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 $\mathbf{C}$ and $X \in \mathbf{C}$ let $\mathbf{C}/X$ denote the slice category of pairs $(A, k : A \rightarrow X)$, and $D_X : \mathbf{C} / X \rightarrow \mathbf{C}$ the forgetful functor. For a presheaf $U \in \presheaves{\mathbf{C}}$, we may form a presheaf $U_X \in \presheaves{(\mathbf{C}/X)}$ by $U_X = U \circ D_X^{\text{op}}$. Notice that if we have a map $f : X \rightarrow Y$ and a natural transformation $\alpha : U_Y \rightarrow V_Y$ having components $\alpha_{A, k}$, we can form a "pullback along $f$" denoted by $f_* \alpha : U_X \rightarrow V_X$ and having components $(f_* \alpha)_{A, k} = \alpha_{A, f \circ k}$.

Theorem: For $S, T \in \presheaves{\mathbf{C}}$, $T^S$ is the presheaf defined on objects by $T^S(X) = \Hom (S_X, T_X)$ and on maps $f : X \rightarrow Y$ by $T^S(f)(\alpha) = f_* \alpha$.

Proof: We define an evaluation $e : T^S \times S \rightarrow T$ with components $e_X : T^S(X) \times S(X) \rightarrow T(X)$ by $e_X(\alpha, x) = \alpha_{X, \id}(x)$. We'll show that this evaluation is universal from $\dash \times S$ to $T$. That is, given a $\gamma : U \times S \rightarrow T$, we must find a unique map $\delta : U \rightarrow T^S$ factoring through the evaluation:

Define $\delta$, having components $\delta_X : U(X) \rightarrow T^S(X)$, by $\delta_X(x)_{A, k} = t \mapsto \gamma_A(\overline{x}_A(k), t)$, where $\overline{x}$ denotes the associated natural transformation $\Hom(\dash, X) \rightarrow U$ 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 $\delta$ is actually natural, that is for $f : X \rightarrow Y$, $\delta_X \circ U(f) = T^S(f) \circ \delta_Y$. Applying the left side to $y \in U(Y)$, we first get $U(f)(y)$ and then the natural transformation $\alpha$ with $\alpha_{A,k} = t \mapsto \gamma_A(\overline{U(f)(y)}_A(k), t)$. Applying the other side to $y$, we first get the natural transformation $\beta$ with $\beta_{A,k} = t \mapsto \gamma_A(\overline{y}_A(k), t)$. We then form $f_*\beta$ with $(f_* \beta)_{A,k} = t \mapsto \alpha'_{A, f \circ k}(t) = \gamma_A(\overline{y}_A(f \circ k), t)$. So naturality comes down to showing that $\overline{U(f)(y)}_A(k) = \overline{y}_A(f \circ k)$. In fact, by the definition of the Yoneda correspondence, we have $\overline{y}_A(f \circ k) = U(f \circ k)(y)$, while $\overline{U(f)(y)}_A(k) = U(k)(U(f)(y))$ -- these expressions are equal precisely because $U$ is a contravariant functor.

Next, we verify that $\delta$ does actually make the triangle commute, that is for any $X$ we have $e_X \circ (\delta_X \times \id) = \gamma_X$. Applying the left side to $(x, x') \in U(X) \times S(X)$, we first obtain the pair $(\alpha, x')$ with $\alpha_{A,k} = t \mapsto \gamma_A(\overline{x}_A(k), t)$. Then we have $e_X(\alpha, x') = \alpha_{X,\id}(x') = \gamma_X(\overline{x}_X(\id), x') = \gamma_X(x, x')$.

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 $\delta'$ be another map making the triangle commute. From the triangle then, the only requirement imposed on $\delta'$ is that for any $X$, $x \in U(X)$, $\delta'_X(x)_{X,\id} = t \mapsto \gamma_X(x, t)$. Essentially, the evaluation forces $\delta'_X(x)$ to be identical to $\delta_X(x)$ at the $(X, \id)$ component. We will evaluate $\delta'_X(x)$ at the $(A, k)$ component, and show that it too is identical to that of $\delta_X(x)$. To find the component at $(A, k)$ we use the fact that $\delta'$ is natural. For a map $f: X \rightarrow Y$, we have the following naturality square, followed by its evaluation at some $x \in U(X)$:

And the two final expressions are of course equal. Expanding the bottom, then, we have $(f_*\delta'_Y(x))_{A, k} = \delta'_Y(x)_{A, f \circ k} = \delta'_X(U(f)(x))_{A, k}$. Then, a particular application of this final equality says that $\delta'_X(x)_{A, k \circ id} = \delta'_A(U(k)(x))_{A, \id}$. The former expression is exactly $\delta'_X(x)_{A, k}$, and the latter's value is fixed from the preceding discussion -- it is required to be precisely $t \mapsto \gamma_A(U(k)(x), t)$. Finally, this is just the expanded form of $t \mapsto \gamma_A(\overline{x}_A(k), t)$, exactly the value of $\delta_X(x)_{A, k}$. So, $\delta'_X(x)$ has identical components to $\delta_X(x)$ and thus $\delta'_X(x) = \delta_X(x)$ for every $X$ and $x \in U(X)$, so ultimately $\delta' = \delta$, completing the proof of uniqueness.

Finally, we have defined $T^S$ and provided a universal evaluation map, exhibiting $T^S$ as the exponent in $\presheaves{\mathbf{C}}$.