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 $\iota:\vert\mathcal{C}\vert\to \mathcal{C}^{op}$!

  • 3
    $\begingroup$ What happens if you take the left Kan extension instead? $\endgroup$ – Andreas Blass Jun 29 '16 at 5:28
  • 2
    $\begingroup$ For every functor $f:\cal D\to\cal C$, the induced functor "precompose with $f$" from presheaves on $\cal C$ to presheaves on $\cal D$ has both adjoints; as @Andreas points out these are given by Kan extensions. Now take $\cal D$ to be discrete category with objects $\cal C$. $\endgroup$ – მამუკა ჯიბლაძე Jun 29 '16 at 6:49

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.