So I want to show that $\hat{\mathbb{C}}$, which is the category of presheaves for $\mathbb{C}$ ([$\mathbb{C}^{\mathrm{op}} \rightarrow \mathbf{Set}$]), is cartesian closed. Using the Yoneda lemma and adjunction, I define $F^G$, for presheaves $F$ and $G$ as follows. $$ F^G(C) \cong \operatorname{Hom}(y(C), F^G) \cong \operatorname{Hom}(y(C) \times G, F) \,. $$ So I will just define, $F^G(C) = \operatorname{Hom}(y(C)\times G, F)$.
But I made a mess checking this definition works; unless there is a easier way, I would have to define an $\varepsilon$, evaluation map and then check that it satisfy the required universal property.
Any help or insight is deeply appreciated.
Cheers
Once you have convinced yourself that $\operatorname{Hom}(y(C) \times G, F)$ is indeed a presheaf, then we have by construction (and the Yoneda lemma) a natural bijection: $$ \operatorname{Hom}(y(C), F^G) \cong F^G(C) = \operatorname{Hom}(y(C) \times G, F). $$ This is almost the universal property for the exponential object, we just need an arbitrary presheaf $Z$ in place of $y(C)$. Since every presheaf is a colimit of representables, there is functor $D: \mathbb{D} \to \mathbb{C}$ such that $Z$ is the colimit of $yD: \mathbb{D} \to \hat{\mathbb{C}}$. Arrows $Z \to F^G$ correspond naturally to cocones on $yD$ with vertex $F^G$. Such cocones consist of arrows $y(C) \to F^G$, so by the above natural bijection this corresponds to cocones with vertex $F$ on the diagram $$ \mathbb{D} \xrightarrow{D} \mathbb{C} \xrightarrow{y} \hat{\mathbb{C}} \xrightarrow{(-) \times G} \hat{\mathbb{C}}. $$ Another basic fact about presheaves is that the functor $(-) \times G$ preserves colimits. So the colimit of the above diagram is $Z \times G$. So cocones with vertex $F$ on the above diagram correspond to arrows $Z \times G \to F$. We thus see that $$ \operatorname{Hom}(Z, F^G) \cong \operatorname{Hom}(Z \times G, F), $$ which is the defining property of the exponential object.