Lemma 76.8.4. In Situation 76.8.1. Let $T \subset S$ be a subset. Let $s \in S$ be in the closure of $T$. For $t \in T$, let $u_ t$ be the pullback of $u$ to $X_ t$ and let $u_ s$ be the pullback of $u$ to $X_ s$. If $X$ is locally of finite presentation over $S$, $\mathcal{G}$ is of finite presentation1, and $u_ t = 0$ for all $t \in T$, then $u_ s = 0$.

Proof. To check whether $u_ s$ is zero, is étale local on the fibre $X_ s$. Hence we may pick a point $x \in |X_ s| \subset |X|$ and check in an étale neighbourhood. Choose

$\xymatrix{ (X, x) \ar[d] & (X', x') \ar[l]^ g \ar[d] \\ (S, s) & (S', s') \ar[l] }$

as in Proposition 76.5.1. Let $T' \subset S'$ be the inverse image of $T$. Observe that $s'$ is in the closure of $T'$ because $S' \to S$ is open. Hence we reduce to the algebra problem described in the next paragraph.

We have an $R$-module map $u : M \to N$ such that $N$ is projective as an $R$-module and such that $u_ t : M \otimes _ R \kappa (t) \to N \otimes _ R \kappa (t)$ is zero for each $t \in T$. Problem: show that $u_ s = 0$. Let $I \subset R$ be the ideal defined in Lemma 76.8.3. Then $I$ maps to zero in $\kappa (t)$ for all $t \in T$. Hence $T \subset V(I)$. Since $s$ is in the closure of $T$ we see that $s \in V(I)$. Hence $u_ s = 0$. $\square$

 It would suffice if $X$ is locally of finite type over $S$ and $\mathcal{G}$ is finitely presented relative to $S$, but this notion hasn't yet been defined in the setting of algebraic spaces. The definition for schemes is given in More on Morphisms, Section 37.56.

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