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.

Lovely little lemmas

Or maybe ‘amusing observations’

Classification of compact objects in Top

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