in \Z, then

    \[ U_{a,n} \cap U_{b,m} = U_{c, \lcm (n,m)}. \]

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

    \[ U_{a,n}\comp = \bigcup_{b \not \equiv a} U_{b,n} \]

are also open.

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

    \[ V = \bigcap_{p \text{ prime}} U_{0,p}\comp \]

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 \{1, -1\}. But all basic open sets are infinite, so this set can never be open. \qedsymbol

Remark. This proof is in essence the usual proof, where we consider p_1 \cdots p_n + 1 and conclude that some prime has to divide it and this prime can be none of the p_i. Indeed, when we showed that the intersection of basic opens is open, we went to a set whose period is \lcm(n,m) (which for V gives period p_1 \cdots p_n). Since 1 is in V, so should p_1 \cdots p_n + 1 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 V: 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 G be a topological group. Then G is T_1 if and only if G is Hausdorff.

Proof. One implication is clear. Conversely, suppose G is T_1. Then the identity element is closed. The map

    \begin{align*} G\times G &\rA G\\ (g,h) &\rM gh^{-1} \end{align*}

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

Exercise. Prove that Hausdorff is in fact equivalent to T_0.