The Stacks project

Proof. Choose $S$ and $\mathcal{S}$ as in Lemma 7.32.8. We may identify $\mathop{\mathit{Sh}}\nolimits (pt) = \mathop{\mathit{Sh}}\nolimits (\mathcal{S})$ as in that lemma, and we may write $p = f : \mathop{\mathit{Sh}}\nolimits (\mathcal{S}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ for the morphism of topoi induced by $u$. By Lemma 7.28.1 we get a commutative diagram of topoi

where $p'$ is given by the functor $u' : \mathcal{C}/U \to \mathcal{S}/u(U)$, $V/U \mapsto u(V)/u(U)$. Consider the functor $j_ x : \mathcal{S} \cong \mathcal{S}/x$ obtained by assigning to a set $E$ the set $E$ endowed with the constant map $E \to u(U)$ with value $x$. Then $j_ x$ is a fully faithful cocontinuous functor which has a continuous right adjoint $v_ x : (\psi : E \to u(U)) \mapsto \psi ^{-1}(\{ x\} )$. Note that $j_{u(U)} \circ j_ x = \text{id}_\mathcal {S}$, and $v_ x \circ u' = v$. These observations imply that we have the following commutative diagram of topoi

Namely:

1. The morphism $a : \mathop{\mathit{Sh}}\nolimits (\mathcal{S}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{S}/u(U))$ is the morphism of topoi associated to the cocontinuous functor $j_ x$, which equals the morphism associated to the continuous functor $v_ x$, see Lemma 7.21.1 and Section 7.22.

2. The composition $p \circ j_{u(U)} \circ a = p$ since $j_{u(U)} \circ j_ x = \text{id}_\mathcal {S}$.

3. The composition $p' \circ a$ gives a morphism of topoi. Moreover, it is the morphism of topoi associated to the continuous functor $v_ x \circ u' = v$. Hence $v$ does indeed define a point $q$ of $\mathcal{C}/U$ which fits into the diagram above by construction.

This ends the proof of the lemma. $\square$

Proof. Let $q$ be a point of $\mathcal{C}/U$ given by the functor $v : \mathcal{C}/U \to \textit{Sets}$ such that $j_ U \circ q = p$ as morphisms of topoi. Recall that $u(V) = p^{-1}(h_ V^\# )$ for any object $V$ of $\mathcal{C}$, see Lemma 7.32.7. Similarly $v(V/U) = q^{-1}(h_{V/U}^\# )$ for any object $V/U$ of $\mathcal{C}/U$. Consider the following two diagrams

The right hand diagram is the sheafification of the diagram of presheaves on $\mathcal{C}/U$ which maps $W/U$ to the left hand diagram of sets. (There is a small technical point to make here, namely, that we have $(j_ U^{-1}h_ V)^\# = j_ U^{-1}(h_ V^\# )$ and similarly for $h_ U$, see Lemma 7.20.4.) Note that the left hand diagram of sets is cartesian. Since sheafification is exact (Lemma 7.10.14) we conclude that the right hand diagram is cartesian.

Apply the exact functor $q^{-1}$ to the right hand diagram to get a cartesian diagram

of sets. Here we have used that $q^{-1} \circ j^{-1} = p^{-1}$. Since $U/U$ is a final object of $\mathcal{C}/U$ we see that $v(U/U)$ is a singleton. Hence the image of $v(U/U)$ in $u(U)$ is an element $x$, and the top horizontal map gives a bijection $v(V/U) \to \{ y \in u(V) \mid y \mapsto x\text{ in }u(U)\} $ as desired. $\square$

Proof. We use the description of $j_{U!}\mathcal{G}$ as the sheaf associated to the presheaf $V \mapsto \coprod \nolimits _{\varphi \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(V, U)} \mathcal{G}(V/_\varphi U)$ of Lemma 7.25.2. Also, the stalk of $j_{U!}\mathcal{G}$ at $p$ is equal to the stalk of this presheaf, see Lemma 7.32.5. Hence we see that

To each element $(V, y, \varphi , s)$ of this colimit, we can assign $x = u(\varphi )(y) \in u(U)$. Hence we obtain

This is equal to the expression of the lemma by our construction of the points $q$. $\square$