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.

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).

Cardinality of fraction field

This is a ridiculous lemma that I came up with.

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

Proof. If R is finite, my previous post shows that R = K. If R is infinite, then K is a subquotient of R^2, hence |K| \leq |R^2| = |R|. But R injects into K, so |R| \leq |K|. \qedsymbol

Corollary. If R is a domain, then |\operatorname{Frac}(R)| = |R|.

Proof. This is a special case of the lemma. \qedsymbol