Lemma 61.25.1. Let $i : Z \to X$ be a closed immersion morphism of affine schemes. Denote $X_{app}$ and $Z_{app}$ the sites introduced in Lemma 61.12.21. The base change functor

$u : X_{app} \to Z_{app},\quad U \longmapsto u(U) = U \times _ X Z$

is continuous and has a fully faithful left adjoint $v$. For $V$ in $Z_{app}$ the morphism $V \to v(V)$ is a closed immersion identifying $V$ with $u(v(V)) = v(V) \times _ X Z$ and every point of $v(V)$ specializes to a point of $V$. The functor $v$ is cocontinuous and sends coverings to coverings.

Proof. The existence of the adjoint follows immediately from Lemma 61.7.7 and the definitions. It is clear that $u$ is continuous from the definition of coverings in $X_{app}$.

Write $X = \mathop{\mathrm{Spec}}(A)$ and $Z = \mathop{\mathrm{Spec}}(A/I)$. Let $V = \mathop{\mathrm{Spec}}(\overline{C})$ be an object of $Z_{app}$ and let $v(V) = \mathop{\mathrm{Spec}}(C)$. We have seen in the statement of Lemma 61.7.7 that $V$ equals $v(V) \times _ X Z = \mathop{\mathrm{Spec}}(C/IC)$. Any $g \in C$ which maps to an invertible element of $C/IC = \overline{C}$ is invertible in $C$. Namely, we have the $A$-algebra maps $C \to C_ g \to C/IC$ and by adjointness we obtain an $C$-algebra map $C_ g \to C$. Thus every point of $v(V)$ specializes to a point of $V$.

Suppose that $\{ V_ i \to V\}$ is a covering in $Z_{app}$. Then $\{ v(V_ i) \to v(V)\}$ is a finite family of morphisms of $Z_{app}$ such that every point of $V \subset v(V)$ is in the image of one of the maps $v(V_ i) \to v(V)$. As the morphisms $v(V_ i) \to v(V)$ are flat (since they are weakly étale) we conclude that $\{ v(V_ i) \to v(V)\}$ is jointly surjective. This proves that $v$ sends coverings to coverings.

Let $V$ be an object of $Z_{app}$ and let $\{ U_ i \to v(V)\}$ be a covering in $X_{app}$. Then we see that $\{ u(U_ i) \to u(v(V)) = V\}$ is a covering of $Z_{app}$. By adjointness we obtain morphisms $v(u(U_ i)) \to U_ i$. Thus the family $\{ v(u(U_ i)) \to v(V)\}$ refines the given covering and we conclude that $v$ is cocontinuous. $\square$

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