Part (3) is trivial from the definition of $L$ and the definition of a sheaf (Definition \ref{definition-sheaf-sets-topology}). Part (4) follows formally from the others. \medskip\noindent We sketch the proof of (1). Suppose $S$ is a covering sieve of the object $U$. Suppose that $\varphi_i \in L\mathcal{F}(U)$, $i = 1, 2$ map to the same element in $\Mor_{\textit{PSh}(\mathcal{C})}(S, L\mathcal{F})$. We may find a single covering sieve $S'$ on $U$ such that both $\varphi_i$ are represented by elements $\varphi_i \in \Mor_{\textit{PSh}(\mathcal{C})}(S', \mathcal{F})$. We may assume that $S' = S$ by replacing both $S$ and $S'$ by $S' \cap S$ which is also a covering sieve, see Lemma \ref{lemma-sieves-set}. Suppose $V\in \Ob(\mathcal{C})$, and $\alpha : V \to U$ in $S(V)$. Then we have $S \times_U V = h_V$, see Lemma \ref{lemma-pullback-sieve-section}. Thus the restrictions of $\varphi_i$ via $V \to U$ correspond to sections $s_{i, V, \alpha}$ of $\mathcal{F}$ over $V$. The assumption is that there exist a covering sieve $S_{V, \alpha}$ of $V$ such that $s_{i, V, \alpha}$ restrict to the same element of $\Mor_{\textit{PSh}(\mathcal{C})}(S_{V, \alpha}, \mathcal{F})$. Consider the sieve $S''$ on $U$ defined by the rule \begin{eqnarray} \label{equation-S-prime-prime} (f : T \to U) \in S''(T) & \Leftrightarrow & \exists\ V , \ \alpha : V \to U, \ \alpha \in S(V), \nonumber \\ & & \exists\ g : T \to V, \ g \in S_{V, \alpha}(T), \\ & & f = \alpha \circ g \nonumber \end{eqnarray} By axiom (2) of a topology we see that $S''$ is a covering sieve on $U$. By construction we see that $\varphi_1$ and $\varphi_2$ restrict to the same element of $\Mor_{\textit{PSh}(\mathcal{C})}(S'', L\mathcal{F})$ as desired. \medskip\noindent We sketch the proof of (2). Assume that $\mathcal{F}$ is a separated presheaf of sets on $\mathcal{C}$ with respect to the topology $J$. Let $S$ be a covering sieve of the object $U$ of $\mathcal{C}$. Suppose that $\varphi \in \Mor_\mathcal{C}(S, L\mathcal{F})$. We have to find an element $s \in L\mathcal{F}(U)$ restricting to $\varphi$. Suppose $V\in \Ob(\mathcal{C})$, and $\alpha : V \to U$ in $S(V)$. The value $\varphi(\alpha) \in L\mathcal{F}(V)$ is given by a covering sieve $S_{V, \alpha}$ of $V$ and a morphism of presheaves $\varphi_{V, \alpha} : S_{V, \alpha} \to \mathcal{F}$. As in the proof above, define a covering sieve $S''$ on $U$ by Equation (\ref{equation-S-prime-prime}). We define $$ \varphi'' : S'' \longrightarrow \mathcal{F} $$ by the following simple rule: For every $f : T \to U$, $f \in S''(T)$ choose $V, \alpha, g$ as in Equation (\ref{equation-S-prime-prime}). Then set $$ \varphi''(f) = \varphi_{V, \alpha}(g). $$ We claim this is independent of the choice of $V, \alpha, g$. Consider a second such choice$ V', \alpha', g'$. The restrictions of $\varphi_{V, \alpha}$ and $\varphi_{V', \alpha'}$ to the intersection of the following covering sieves on $T$ $$ (S_{V, \alpha} \times_{V, g} T) \cap (S_{V', \alpha'} \times_{V', g'} T) $$ agree. Namely, these restrictions both correspond to the restriction of $\varphi$ to $T$ (via $f$) and the desired equality follows because $\mathcal{F}$ is separated. Denote the common restriction $\psi$. The independence of choice follows because $\varphi_{V, \alpha}(g) = \psi(\text{id}_T) = \varphi_{V', \alpha'}(g')$. OK, so now $\varphi''$ gives an element $s \in L\mathcal{F}(U)$. We leave it to the reader to check that $s$ restricts to $\varphi$.