Lemma 7.40.1. Let $\mathcal{C}$ be a site. Let $U$ be an object of $\mathcal{C}$. The following conditions are equivalent
For every covering $\{ U_ i \to U\} $ there exists a map of sheaves $h_ U^\# \to \coprod h_{U_ i}^\# $ right inverse to the sheafification of $\coprod h_{U_ i} \to h_ U$.
For every surjection of sheaves of sets $\mathcal{F} \to \mathcal{G}$ the map $\mathcal{F}(U) \to \mathcal{G}(U)$ is surjective.
Proof. Assume (1) and let $\mathcal{F} \to \mathcal{G}$ be a surjective map of sheaves of sets. For $s \in \mathcal{G}(U)$ there exists a covering $\{ U_ i \to U\} $ and $t_ i \in \mathcal{F}(U_ i)$ mapping to $s|_{U_ i}$, see Definition 7.11.1. Think of $t_ i$ as a map $t_ i : h_{U_ i}^\# \to \mathcal{F}$ via (7.12.3.1). Then precomposing $\coprod t_ i : \coprod h_{U_ i}^\# \to \mathcal{F}$ with the map $h_ U^\# \to \coprod h_{U_ i}^\# $ we get from (1) we obtain a section $t \in \mathcal{F}(U)$ mapping to $s$. Thus (2) holds.
Assume (2) holds. Let $\{ U_ i \to U\} $ be a covering. Then $\coprod h_{U_ i}^\# \to h_ U^\# $ is surjective (Lemma 7.12.4). Hence by (2) there exists a section $s$ of $\coprod h_{U_ i}^\# $ mapping to the section $\text{id}_ U$ of $h_ U^\# $. This section corresponds to a map $h_ U^\# \to \coprod h_{U_ i}^\# $ which is right inverse to the sheafification of $\coprod h_{U_ i} \to h_ U$ which proves (1). $\square$
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Comment #1193 by JuanPablo on