Lemma 76.11.6. In Situation 76.11.4. Let $h : X' \to X$ be an étale morphism. Set $\mathcal{F}' = h^*\mathcal{F}$ and $f' = f \circ h$. Let $F_ n'$ be (76.11.4.1) associated to $(f' : X' \to Y, \mathcal{F}')$. Then $F_ n$ is a subfunctor of $F_ n'$ and if $h(X') \supset \text{Ass}_{X/Y}(\mathcal{F})$, then $F_ n = F'_ n$.

Proof. Choose $U \to X$, $V \to Y$, $U \to V$ as in part (1) of Lemma 76.11.2. Choose a surjective étale morphism $U' \to U \times _ X X'$ where $U'$ is a scheme. Then we have the lemma for the two functors $F_{U, n}$ and $F_{U', n}$ determined by $U' \to U$ and $\mathcal{F}|_ U$ over $V$, see More on Flatness, Lemma 38.27.2. On the other hand, Lemma 76.11.2 tells us that given $T \to Y$ we have $F_ n(T) = F_{U, n}(V \times _ Y T)$ and $F'_ n(T) = F_{U', n}(V \times _ Y T)$. This proves the lemma. $\square$

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