John wrote:

> An object of \$$\mathrm{Gr}(I)\$$ is an object \$$d\$$ of \$$\mathcal{D}\$$ together with an element of the set \$$I(d)\$$. So, a functor \$$\mathrm{Gr}(I) \to \mathcal{C}\$$ takes an object \$$d\$$ of \$$\mathcal{D}\$$ together with an element of \$$I(d)\$$ and spits out an object in \$$\mathcal{C}\$$. On the other hand, a functor \$$\mathcal{C}\to \mathbf{Set}\$$ takes an object of \$$\mathcal{C}\$$ and spits out a set. These are pretty darn different things to do!

That's true, but there's a way to at least make the types match up: We can ask if the following square commutes (for a suitable choice of left or right Kan extension):

$\begin{array}{ccccc} & & \text{Grothendieck construction} & & \\\\ \text{Left/right} & \text{Functors }\mathcal{D}\to\mathbf{Set} & \to & \text{Functors to }\mathcal{D} & \\\\ \text{Kan extension along } & \downarrow & & \downarrow & \text{Compose with }\\\\ \mathcal{D}\to\mathcal{C} & \text{Functors }\mathcal{C}\to\mathbf{Set} & \to & \text{Functors to }\mathcal{C} & \mathcal{D}\to\mathcal{C} \\\\ & & \text{Grothendieck construction} & & \end{array}$

It's bizarre to me that (certain types of) functors to \$$\mathcal{C}\$$ correspond to functors \$$\mathcal{C}\to\mathbf{Set}\$$. The former can be easily pushed forward along functors \$$\mathcal{C}\to\mathcal{C}'\$$; the latter can be easily pulled *back* along functors \$$\mathcal{C}'\to\mathcal{C}\$$. Usually I expect at most one of these to be easy!