The Stacks project

Proof. The final assertion means that $\mathcal{F}_ Z \to (g_ Z)_*g_ Z^*\mathcal{F}_ Z$ is injective for any morphism $Z \to Y$. Since the assumption on the relative assassin is preserved by base change (Lemma 77.6.2) 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 on Spaces, Lemma 71.2.5. We have $\text{WeakAss}_ X(\mathcal{K}) \subset \text{WeakAss}_ X(\mathcal{F})$, see Divisors on Spaces, Lemma 71.2.4. As $\mathcal{F}$ is flat we see from Lemma 77.4.4 that $\text{WeakAss}_ X(\mathcal{F}) \subset \text{Ass}_{X/Y}(\mathcal{F})$. By assumption any point $x$ of $\text{Ass}_{X/Y}(\mathcal{F})$ is the image of some $x' \in |X'|$. Since $g$ is flat the local ring map $\mathcal{O}_{X, \overline{x}} \to \mathcal{O}_{X', \overline{x}'}$ is faithfully flat, hence the map

is injective (see Algebra, Lemma 10.82.11). Since the displayed arrow factors through $\mathcal{F}_{\overline{x}} \to (g_*g^*\mathcal{F})_{\overline{x}}$, we conclude that $\mathcal{K}_{\overline{x}} = 0$. Hence $x$ cannot be a weakly associated point of $\mathcal{K}$ and we win. $\square$