# What are compact objects in the category of topological spaces?

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 $\operatorname{\underline{Top}}$?

• 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". Dec 18, 2017 at 6:47
• 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 $\lambda=\aleph_0$. Dec 18, 2017 at 8:39
• @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. Dec 18, 2017 at 15:16
• @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? Dec 18, 2017 at 23:19
• @MikeShulman: ah, you're absolutely right. I was thinking of this definition, but this is characterising a different property. Dec 19, 2017 at 0:37

## 1 Answer

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 $$\operatorname{colim}_i \operatorname{Hom}(X,X_i) \to \operatorname{Hom}(X,\operatorname{colim}_i X_i).\label{1}\tag{1}$$ 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.

• 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)). Dec 17, 2017 at 12:25
• 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? Dec 17, 2017 at 17:12
• 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. Dec 18, 2017 at 3:06
• @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. Dec 18, 2017 at 9:31
• @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. Dec 18, 2017 at 15:23