The Stacks project

Proof. Assume (1). Let $p$ be a point of $\mathcal{C}$ given by a functor $u : \mathcal{C} \to \textit{Sets}$. Let $f_ p \in \mathcal{O}_ p$. Since $\mathcal{O}_ p$ is computed by Sites, Equation (7.32.1.1) we may represent $f_ p$ by a triple $(U, x, f)$ where $x \in U(U)$ and $f \in \mathcal{O}(U)$. By assumption there exists a covering $\{ U_ i \to U\}$ such that for each $i$ either $f$ or $1 - f$ is invertible on $U_ i$. Because $u$ defines a point of the site we see that for some $i$ there exists an $x_ i \in u(U_ i)$ which maps to $x \in u(U)$. By the discussion surrounding Sites, Equation (7.32.1.1) we see that $(U, x, f)$ and $(U_ i, x_ i, f|_{U_ i})$ define the same element of $\mathcal{O}_ p$. Hence we conclude that either $f_ p$ or $1 - f_ p$ is invertible. Thus $\mathcal{O}_ p$ is a ring such that for every element $a$ either $a$ or $1 - a$ is invertible. This means that $\mathcal{O}_ p$ is either zero or a local ring, see Algebra, Lemma 10.18.3.

Assume (2) and assume that $\mathcal{C}$ has enough points. Consider the map of sheaves of sets

of Lemma 18.40.1 part (3). For any local ring $R$ the corresponding map $(R \times R) \amalg (R \times R) \to R \times R$ is surjective, see for example Algebra, Lemma 10.18.3. Since each $\mathcal{O}_ p$ is a local ring or zero the map is surjective on stalks. Hence, by our assumption that $\mathcal{C}$ has enough points it is surjective and we win. $\square$