# Why finitely presentable objects in Top need to be discrete?

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

A topological space is finitely presentable in $$\mathbf{Top}$$, the category of topological spaces and continuous functions, iff it is finite and discrete.

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.

$$\newcommand{\colim}{\operatorname{colim}}$$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.