Anindya wrote:

> The adjunction tells us that there's a unit natural transformation \\(\eta_H : H \Rightarrow \textrm{Lan}_G(H) \circ G\\) but there doesn't seem to be in general any reason to think this is a natural isomorphism...

Probably not in general, but sometimes! In the example of this lecture, it is. And I'm pretty sure we can generalize this fact to a bunch of similar situations. But to find these situations, we need to think about what goes wrong:

**Puzzle.** Find an functor \\(G : \mathcal{D} \to \mathcal{C}\\) where \\(\mathcal{C}\\) and \\(\mathcal{D}\\) have just one object and very few morphisms, and it's _not_ true that left Kan extending a functor \\(H : \mathcal{D} \to \mathbf{Set} \\) along \\(G\\) and then composing it with \\(G\\) gets us back where started, up to natural isomorphism.

> ... let alone an identity of functors.

We should never care whether \\(\textrm{Lan}\_G(H) \circ G\\) is _equal_ to \\(H\\). The reason is that any functor naturally isomorphic to a left (or right) adjoint is again a left (or right) adjoint. So there's not really such a thing as 'the' left Kan extension functor \\(\textrm{Lan}_G\\): any functor naturally isomorphic to this will also have all the properties in the definition of left Kan extension.