The Stacks project

Proof. The final assertion means that $\mathcal{F}_ T \to (g_ T)_*g_ T^*\mathcal{F}_ T$ is injective for any morphism $T \to S$. The assumption $\text{Ass}_{X/S}(\mathcal{F}) \subset g(X')$ is preserved by base change, see Divisors, Lemma 31.7.3 and Remark 31.7.4. The same holds for the assumption of flatness and finite type. Hence it suffices to prove the injectivity of the displayed arrow. Let $\mathcal{K} = \mathop{\mathrm{Ker}}(\mathcal{F} \to g_*g^*\mathcal{F})$. Our goal is to prove that $\mathcal{K} = 0$. In order to do this it suffices to prove that $\text{WeakAss}_ X(\mathcal{K}) = \emptyset $, see Divisors, Lemma 31.5.5. We have $\text{WeakAss}_ X(\mathcal{K}) \subset \text{WeakAss}_ X(\mathcal{F})$, see Divisors, Lemma 31.5.4. As $\mathcal{F}$ is flat we see from Lemma 38.13.5 that $\text{WeakAss}_ X(\mathcal{F}) \subset \text{Ass}_{X/S}(\mathcal{F})$. By assumption any point $x$ of $\text{Ass}_{X/S}(\mathcal{F})$ is the image of some $x' \in X'$. Since $g$ is flat the local ring map $\mathcal{O}_{X, x} \to \mathcal{O}_{X', x'}$ is faithfully flat, hence the map

is injective (see Algebra, Lemma 10.82.11). This implies that $\mathcal{K}_ x = 0$ as desired. $\square$