# Exponential of presheaves

Let $\mathbb{C}$ be a small category, and $Y,Z$ presheaves on $\mathbb{C}$. We showed in class that if an exponential $Z^Y$ exists in $\hat{C}$,then it must be given by the formula \begin{equation} Z^Y(c)\cong \hat{C}(y(c)\times Y, Z)\end{equation} For each $c\in \mathbb{C}$ where $y$ is the yoneda embedding. Work this out carefully, that is, (a) describe fully the functor $\mathbb{C}^{op}\to Set$ whose action on objects is given by the above formula, and (b) show that it does indeed give an exponential for $Y$ and $Z$ in $\mathbb{C}$

This is a question from an old class at my university that I am now independently studying, but did not attend the lectures, and we are working out of Awodey. I understand the basic notion of presheaves and exponentials, and I feel like I could tackle part (b) of this problem, but I am stumped on part (a). First of all, I do not understand what \begin{equation} Z^Y(c)\cong \hat{C}(y(c)\times Y, Z)\end{equation} means... I thought $\hat{C}=Set^{C^{op}}$, the functor category of presheaves. So i dont understand the terminology being used here when it seems to use $\hat{C}$ as a function.

Can someone give me a good way of approaching this problem and perhaps a bit more insight on the motivation behind the problem? I know its a vague question, but I appreciate whatever I can get!

• It is common in textbook to write $\mathcal A(a,b)$ for the set of morphisms between $a$ and $b$ in the category $\mathcal A$. You might be more familiar with the notation $\hom(a,b)$ or $\hom_{\mathcal A}(a,b)$. – Pece Aug 22 '17 at 14:55

I will use brackets for the $\hom$-sets, to simplify notation. $E,Y$ and $Z$ are presheaves on the category $C$. For the intuition of this, note the following: for arbitrary $c\in C,$ we have, by Yoneda, a natural bijection $\tag 1 [yc, Z^Y]\cong Z^Yc.$

Furthermore, if an exponential object exists, then owing to the exponential/product adjunction, we must have $\tag 2[E \times Y, Z] \cong [E, Z^Y].$ Setting $E=yc$ we have now $\tag 3[yc \times Y, Z] \cong [yc, Z^Y].$

Now, combine $(1)$ and $(3)$ to obtain your result.

Of course, you need to show that this assignment actually defines an exponential object.

When $\mathcal{C}$ is a category, this is one of the common notations for the "hom-set"; that is, in five different notation styles,

$$\mathcal{C}(x, y) = \hom(x, y) = \hom_{\mathcal{C}}(x, y) = [x,y] = [x,y]_{\mathcal{C}}$$

All three, of course, are notation for the values of a functor $\mathcal{C}^\circ \times \mathcal{C} \to \mathbf{Set}$, so the same formula apples when $x,y$ are objects, arrows, or a mix of the two.

As for motivation, exponentials are important, so we want to be able to compute them! And it turns out we can do so by combining two of our basic facts:

$$Z^Y(c) \cong \hat{\mathcal{C}}(y(c), Z^Y) \cong \hat{\mathcal{C}}(y(c) \times Y, Z)$$

This formula is functoral in all three variables, expressing the values of isomorphic functors $\mathcal{C}^\circ \times \hat{\mathcal{C}}^\circ \times \hat{\mathcal{C}}$, so $c$ can either be an object or arrow of $\mathcal{C}$, and $Y,Z$ can each either be presheaves or morphisms of presheaves. Although, this problem is only asking you to work through the isomorphisms for a fixed $Y,Z$.

Being able to plug in an arrow $f : c \to d$ and work through the calculation to get an explicit formula for how $Z^Y(f)$ acts to relate $Z^Y(c)$ and $Z^Y(d)$ is a good exercise to help one become comfortable with all of the relevant basics.

• This is great ! – jgcello Aug 24 '17 at 6:28