# Finally dense implies dense

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?

• 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. – Ivan Di Liberti Oct 22 '18 at 21:28
• 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. – Philippe Gaucher Oct 24 '18 at 7:28
• @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. – Stefano Nicotra Oct 25 '18 at 12:01

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.