**Proof.**
Proof of (1). Suppose that $s, s' \in \mathcal{F}^+(U)$ and suppose that there exists some covering $\{ U_ i \to U\} $ such that $s|_{U_ i} = s'|_{U_ i}$ for all $i$. We now have three coverings of $U$: the covering $\{ U_ i \to U\} $ above, a covering $\mathcal{U}$ for $s$ as in Lemma 7.10.8, and a similar covering $\mathcal{U}'$ for $s'$. By Lemma 7.10.5, we can find a common refinement, say $\{ W_ j \to U\} $. This means we have $s_ j, s'_ j \in \mathcal{F}(W_ j)$ such that $s|_{W_ j} = \theta (s_ j)$, similarly for $s'|_{W_ j}$, and such that $\theta (s_ j) = \theta (s'_ j)$. This last equality means that there exists some covering $\{ W_{jk} \to W_ j\} $ such that $s_ j|_{W_{jk}} = s'_ j|_{W_{jk}}$. Then since $\{ W_{jk} \to U\} $ is a covering we see that $s, s'$ map to the same element of $H^0(\{ W_{jk} \to U\} , \mathcal{F})$ as desired.

Proof of (2). It is clear that $\mathcal{F} \to \mathcal{F}^+$ is injective because all the maps $\mathcal{F}(U) \to H^0(\mathcal{U}, \mathcal{F})$ are injective. It is also clear that, if $\mathcal{U} \to \mathcal{U}'$ is a refinement, then $H^0(\mathcal{U}', \mathcal{F}) \to H^0(\mathcal{U}, \mathcal{F})$ is injective. Now, suppose that $\{ U_ i \to U\} $ is a covering, and let $(s_ i)$ be a family of elements of $\mathcal{F}^+(U_ i)$ satisfying the sheaf condition $s_ i|_{U_ i \times _ U U_{i'}} = s_{i'}|_{U_ i \times _ U U_{i'}}$ for all $i, i' \in I$. Choose coverings (as in Lemma 7.10.8) $\{ U_{ij} \to U_ i\} $ such that $s_ i|_{U_{ij}}$ is the image of the (unique) element $s_{ij} \in \mathcal{F}(U_{ij})$. The sheaf condition implies that $s_{ij}$ and $s_{i'j'}$ agree over $U_{ij} \times _ U U_{i'j'}$ because it maps to $U_ i \times _ U U_{i'}$ and we have the equality there. Hence $(s_{ij}) \in H^0(\{ U_{ij} \to U\} , \mathcal{F})$ gives rise to an element $s \in \mathcal{F}^+(U)$. We leave it to the reader to verify that $s|_{U_ i} = s_ i$.

Proof of (3). This is immediate from the definitions because the sheaf property says exactly that every map $\mathcal{F} \to H^0(\mathcal{U}, \mathcal{F})$ is bijective (for every covering $\mathcal{U}$ of $U$).

Statement (4) is now obvious.
$\square$

## Comments (2)

Comment #8575 by Alejandro González Nevado on

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