**Exercise 6.2.24**

**Theorem**:
Every slice of a presheaf category is equivalent to another presheaf category.
In particular, if , then

where is the category of elements of .

*Proof*:
We will define a functor .
Take a pair .
is a contravariant functor .
Given an , we let – that is, the inverse image of under .
On a map let simply be the restriction .
Consider a map in ; that is with .
is a natural transformation with components .
Simply let , again the restriction.
This is well defined – notice that if , then , thus , and thus .
Naturality is also easy to verify – since components of are just restrictions of components of , the appropriate squares still commute.

We claim that is full, faithful, and essentially surjective, making it an equivalence. Let . Suppose we have maps such that . This means that for all . Then and are equal when restricted to , for all . But for every , , so we must have that and are equal on all of – thus for all and . Thus is faithful.

Suppose that is some natural transformation , having components . Consider each as a map . Again since , we have the unique map . Indeed, these ’s form the components of a natural transformation . To verify naturality, let be an arrow. We want the following to commute:

Indeed, for every , we can form the diagram given by restricting each side of the above diagram:

which is actually equivalent to

which commutes precisely because is natural. Since every such restriction diagram commutes, the previous diagram does as well. Finally, it’s clear that and . So, is full.

Finally, let’s show is essentially surjective. Let be a presheaf. We first define a presheaf . Let . For a map , let , where is being considered as a map . We now need a natural transformation . Let be the natural projection sending a member of to its index in the sum. To check naturality, we need the following to commute for an :

but if we restrict to a single the constituent part of the diagram becomes:

which trivially commutes (since and are the natural projections). Since the objects and arrows in the whole diagrams are all disjoint unions of these parts, the whole diagram commutes. Finally, we calculate . Recall that , which is precisely (a subset canonically isomorphic to) . So , and its easily checked that this isomorphism is natural (essentially because it is canonical).

Finally then, is full, faithful, and essentially surjective on objects making it an equivalence.