The Stacks project

Proof. Let $T \to S$ be any morphism. Then $h_ T : X'_ T \to X_ T$ is étale as a base change of the étale morphism $g$. For $t \in T$ denote $Z \subset X_ t$ the set of points where $\mathcal{F}_ T$ is not flat over $T$, and similarly denote $Z' \subset X'_ t$ the set of points where $\mathcal{F}'_ T$ is not flat over $T$. As $\mathcal{F}'_ T = h_ T^*\mathcal{F}_ T$ we see that $Z' = h_ t^{-1}(Z)$, see Morphisms, Lemma 29.25.13. Hence $Z' \to Z$ is an étale morphism, so $\dim (Z') \leq \dim (Z)$ (for example by Descent, Lemma 35.21.2 or just because an étale morphism is smooth of relative dimension $0$). This implies that $F_ n \subset F_ n'$.

Finally, suppose that $h(X') \supset \text{Ass}_{X/S}(\mathcal{F})$ and that $T \to S$ is a morphism such that $F_ n'(T)$ is nonempty, i.e., such that $\mathcal{F}'_ T$ is flat in dimensions $\geq n$ over $T$. Pick a point $t \in T$ and let $Z \subset X_ t$ and $Z' \subset X'_ t$ be as above. To get a contradiction assume that $\dim (Z) \geq n$. Pick a generic point $\xi \in Z$ corresponding to a component of dimension $\geq n$. Let $x \in \text{Ass}_{X_ t}(\mathcal{F}_ t)$ be a generalization of $\xi $. Then $x$ maps to a point of $\text{Ass}_{X/S}(\mathcal{F})$ by Divisors, Lemma 31.7.3 and Remark 31.7.4. Thus we see that $x$ is in the image of $h_ T$, say $x = h_ T(x')$ for some $x' \in X'_ T$. But $x' \not\in Z'$ as $x \leadsto \xi $ and $\dim (Z') < n$. Hence $\mathcal{F}'_ T$ is flat over $T$ at $x'$ which implies that $\mathcal{F}_ T$ is flat at $x$ over $T$ (by Morphisms, Lemma 29.25.13). Since this holds for every such $x$ we conclude that $\mathcal{F}_ T$ is flat over $T$ at $\xi $ by Theorem 38.26.1 which is the desired contradiction. $\square$