**Proof.**
Assume (1) and let $\coprod _{i \in I} \mathcal{F}_ i \to h_ U^\# $ be a surjection. Then $\text{id}_ U$ is a section of $h_ U^\# $ over $U$. Hence there exists a covering $\{ U_ a \to U\} _{a \in A}$ and for each $a \in A$ a section $s_ a$ of $\coprod _{i \in I} \mathcal{F}_ i$ over $U_ a$ mapping to $\text{id}_ U$. By the construction of coproducts as sheafification of coproducts of presheaves (Lemma 7.10.13), for each $a$ there exists a covering $\{ U_{ab} \to U_ a\} _{b \in B_ a}$ and for all $b \in B_ a$ an $\iota (b) \in I$ and a section $s_{b}$ of $\mathcal{F}_{\iota (b)}$ over $U_{ab}$ mapping to $\text{id}_ U|_{U_{ab}}$. Thus after replacing the covering $\{ U_ a \to U\} _{a \in A}$ by $\{ U_{ab} \to U\} _{a \in A, b \in B_ a}$ we may assume we have a map $\iota : A \to I$ and for each $a \in A$ a section $s_ a$ of $\mathcal{F}_{\iota (a)}$ over $U_ a$ mapping to $\text{id}_ U$. Since $U$ is quasi-compact, there is a covering $\{ V_ c \to U\} _{c \in C}$, a map $\alpha : C \to A$ with finite image, and $V_ c \to U_{\alpha (c)}$ over $U$. Then we see that $J = \mathop{\mathrm{Im}}(\iota \circ \alpha ) \subset I$ works because $\coprod _{c \in C} h_{V_ c}^\# \to h_ U^\# $ is surjective (Lemma 7.12.4) and factors through $\coprod _{i \in J} \mathcal{F}_ i \to h_ U^\# $. (Here we use that the composition $h_{V_ c}^\# \to h_{U_{\alpha (c)}} \xrightarrow {s_{\alpha (c)}} \mathcal{F}_{\iota (\alpha (c))} \to h_ U^\# $ is the map $h_{V_ c}^\# \to h_ U^\# $ coming from the morphism $V_ c \to U$ because $s_{\alpha (c)}$ maps to $\text{id}_ U|_{U_{\alpha (c)}}$.)

Assume (2). Let $\{ U_ i \to U\} _{i \in I}$ be a covering. By Lemma 7.12.4 we see that $\coprod _{i \in I} h_{U_ i}^\# \to h_ U^\# $ is surjective. Thus we find a finite subset $J \subset I$ such that $\coprod _{j \in J} h_{U_ j}^\# \to h_ U^\# $ is surjective. Then arguing as above we find a covering $\{ V_ c \to U\} _{c \in C}$ of $U$ in $\mathcal{C}$ and a map $\iota : C \to J$ such that $\text{id}_ U$ lifts to a section of $s_ c$ of $h_{U_{\iota (c)}}^\# $ over $V_ c$. Refining the covering even further we may assume $s_ c \in h_{U_{\iota (c)}}(V_ c)$ mapping to $\text{id}_ U$. Then $s_ c : V_ c \to U_{\iota (c)}$ is a morphism over $U$ and we conclude.
$\square$

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