Lemma 7.42.4. Let $\mathcal{C}$, $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. Assume that $u$ is continuous and almost cocontinuous. Let $\mathcal{G}$ be a presheaf on $\mathcal{D}$ such that $\mathcal{G}(V)$ is a singleton whenever $V$ is sheaf theoretically empty. Then $(u^ p\mathcal{G})^\# = u^ p(\mathcal{G}^\# )$.

**Proof.**
Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. We have to show that $(u^ p\mathcal{G})^\# (U) = u^ p(\mathcal{G}^\# )(U)$. It suffices to show that $(u^ p\mathcal{G})^+(U) = u^ p(\mathcal{G}^+)(U)$ since $\mathcal{G}^+$ is another presheaf for which the assumption of the lemma holds. We have

where the colimit is over the coverings $\mathcal{V}$ of $u(U)$ in $\mathcal{D}$. On the other hand, we see that

where the colimit is over the category of coverings $\mathcal{U} = \{ U_ i \to U\} _{i \in I}$ of $U$ in $\mathcal{C}$ and $u(\mathcal{U}) = \{ u(U_ i) \to u(U)\} _{i \in I}$. The condition that $u$ is continuous means that each $u(\mathcal{U})$ is a covering. Write $I = I_1 \amalg I_2$, where

Then $u(\mathcal{U})' = \{ u(U_ i) \to u(U)\} _{i \in I_1}$ is still a covering of because each of the other pieces can be covered by the empty family and hence can be dropped by Axiom (2) of Definition 7.6.2. Moreover, $\check H^0(u(\mathcal{U}), \mathcal{G}) = \check H^0(u(\mathcal{U})', \mathcal{G})$ by our assumption on $\mathcal{G}$. Finally, the condition that $u$ is almost cocontinuous implies that for every covering $\mathcal{V}$ of $u(U)$ there exists a covering $\mathcal{U}$ of $U$ such that $u(\mathcal{U})'$ refines $\mathcal{V}$. It follows that the two colimits displayed above have the same value as desired. $\square$

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