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. $\square$