Exercise 8

We determine the exponent in a presheaf category. This is a slightly different formulation than the exponent defined in Proposition 1 of the text, but the proof contains exactly the same concepts and ideas that are key to the original. In the course of this proof, however, we provide a “few” more details than the authors’ proof of Proposition 1.

To set the stage, consider a small category and let denote the slice category of pairs , and the forgetful functor. For a presheaf , we may form a presheaf by . Notice that if we have a map and a natural transformation having components , we can form a “pullback along ” denoted by and having components .

Theorem: For , is the presheaf defined on objects by and on maps by .

Proof: We define an evaluation with components by . We’ll show that this evaluation is universal from to . That is, given a , we must find a unique map factoring through the evaluation:

Define , having components , by , where denotes the associated natural transformation under the Yoneda correspondence. After staring at this expression for a very long time, one can be convinced that it does indeed “type-check”.

It’s straightfoward if a bit messy to verify that is actually natural, that is for , . Applying the left side to , we first get and then the natural transformation with . Applying the other side to , we first get the natural transformation with . We then form with . So naturality comes down to showing that . In fact, by the definition of the Yoneda correspondence, we have , while – these expressions are equal precisely because is a contravariant functor.

Next, we verify that does actually make the triangle commute, that is for any we have . Applying the left side to , we first obtain the pair with . Then we have .

Finally, we show uniqueness, which is probably the trickiest part of this proof and also the only part kindly glossed over completely by the authors of the book. Let be another map making the triangle commute. From the triangle then, the only requirement imposed on is that for any , , . Essentially, the evaluation forces to be identical to at the component. We will evaluate at the component, and show that it too is identical to that of . To find the component at we use the fact that is natural. For a map , we have the following naturality square, followed by its evaluation at some :

And the two final expressions are of course equal. Expanding the bottom, then, we have . Then, a particular application of this final equality says that . The former expression is exactly , and the latter’s value is fixed from the preceding discussion – it is required to be precisely . Finally, this is just the expanded form of , exactly the value of . So, has identical components to and thus for every and , so ultimately , completing the proof of uniqueness.

Finally, we have defined and provided a universal evaluation map, exhibiting as the exponent in .