In *Locally Presentable and Accessible Categories*, page 12 (10),



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> A topological space is finitely presentable in $\mathbf{Top}$, the category of topological spaces and continuous functions, iff it is finite and discrete.
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But the explanation after this sentence makes little sense to me. In particular, I want a rigorous proof that every finitely presentable object in $\mathbf{Top}$ is finite and discrete.



Why finitely presentable objects in Top need to be discrete?
$\newcommand{\colim}{\operatorname{colim}}$I believe there is a mistake in the argument - as written, the obvious maps $A\to D\_n$ are continuous, so the map $A\to\colim D\_n$ certainly factors through the maps $D\_n\to\colim D\_n$. I also don't believe the claim that $\colim D\_n$ is indiscrete is accurate - all open subsets of $A$ are also open in that colimit.


Instead, the argument should work if instead you modify $D\_n$ so that a subset $U\subseteq A\sqcup\mathbb N$ is open iff $U$ is empty, the whole space or $U$ is *cofinite* and disjoint from $\{0,1,\dots,n-1\}$, with no condition on elemens of $A$.


Now, because $A$ is not discrete, but the subset of $D\_n$ on underlying set of $A$ *is* discrete, we see that the maps $A\to D\_n$ are not continuous. On the other hand, you can check the colimit in this case is indeed indiscrete - since no cofinite subset of $\mathbb N$ can be disjoint from all $\{0,\dots,n-1\}$, the only two sets which are open in each $D\_n$ are the empty set and all of $A\sqcup\mathbb N$. Thus we have a map $A\to\colim D\_n$ which doesn't factor through any of the $D\_n$.


Finiteness is an easier part - any infinite discrete space $X$ can be written as a colimit of its finite subspaces, $X=\colim X\_i$. But $X\to X$ certainly doesn't factor through any of the $X\_i$ unless $X$ itself is finite.


