Consider the inclusion of presheaves on $\mathbb{C}$ into families of sets indexed by $\mathbb{C}$-objects (which proceeds by forgetting the action on morphisms). Is there a left adjoint to this inclusion functor?


I thought that I had constructed something that seemed reasonable, but now I am doubtful whether it is in fact adjoint.


EDIT: I have just realized that whilst it may not have a left adjoint (does anyone know?), it at least has a right adjoint (which is in fact the thing that I had originally constructed, hoping it was the left adjoint). Basically, take the right kan extension of a presheaf along the inclusion $\iota:\vert\mathcal{C}\vert\to \mathcal{C}^{op}$!



Does the inclusion of presheaves into families of sets have a left adjoint?
Although I wouldn't call it an inclusion functor, the answer is yes and in fact the forgetful functor $Set^{C^{op}} \to Set/C\_0$ is monadic (as well as comonadic). 


I think the most illuminating way to see this is to regard a presheaf as a set $F: X \to C\_0$ over $C\_0$, equipped with a $C$-action which is a map $C\_1 \times\_{C\_0} F \to F$ satisfying suitable axioms. If $c \in C\_0$ and $F c$ represents the fiber, then over $c$ the action is a function of the form 


$$\sum\_d \hom(c, d) \times F d = \sum\_d \sum\_{f: c \to d} F d \to F c,$$ 


a many-object generalization of the way that an $M$-set over a monoid $M$ is given by a map $M \times X \to X$ satisfying suitable conditions. Just as $M \times X$ is the (underlying set of the) free $M$-set on $X$, so the domain described above, or more precisely the pullback $C\_1 \times\_{C\_0} X$ as in the diagram 


$$\begin{array}{ccc}
C\_1 \times\_{C\_0} X & \to & X \\
\downarrow & & \downarrow F \\ 
C\_1 & \underset{d\_1}{\to} & C\_0,
\end{array}
$$ 


is the (underlying indexed set of the) free $C$-module on $F: X \to C\_0$. This is a presheaf over $C$ in an evident way: the requisite $C$-action is given by 


$$C\_1 \times\_{C\_0} C\_1 \times\_{C\_0} X \stackrel{\text{comp} \times\_{C\_0} X}{\longrightarrow} C\_1 \times\_{C\_0} X$$ 


Summarizing: the functor given on objects by $(F: X \to C\_0) \mapsto C\_1 \times\_{C\_0} X$, with this $C$-action just described, gives the left adjoint $Set/C\_0 \to Set^{C^{op}}$ to the forgetful functor. 


This is covered in lots of places, for example Mac Lane and Moerdijk's Sheaves in Geometry and Logic. 


