Let $\mathscr C$ be a locally small category that has filtered colimits. Then an object $X$ in $\mathscr C$ is compact if $\operatorname{Hom}(X,-)$ commutes with filtered colimits.

On the other hand, the category of topological spaces has a competing notion of compactness. Not every compact topological space is a compact object in $\operatorname{\underline{Top}}$, as is explained here. Todd Trimble asked (in the $n$-category café) if the situation is any better if $X$ is assumed compact Hausdorff.

More generally, is there some sort of classification of compact objects in $\operatorname{\underline{Top}}$?

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    $\begingroup$ This is why I've always disliked the use of the word "compact" for the property in your first paragraph. An alternative, which also has a longer weight of tradition behind it, is "finitely presentable". $\endgroup$ Dec 18, 2017 at 6:47
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    $\begingroup$ The presentable objects in the category of general topological spaces are the discrete ones (it's a very known joke). A topological space is $\lambda$-presentable if and only if it is discrete of cardinal less than $\lambda$. Your question is the case $\lambda=\aleph_0$. $\endgroup$ Dec 18, 2017 at 8:39
  • $\begingroup$ @MikeShulman: for people working with derived categories (say of $A$-modules), the word finitely presentable is not great because it already has a concrete meaning for $A$-modules. This might be why different terminology was chosen, although admittedly I don't know the history. $\endgroup$ Dec 18, 2017 at 15:16
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    $\begingroup$ @R.vanDobbendeBruyn In almost all cases I'm aware of, the abstract meaning coincides with the concrete meaning. In derived categories in the homotopy-category sense (e.g. triangulated categories) directed colimits don't generally exist, so you're talking about a different notion anyway. Are you saying there is a category that has real honest directed colimits and a concrete notion of "finitely presentable" that doesn't coincide with this abstract one? $\endgroup$ Dec 18, 2017 at 23:19
  • $\begingroup$ @MikeShulman: ah, you're absolutely right. I was thinking of this definition, but this is characterising a different property. $\endgroup$ Dec 19, 2017 at 0:37

1 Answer 1


Proposition. Let $X$ be a topological space. Then $X$ is a compact object if and only if $X$ is a finite discrete space.

Before giving the proof, we state an easy lemma.

Lemma. Suppose $X$ is a compact object in $\operatorname{\underline{Top}}$. Then $X$ is finite.

Proof. Let $Y$ be the indiscrete topological space with underlying set $X$; i.e. $\mathcal T_Y = \{\varnothing,Y\}$. It is the union of 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 every finite subset is. Indeed, if $U$ were neither $\varnothing$ nor $Y$, then there exist $y_1, y_2 \in Y$ such that $y_1 \in U$ 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 $Y$.

Since $X$ is compact, the identity map $X \to Y$ factors through a finite subset of $Y$. This forces $X$ to be finite. $\square$

For the remainder of the proof, we will use the auxiliary construction of a specific colimit given here. We will recall the notation. For reasons that become clear later, we have swapped the roles of $0$ and $1$.

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 Proposition. It's easy to check that finite discrete spaces are compact: any map out of them is continuous, and finite sets are compact in $\operatorname{\underline{Set}}$.

Conversely, if $X$ is compact, then $X$ is finite by the Lemma. Let $U \subseteq X$ be any subset, and let $f \colon X \to X_\infty = \{0,1\}$ be indicator function $\mathbb 1_U$; this is continuous because $X_\infty$ has the indiscrete topology. Since $X$ is a compact object, there exists $n \in \mathbb N$ such that $f$ comes from a map $g \colon X \to X_n$. Let $h \colon X \to X_n \to \mathbb N_{\geq n}$ 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 $m \in \mathbb N_{\geq n}$ be a number larger than $h(x)$ for all $x \in X\setminus U$ (we can do this because $X$ is finite). Then $g^{-1}(U_{n,m}) = U$, hence $U$ is open. Since $U$ was arbitrary, we conclude that $X$ is discrete. $\square$

Remark. We have only used the surjectivity part of the natural map \begin{equation} \operatorname{colim}_i \operatorname{Hom}(X,X_i) \to \operatorname{Hom}(X,\operatorname{colim}_i X_i).\label{1}\tag{1} \end{equation} In particular, we see that in $\operatorname{\underline{Top}}$, surjectivity for all systems $X_i$ implies injectivity for all systems. This is not completely formal; for example for the system $X_n$ above, the maps $$f,g \colon \mathbb N^{\operatorname{disc}} \to X_0$$ given by $f(x) = (x,0)$ and $g(x) = (x+1,0)$ give the same morphism $\mathbb N^{\operatorname{disc}} \to X_\infty$, but they don't give the same morphism $\mathbb N^{\operatorname{disc}} \to X_n$ for any $n$. Thus, the map in (\ref{1}) is not injective in general. I don't know in what generality surjectivity implies injectivity.

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    $\begingroup$ This is due to Gabriel and Ulmer, Lokal prasentierbare Kategorien (see 6.4). It is also in my book with Adámek Locally Presentable and Accessible Categories (see 1.2(10)). $\endgroup$ Dec 17, 2017 at 12:25
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    $\begingroup$ A natural question arises then (whether it deserves a separate MO entry I don't know) - can compact spaces be characterized by an abstract categorical property inside Top? $\endgroup$ Dec 17, 2017 at 17:12
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    $\begingroup$ A related question: I've always wondered what the compact objects in the category of locales are. The proof above fails in the case of locales already in the first step (Lemma): there is no such thing as an indiscrete locale. $\endgroup$ Dec 18, 2017 at 3:06
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    $\begingroup$ @მამუკაჯიბლაძე A space $X$ is compact if and only if it is a compact object in the category of open subsets of $X$: it's Proposition 2.6 of ncatlab.org/nlab/show/compact+space. $\endgroup$ Dec 18, 2017 at 9:31
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    $\begingroup$ @PhilippeGaucher: ah, and a way to find the Sierpiński space $S$ categorically is as the unique two-point space for which the transposition is not continuous (this is a purely categorical characterisation). Then the set of open subsets of $X$ can be found as the set of morphisms $X \to S$, and you can also find the inclusion order on this set from this characterisation. $\endgroup$ Dec 18, 2017 at 15:23

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