Lemma 21.38.2. Assumptions and notation as in Situation 21.38.1. For $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ consider the induced morphism of topoi

$\pi _ U : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \longrightarrow \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/p(U))$

Then there exists a morphism of topoi

$\sigma : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}/p(U)) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)$

such that $\pi _ U \circ \sigma = \text{id}$ and $\sigma ^{-1} = \pi _{U, *}$.

Proof. Observe that $\pi _ U$ is the restriction of $\pi$ to the localizations, see Sites, Lemma 7.28.4. For an object $V \to p(U)$ of $\mathcal{D}/p(U)$ denote $V \times _{p(U)} U \to U$ the strongly cartesian morphism of $\mathcal{C}$ over $\mathcal{D}$ which exists as $p$ is a fibred category. The functor

$v : \mathcal{D}/p(U) \to \mathcal{C}/U,\quad V/p(U) \mapsto V \times _{p(U)} U/U$

is continuous by the definition of the topology on $\mathcal{C}$. Moreover, it is a right adjoint to $p$ by the definition of strongly cartesian morphisms. Hence we are in the situation discussed in Sites, Section 7.22 and we see that the sheaf $\pi _{U, *}\mathcal{F}$ is equal to $V \mapsto \mathcal{F}(V \times _{p(U)} U)$ (see especially Sites, Lemma 7.22.2).

But here we have more. Namely, the functor $v$ is also cocontinuous (as all morphisms in coverings of $\mathcal{C}$ are strongly cartesian). Hence $v$ defines a morphism $\sigma$ as indicated in the lemma. The equality $\sigma ^{-1} = \pi _{U, *}$ is immediate from the definition. Since $\pi _ U^{-1}\mathcal{G}$ is given by the rule $U'/U \mapsto \mathcal{G}(p(U')/p(U))$ it follows that $\sigma ^{-1} \circ \pi _ U^{-1} = \text{id}$ which proves the equality $\pi _ U \circ \sigma = \text{id}$. $\square$

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