Lemma 26.11.7. Let $X$ be a scheme. Let $\mathcal{B}$ be the set of affine opens of $X$. Let $\mathcal{F}$ be a presheaf of sets on $\mathcal{B}$, see Sheaves, Definition 6.30.1. The following are equivalent

$\mathcal{F}$ is the restriction of a sheaf on $X$ to $\mathcal{B}$,

$\mathcal{F}$ is a sheaf on $\mathcal{B}$, and

$\mathcal{F}(\emptyset )$ is a singleton and whenever $U = V \cup W$ with $U, V, W \in \mathcal{B}$ and $V, W \subset U$ standard open (Algebra, Definition 10.17.3) the map

\[ \mathcal{F}(U) \longrightarrow \mathcal{F}(V) \times \mathcal{F}(W) \]

is injective with image the set of pairs $(s, t)$ such that $s|_{V \cap W} = t|_{V \cap W}$.

**Proof.**
The equivalence of (1) and (2) is Sheaves, Lemma 6.30.7. It is clear that (2) implies (3). Hence it suffices to prove that (3) implies (2). By Sheaves, Lemma 6.30.4 and Lemma 26.5.1 it suffices to prove the sheaf condition holds for standard open coverings (Definition 26.5.2) of elements of $\mathcal{B}$. Let $U = U_1 \cup \ldots \cup U_ n$ be a standard open covering with $U \subset X$ affine open. We will prove the sheaf condition for this covering by induction on $n$. If $n = 0$, then $U$ is empty and we get the sheaf condition by assumption. If $n = 1$, then there is nothing to prove. If $n = 2$, then this is assumption (3). If $n > 2$, then we write $U_ i = D(f_ i)$ for $f_ i \in A = \mathcal{O}_ X(U)$. Suppose that $s_ i \in \mathcal{F}(U_ i)$ are sections such that $s_ i|_{U_ i \cap U_ j} = s_ j|_{U_ i \cap U_ j}$ for all $1 \leq i < j \leq n$. Since $U = U_1 \cup \ldots \cup U_ n$ we have $1 = \sum _{i = 1, \ldots , n} a_ i f_ i$ in $A$ for some $a_ i \in A$, see Algebra, Lemma 10.17.2. Set $g = \sum _{i = 1, \ldots , n - 1} a_ if_ i$. Then $U = D(g) \cup D(f_ n)$. Observe that $D(g) = D(gf_1) \cup \ldots \cup D(gf_{n - 1})$ is a standard open covering. By induction there is a unique section $s' \in \mathcal{F}(D(g))$ which agrees with $s_ i|_{D(gfi)}$ for $i = 1, \ldots , n - 1$. We claim that $s'$ and $s_ n$ have the same restriction to $D(gf_ n)$. This is true by induction and the covering $D(gf_ n) = D(gf_ nf_1) \cup \ldots \cup D(gf_ nf_{n - 1})$. Thus there is a unique section $s \in \mathcal{F}(U)$ whose restriction to $D(g)$ is $s'$ and whose restriction to $D(f_ n)$ is $s_ n$. We omit the verification that $s$ restricts to $s_ i$ on $D(f_ i)$ for $i = 1, \ldots , n - 1$ and we omit the verification that $s$ is unique.
$\square$

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