OK, 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.

Now consider a very simple database \\(H : \mathcal{D} \rightarrow \textbf{Set}\\) that sends the object of \\(\mathcal{D}\\) to the set \\(\\{\textrm{Alice}\\}\\). What is its left Kan extension \\(\textrm{Lan}_G (H) : \mathcal{D} \rightarrow \textbf{Set}\\)?

It has to send the object of \\(\mathcal{C}\\) to a set \\(P\\) of people, but also send the friend-of arrow in \\(\mathcal{C}\\) to an endomap on \\(P\\) – and do this in a "free" way.

That means we need \\(\textrm{Alice} \in P\\), but then we need to invent a friend for her, say \\(\textrm{Friend}_0 \in P\\), and a friend for him \\(\textrm{Friend}_1 \in P\\), and a friend for her \\(\textrm{Friend}_2 \in P\\)...

So \\(P\\) is a countably infinite set (isomorphic to the natural numbers). There's a natural inclusion map \\(\\{ \textrm{Alice} \\} \rightarrow P\\) but these sets certainly aren't iso.

Hence \\(\textrm{Lan}_G (H) \circ G\\) is *not* iso to \\(H\\).

I'm guessing that the issue here is that the inclusion functor \\(G\\) is not full.