I am reading the article "A convenient category for directed homotopy" by Fajstrup and Rosicky and I have a doubt about the proof of Proposition 3.5. The setting is the following: let $\cal{C}$ be a concrete category, with forgetful functor $U\colon \cal{C}\to \mathsf{Set}$.

Definition 1. A full subcategory $\cal{I}$ of $\cal{C}$ is said to be finally dense in $\cal{C}$ if, for every object $X$ in $\cal{C}$ there exists a family of maps $(f_\lambda\colon C_{\lambda}\to X)_{\lambda\in \Lambda}$ such that a map $g\colon UX\to UY$ lifts to a morphism $\bar{g}\colon X\to Y$ iff $g\circ Uf_\lambda$ lifts to a morphism $\bar{g_\lambda}$ for every $\lambda$.

Definition 2. A full subcategory $\cal{I}$ of $\cal{C}$ is said to be dense in $\cal{C}$ if every object $X$ in $\cal{C}$ is isomorphic to the colimit of the canonical functor $\mathcal{I}/X\to \cal{C}$.

In Proposition 3.5 of the aforementioned paper, $\cal{C}$ is actually the category $\cal{K}_{\cal{I}}$ of $\cal{I}$-generated objects, for a topological construct $\cal{K}$ and a full subcategory $\cal{I}$ of $\cal{K}$. In particular, this means that we have a forgetful functor $U\colon\cal{K}\to \mathsf{Set}$ which has a right and a left adjoint, that associate to every set $S$ the discrete and indiscrete object on $S$, respectively. The statement I'm a bit puzzled about is that, "since $\cal{I}$ is finally dense in $\cal{C}$, then it is dense."

I can see why this should be true if we could show that, for any $X\in \cal{K}$ the natural map $$\mathrm{colim}_{C\to X}UC\to UX$$ is a bijection. If either $\cal{I}$ contains the discrete object on the terminal object, or the discrete object and the indiscrete object on the terminal object conincide, then I know how to conclude, since the aformentioned cocone would contain all the points of $UX$ (In the second case it is enough to consider all constant maps $C\to *_I=*_D\to X$. However, in a more general setting I don't know why such a statement should hold.

Here comes the question:

Am I missing something? Is there a reason why the above statement should hold in a more general setting?

If the answer is no, do you have any example of a topological category where UX and the canonical colimit displayed above do not coincide?

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    $\begingroup$ This is not a proof, nor a formal argument, but I see an analogy. When in a cocomplete category you have a strong generator made by $\lambda$-presentable objects, its closure under $\lambda$-small colimits is dense. This appears as 1.11 in LPAC. $\endgroup$ Oct 22, 2018 at 21:28
  • $\begingroup$ As far as I understand your post, $\mathcal{K}_\mathcal{I}$ is a full coreflective subcategory of $\mathcal{K}$. The coreflector is given by the canonical colimit, which is the identity on $\mathcal{K}_\mathcal{I}$. Since $\mathcal{I}$ is finally dense, $\mathcal{K}_\mathcal{I}=\mathcal{K}$, and therefore $\mathcal{I}$ is dense. $\endgroup$ Oct 24, 2018 at 7:28
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    $\begingroup$ @PhilippeGaucher I don't think $\cal{K}_{\cal{I}}$ is going to be equal to $\cal{K}$ in general, here $\cal{I}$ is finally dense in $\cal{K}_{\cal{I}}$, not in $\cal{K}$. My problem is really about why the coreflector is given by the canonical colimit. Or better, if you want, how to prove that the underlying set of $X$ is equal to the underlying set of the colimit, when we don't require $\cal{K}$ to be well fibered. $\endgroup$ Oct 25, 2018 at 12:01

1 Answer 1


You are right, in general $\mathcal I$ should contain the discrete object $D_1$ on the singleton. A general result is in my paper Codensity and binding categories (Theorem 1.3). In Proposition 3.5 of the aforementioned paper, $D_1$ is $\mathcal I$-generated and thus it can be added to $\mathcal I$ without changing $\mathcal I$-generated objects. Then Proposition 3.5 follows.


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