Puzzle 124: I don't think there's just one functor from D to C. There is just one graph homomorphism, though. (Feel free to delete this paragraph after reading it.)
I'm excited about Kan extensions! I've come across them before but I've never been able to make a gut connection with them on the level of adjoints and limits, which they apparently generalize.
In the discussion after [Lecture 38](https://forum.azimuthproject.org/discussion/comment/18986/#Comment_18986), we were looking at some slides from a talk by David Spivak about databases, and he mentioned the Grothendieck construction that converts an instance \\(I:\mathcal{D}\to \mathbf{Set}\\) to another category \\(\mathrm{Gr}(I)\\) equipped with a functor \\(\mathrm{Gr}(I)\to \mathcal{D}\\). It would be easy to compose this with the functor \\(\mathcal{D}\to\mathcal{C}\\) and get at least what he calls a "semi-instance" \\(\mathrm{Gr}(I) \to \mathcal{C}\\). Does this semi-instance correspond to the actual instance \\(\mathcal{C}\to \mathbf{Set}\\) we'll get with Kan extensions later?
I'm excited about Kan extensions! I've come across them before but I've never been able to make a gut connection with them on the level of adjoints and limits, which they apparently generalize.
In the discussion after [Lecture 38](https://forum.azimuthproject.org/discussion/comment/18986/#Comment_18986), we were looking at some slides from a talk by David Spivak about databases, and he mentioned the Grothendieck construction that converts an instance \\(I:\mathcal{D}\to \mathbf{Set}\\) to another category \\(\mathrm{Gr}(I)\\) equipped with a functor \\(\mathrm{Gr}(I)\to \mathcal{D}\\). It would be easy to compose this with the functor \\(\mathcal{D}\to\mathcal{C}\\) and get at least what he calls a "semi-instance" \\(\mathrm{Gr}(I) \to \mathcal{C}\\). Does this semi-instance correspond to the actual instance \\(\mathcal{C}\to \mathbf{Set}\\) we'll get with Kan extensions later?
Version 2.1.8p2
