**Proof.**
Proof of (1). By the above the functors $p_*$ and $p^{-1}$ are adjoint. The functor $p^{-1}$ is required to be exact by Definition 7.32.2. Hence the conditions imposed in Definition 7.15.1 are all satisfied and we see that (1) holds.

Proof of (2). Let $\{ U_ i \to U\} $ be a covering of $\mathcal{C}$. Then $\coprod (h_{U_ i})^\# \to h_ U^\# $ is surjective, see Lemma 7.12.4. Since $p^{-1}$ is exact (by definition of a morphism of topoi) we conclude that $\coprod u(U_ i) \to u(U)$ is surjective. This proves part (1) of Definition 7.32.2. Sheafification is exact, see Lemma 7.10.14. Hence if $U \times _ V W$ exists in $\mathcal{C}$, then

\[ h_{U \times _ V W}^\# = h_ U^\# \times _{h_ V^\# } h_ W^\# \]

and we see that $u(U \times _ V W) = u(U) \times _{u(V)} u(W)$ since $p^{-1}$ is exact. This proves part (2) of Definition 7.32.2. Let $p' = u$, and let $\mathcal{F}_{p'}$ be the stalk functor defined by Equation (7.32.1.1) using $u$. There is a canonical comparison map $c : \mathcal{F}_{p'} \to \mathcal{F}_ p = p^{-1}\mathcal{F}$. Namely, given a triple $(U, x, \sigma )$ representing an element $\xi $ of $\mathcal{F}_{p'}$ we think of $\sigma $ as a map $\sigma : h_ U^\# \to \mathcal{F}$ and we can set $c(\xi ) = p^{-1}(\sigma )(x)$ since $x \in u(U) = p^{-1}(h_ U^\# )$. By Lemma 7.32.3 we see that $(h_ U)_{p'} = u(U)$. Since conditions (1) and (2) of Definition 7.32.2 hold for $p'$ we also have $(h_ U^\# )_{p'} = (h_ U)_{p'}$ by Lemma 7.32.5. Hence we have

\[ (h_ U^\# )_{p'} = (h_ U)_{p'} = u(U) = p^{-1}(h_ U^\# ) \]

We claim this bijection equals the comparison map $c : (h_ U^\# )_{p'} \to p^{-1}(h_ U^\# )$ (verification omitted). Any sheaf on $\mathcal{C}$ is a coequalizer of maps of coproducts of sheaves of the form $h_ U^\# $, see Lemma 7.12.5. The stalk functor $\mathcal{F} \mapsto \mathcal{F}_{p'}$ and the functor $p^{-1}$ commute with arbitrary colimits (as they are both left adjoints). We conclude $c$ is an isomorphism for every sheaf $\mathcal{F}$. Thus the stalk functor $\mathcal{F} \mapsto \mathcal{F}_{p'}$ is isomorphic to $p^{-1}$ and we in particular see that it is exact. This proves condition (3) of Definition 7.32.2 holds and $p'$ is a point. The final assertion has already been shown above, since we saw that $p^{-1} = (p')^{-1}$.
$\square$

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