**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

\[ \varphi \in \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(S', LS). \]

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.
$\square$

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