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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