The Stacks project

Proof. The second formula follows formally from the first, since if $\mathcal{F}$ and $\mathcal{G}$ are abelian sheaves then

by the universal property of sheafification.

To prove the first formula, let $\mathcal{F}$, $\mathcal{G}$ be abelian presheaves. To prove the lemma we will construct maps from the group on the left to the group on the right and omit the verification that these are mutually inverse.

Note that there is a canonical map of abelian presheaves $\mathcal{F} \to u^ pg_{p!}\mathcal{F}$ which on sections over $U$ is the natural map $\mathcal{F}(U) \to \mathop{\mathrm{colim}}\nolimits _{u(U) \to u(U')} \mathcal{F}(U')$, see Sites, Lemma 7.5.3. Given a map $\alpha : g_{p!}\mathcal{F} \to \mathcal{G}$ we get $u^ p\alpha : u^ pg_{p!}\mathcal{F} \to u^ p\mathcal{G}$. which we can precompose by the map $\mathcal{F} \to u^ pg_{p!}\mathcal{F}$.

Note that there is a canonical map of abelian presheaves $g_{p!}u^ p\mathcal{G} \to \mathcal{G}$ which on sections over $V$ is the natural map $\mathop{\mathrm{colim}}\nolimits _{V \to u(U)} \mathcal{G}(u(U)) \to \mathcal{G}(V)$. It maps a section $s \in u(U)$ in the summand corresponding to $t : V \to u(U)$ to $t^*s \in \mathcal{G}(V)$. Hence, given a map $\beta : \mathcal{F} \to u^ p\mathcal{G}$ we get a map $g_{p!}\beta : g_{p!}\mathcal{F} \to g_{p!}u^ p\mathcal{G}$ which we can postcompose with the map $g_{p!}u^ p\mathcal{G} \to \mathcal{G}$ above. $\square$