**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.

finitely presentableis 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. – R. van Dobben de Bruyn Dec 18 '17 at 15:16