It is convenient to list some consequences of having enough weakly contractible objects here.

**Proof.**
Proof of (1). If the sequence is exact, then evaluating at any weakly contractible element of $\mathcal{C}$ gives an exact sequence by Lemma 21.51.1. Conversely, assume that $\mathcal{F}_1(U) \to \mathcal{F}_2(U) \to \mathcal{F}_3(U)$ is exact for all $U \in \mathcal{B}$. Let $V$ be an object of $\mathcal{C}$ and let $s \in \mathcal{F}_2(V)$ be an element of the kernel of $\mathcal{F}_2 \to \mathcal{F}_3$. By assumption there exists a covering $\{ U_ i \to V\} $ with $U_ i \in \mathcal{B}$. Then $s|_{U_ i}$ lifts to a section $s_ i \in \mathcal{F}_1(U_ i)$. Thus $s$ is a section of the image sheaf $\mathop{\mathrm{Im}}(\mathcal{F}_1 \to \mathcal{F}_2)$. In other words, the sequence $\mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3$ is exact.

Proof of (2). This holds by Lemma 21.23.10 with $d = 0$.

Proof of (3). Let $(\mathcal{F}_ n)$ be a system as in (2) and set $\mathcal{F} = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. If $U \in \mathcal{B}$, then $\mathcal{F}(U) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U)$ surjects onto $\mathcal{F}_1(U)$ as all the transition maps $\mathcal{F}_{n + 1}(U) \to \mathcal{F}_ n(U)$ are surjective. Thus $\mathcal{F} \to \mathcal{F}_1$ is surjective by Sites, Definition 7.11.1 and the assumption that every object has a covering by elements of $\mathcal{B}$.

Proof of (4). Let $\mathcal{F}_{i, 1} \to \mathcal{F}_{i, 2} \to \mathcal{F}_{i, 3}$ be a family of exact sequences of $\mathcal{O}$-modules. We want to show that $\prod \mathcal{F}_{i, 1} \to \prod \mathcal{F}_{i, 2} \to \prod \mathcal{F}_{i, 3}$ is exact. We use the criterion of (1). Let $U \in \mathcal{B}$. Then

\[ (\prod \mathcal{F}_{i, 1})(U) \to (\prod \mathcal{F}_{i, 2})(U) \to (\prod \mathcal{F}_{i, 3})(U) \]

is the same as

\[ \prod \mathcal{F}_{i, 1}(U) \to \prod \mathcal{F}_{i, 2}(U) \to \prod \mathcal{F}_{i, 3}(U) \]

Each of the sequences $\mathcal{F}_{i, 1}(U) \to \mathcal{F}_{i, 2}(U) \to \mathcal{F}_{i, 3}(U)$ are exact by (1). Thus the displayed sequences are exact by Homology, Lemma 12.32.1. We conclude by (1) again.

Proof of (5). Follows from (4) and (slightly generalized) Derived Categories, Lemma 13.34.2.

Proof of (6) and (7). We refer to Section 21.23 for a discussion of derived and homotopy limits and their relationship. By Derived Categories, Definition 13.34.1 we have a distinguished triangle

\[ R\mathop{\mathrm{lim}}\nolimits K_ n \to \prod K_ n \to \prod K_ n \to R\mathop{\mathrm{lim}}\nolimits K_ n[1] \]

Taking the long exact sequence of cohomology sheaves we obtain

\[ H^{p - 1}(\prod K_ n) \to H^{p - 1}(\prod K_ n) \to H^ p(R\mathop{\mathrm{lim}}\nolimits K_ n) \to H^ p(\prod K_ n) \to H^ p(\prod K_ n) \]

Since products are exact by (4) this becomes

\[ \prod H^{p - 1}(K_ n) \to \prod H^{p - 1}(K_ n) \to H^ p(R\mathop{\mathrm{lim}}\nolimits K_ n) \to \prod H^ p(K_ n) \to \prod H^ p(K_ n) \]

Now we first apply this to the case $K_ n = \mathcal{F}_ n[0]$ where $(\mathcal{F}_ n)$ is as in (6). We conclude that (6) holds. Next we apply it to $(K_ n)$ as in (7) and we conclude (7) holds.
$\square$

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