The Stacks project

Proof. To prove the first assertion we just note that axioms (1), (2) and (3) of the definition of a site (Definition 7.6.2) directly imply the axioms (3), (2) and (1) of the definition of a topology (Definition 7.47.6). As an example we prove $J$ has property (2). Namely, let $U$ be an object of $\mathcal{C}$, let $S, S'$ be sieves on $U$ such that $S \in J(U)$, and such that for every $V \to U$ in $S(V)$ we have $S' \times _ U V \in J(V)$. By definition of $J(U)$ we can find a covering $\{ f_ i : U_ i \to U\}$ of the site such that $S$ the image of $h_{U_ i} \to h_ U$ is contained in $S$. Since each $S'\times _ U U_ i$ is in $J(U_ i)$ we see that there are coverings $\{ U_{ij} \to U_ i\}$ of the site such that $h_{U_{ij}} \to h_{U_ i}$ is contained in $S' \times _ U U_ i$. By definition of the base change this means that $h_{U_{ij}} \to h_ U$ is contained in the subpresheaf $S' \subset h_ U$. By axiom (2) for sites we see that $\{ U_{ij} \to U\}$ is a covering of $U$ and we conclude that $S' \in J(U)$ by definition of $J$.

Let $\mathcal{F}$ be a presheaf. Suppose that $\mathcal{F}$ is a sheaf in the topology $J$. We will show that $\mathcal{F}$ is a sheaf on the site as well. Let $\{ f_ i : U_ i \to U\} _{i\in I}$ be a covering of the site. Let $s_ i \in \mathcal{F}(U_ i)$ be a family of sections such that $s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j}$ for all $i, j$. We have to show that there exists a unique section $s \in \mathcal{F}(U)$ restricting back to the $s_ i$ on the $U_ i$. Let $S \subset h_ U$ be the sieve generated by the $f_ i$. Note that $S \in J(U)$ by definition. In stead of constructing $s$, by the sheaf condition in the topology, it suffices to construct an element

Take $\alpha \in S(T)$ for some object $T \in \mathcal{U}$. This means exactly that $\alpha : T \to U$ is a morphism which factors through $f_ i$ for some $i\in I$ (and maybe more than $1$). Pick such an index $i$ and a factorization $\alpha = f_ i \circ \alpha _ i$. Define $\varphi (\alpha ) = \alpha _ i^* s_ i$. If $i'$, $\alpha = f_ i \circ \alpha _{i'}'$ is a second choice, then $\alpha _ i^* s_ i = (\alpha _{i'}')^* s_{i'}$ exactly because of our condition $s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j}$ for all $i, j$. Thus $\varphi (\alpha )$ is well defined. We leave it to the reader to verify that $\varphi$, which in turn determines $s$ is correct in the sense that $s$ restricts back to $s_ i$.

Let $\mathcal{F}$ be a presheaf. Suppose that $\mathcal{F}$ is a sheaf on the site $(\mathcal{C}, \text{Cov}(\mathcal{C}))$. We will show that $\mathcal{F}$ is a sheaf for the topology $J$ as well. Let $U$ be an object of $\mathcal{C}$. Let $S$ be a covering sieve on $U$ with respect to the topology $J$. Let

We have to show there is a unique element in $\mathcal{F}(U) = \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, \mathcal{F})$ which restricts back to $\varphi$. By definition there exists a covering $\{ f_ i : U_ i \to U\} _{i\in I} \in \text{Cov}(\mathcal{C})$ such that $f_ i : U_ i \in U$ belongs to $S(U_ i)$. Hence we can set $s_ i = \varphi (f_ i) \in \mathcal{F}(U_ i)$. Then it is a pleasant exercise to see that $s_ i|_{U_ i \times _ U U_ j} = s_ j|_{U_ i \times _ U U_ j}$ for all $i, j$. Thus we obtain the desired section $s$ by the sheaf condition for $\mathcal{F}$ on the site $(\mathcal{C}, \text{Cov}(\mathcal{C}))$. Details left to the reader. $\square$