Anindya wrote:
> K, let's take \\(\mathcal{C}\\) to be the people/friend-of schema from [Lecture 42](https://forum.azimuthproject.org/discussion/2236/lecture-42-chapter-3-transforming-databases).
> 
> And \\(\mathcal{D}\\) is the same category but without the friend-of arrow. There's only one choice for \\(G : \mathcal{D} \rightarrow \mathcal{C}\\), which is the obvious inclusion functor.
Yes, that's the example I had in mind! \\(\mathrm{Lan}_G\\) generates an infinite list of friends for any person, so if start with a database \\(H\\) with just one person, then form \\(\mathrm{Lan}_G(H)\\), then form \\(\mathrm{Lan}_G(H) \circ G\\), we've got an enormous database with that person and an infinite list of others.
> I'm guessing that the issue here is that the inclusion functor \\(G\\) is not full.
That sounds about right. It would be fun to work out necessary and sufficient conditions.
> K, let's take \\(\mathcal{C}\\) to be the people/friend-of schema from [Lecture 42](https://forum.azimuthproject.org/discussion/2236/lecture-42-chapter-3-transforming-databases).
>
> And \\(\mathcal{D}\\) is the same category but without the friend-of arrow. There's only one choice for \\(G : \mathcal{D} \rightarrow \mathcal{C}\\), which is the obvious inclusion functor.
Yes, that's the example I had in mind! \\(\mathrm{Lan}_G\\) generates an infinite list of friends for any person, so if start with a database \\(H\\) with just one person, then form \\(\mathrm{Lan}_G(H)\\), then form \\(\mathrm{Lan}_G(H) \circ G\\), we've got an enormous database with that person and an infinite list of others.
> I'm guessing that the issue here is that the inclusion functor \\(G\\) is not full.
That sounds about right. It would be fun to work out necessary and sufficient conditions.
Version 2.1.8p2
