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.