The Stacks project

Proof. First we make a couple of general remarks. We will use that $S^\# = LLS$, and $h_ U^\# = LLh_ U$. In particular, by Lemma 7.49.1, we see that $S^\# \to h_ U^\# $ is injective. Note that $\text{id}_ U \in h_ U(U)$. Hence it gives rise to sections of $Lh_ U$ and $h_ U^\# = LLh_ U$ over $U$ which we will also denote $\text{id}_ U$.

Suppose $S$ is a covering sieve. It clearly suffices to find a morphism $h_ U \to S^\# $ such that the composition $h_ U \to h_ U^\# $ is the canonical map. To find such a map it suffices to find a section $s \in S^\# (U)$ which restricts to $\text{id}_ U$. But since $S$ is a covering sieve, the element $\text{id}_ S \in \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(S, S)$ gives rise to a section of $LS$ over $U$ which restricts to $\text{id}_ U$ in $Lh_ U$. Hence we win.

Suppose that $S^\# \to h_ U^\# $ is an isomorphism. Let $1 \in S^\# (U)$ be the element corresponding to $\text{id}_ U$ in $h_ U^\# (U)$. Because $S^\# = LLS$ there exists a covering sieve $S'$ on $U$ such that $1$ comes from a

This in turn means that for every $\alpha : V \to U$, $\alpha \in S'(V)$ there exists a covering sieve $S_{V, \alpha }$ on $V$ such that $\varphi (\alpha )$ corresponds to a morphism of presheaves $S_{V, \alpha } \to S$. In other words $S_{V, \alpha }$ is contained in $S \times _ U V$. By the second axiom of a topology we see that $S$ is a covering sieve. $\square$

Proof. It is a tautology that if $J = J'$ then the notions of sheaves are the same. Conversely, Lemma 7.50.1 characterizes covering sieves in terms of the sheafification functor. But the sheafification functor $\textit{PSh}(\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J)$ is the left adjoint of the inclusion functor $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J) \to \textit{PSh}(\mathcal{C})$. Hence if the subcategories $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J)$ and $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J')$ are the same, then the sheafification functors are the same and hence the collections of covering sieves are the same. $\square$

Proof. One direction is clear. For the other direction suppose that $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J') \subset \mathop{\mathit{Sh}}\nolimits (\mathcal{C}, J)$. By formal nonsense this implies that if $\mathcal{F}$ is a presheaf of sets, and $\mathcal{F} \to \mathcal{F}^\# $, resp. $\mathcal{F} \to \mathcal{F}^{\# , \prime }$ is the sheafification wrt $J$, resp. $J'$ then there is a canonical map $\mathcal{F}^\# \to \mathcal{F}^{\# , \prime }$ such that $\mathcal{F} \to \mathcal{F}^\# \to \mathcal{F}^{\# , \prime }$ equals the canonical map $\mathcal{F} \to \mathcal{F}^{\# , \prime }$. Of course, $\mathcal{F}^\# \to \mathcal{F}^{\# , \prime }$ identifies the second sheaf as the sheafification of the first with respect to the topology $J'$. Apply this to the map $S \to h_ U$ of Lemma 7.50.1. We get a commutative diagram

And clearly, if $S$ is a covering sieve for the topology $J$ then the middle vertical map is an isomorphism (by the lemma) and we conclude that the right vertical map is an isomorphism as it is the sheafification of the one in the middle wrt $J'$. By the lemma again we conclude that $S$ is a covering sieve for $J'$ as well. $\square$