The Stacks project

Proof. Let $\mathcal{G}$ be a presheaf on $\mathcal{D}$ and let $\mathcal{F}$ be a presheaf on $\mathcal{C}$. We will show that the displayed formula holds by constructing maps either way. We will leave it to the reader to verify they are each others inverse.

Given a map $\alpha : u_ p \mathcal{F} \to \mathcal{G}$ we get $u^ p\alpha : u^ p u_ p \mathcal{F} \to u^ p \mathcal{G}$. Lemma 7.5.3 says that there is a map $\mathcal{F} \to u^ p u_ p \mathcal{F}$. The composition of the two gives the desired map. (The good thing about this construction is that it is clearly functorial in everything in sight.)

Conversely, given a map $\beta : \mathcal{F} \to u^ p\mathcal{G}$ we get a map $u_ p\beta : u_ p\mathcal{F} \to u_ p u^ p\mathcal{G}$. Let $V \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$. We claim that the functor $u^ p\mathcal{G}_ V$ on $\mathcal{I}_ V$ has a canonical map to the constant functor with value $\mathcal{G}(V)$. Namely, for every object $(U, \phi )$ of $\mathcal{I}_ V$, the value of $u^ p\mathcal{G}_ V$ on this object is $\mathcal{G}(u(U))$ which maps to $\mathcal{G}(V)$ by $\mathcal{G}(\phi ) = \phi ^*$. This is a transformation of functors because $\mathcal{G}$ is a functor itself. This leads to a map $u_ p u^ p \mathcal{G}(V) \to \mathcal{G}(V)$. Another trivial verification shows that this is functorial in $V$ leading to a map of presheaves $u_ p u^ p \mathcal{G} \to \mathcal{G}$. The composition $u_ p\mathcal{F} \to u_ p u^ p\mathcal{G} \to \mathcal{G}$ is the desired map. $\square$