Lemma 84.15.2. Let $\mathcal{C}$ be a site and $K$ in $\text{SR}(\mathcal{C})$. For $\mathcal{F}$ in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ we have

where $F$ is as in Hypercoverings, Definition 25.2.2.

Lemma 84.15.2. Let $\mathcal{C}$ be a site and $K$ in $\text{SR}(\mathcal{C})$. For $\mathcal{F}$ in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ we have

\[ j_*j^{-1}\mathcal{F} = \mathop{\mathcal{H}\! \mathit{om}}\nolimits (F(K)^\# , \mathcal{F}) \]

where $F$ is as in Hypercoverings, Definition 25.2.2.

**Proof.**
Say $K = \{ U_ i\} _{i \in I}$. Using the description of the functors $j^{-1}$ and $j_*$ given above we see that

\[ j_*j^{-1}\mathcal{F} = \prod \nolimits _{i \in I} j_{i, *}(\mathcal{F}|_{\mathcal{C}/U_ i}) = \prod \nolimits _{i \in I} \mathop{\mathcal{H}\! \mathit{om}}\nolimits (h_{U_ i}^\# , \mathcal{F}) \]

The second equality by Sites, Lemma 7.26.3. Since $F(K) = \coprod h_{U_ i}$ in $\textit{PSh}(\mathcal{C}$, we have $F(K)^\# = \coprod h_{U_ i}^\# $ in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ and since $\mathop{\mathcal{H}\! \mathit{om}}\nolimits (-, \mathcal{F})$ turns coproducts into products (immediate from the construction in Sites, Section 7.26), we conclude. $\square$

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