Lemma 75.2.4. In Situation 75.2.1. Let $(Y_1, y_1) \to (Y, y)$ be a morphism of pointed algebraic spaces over $S$. Assume $Y_1 \to Y$ is flat at $y_1$. If $(T \to Y, t' \leadsto t, \xi )$ is an impurity of $\mathcal{F}$ above $y$, then there exists an impurity $(T_1 \to Y_1, t_1' \leadsto t_1, \xi _1)$ of the pullback $\mathcal{F}_1$ of $\mathcal{F}$ to $X_1 = Y_1 \times _ Y X$ over $y_1$ such that $T_1$ is étale over $Y_1 \times _ Y T$.

Proof. Choose an étale morphism $T_1 \to Y_1 \times _ Y T$ where $T_1$ is a scheme and let $t_1 \in T_1$ be a point mapping to $y_1$ and $t$. It is possible to find a pair $(T_1, t_1)$ like this by Properties of Spaces, Lemma 64.4.3. The morphism of schemes $T_1 \to T$ is flat at $t_1$ (use Morphisms of Spaces, Lemma 65.30.4 and the definition of flat morphisms of algebraic spaces) there exists a specialization $t'_1 \leadsto t_1$ lying over $t' \leadsto t$, see Morphisms, Lemma 29.25.9. Choose a point $\xi _1 \in |X_{T_1}|$ mapping to $t'_1$ and $\xi$ with $\xi _1 \in \text{Ass}_{X_{T_1}/T_1}(\mathcal{F}_{T_1})$. point of $\mathop{\mathrm{Spec}}(\kappa (t'_1) \otimes _{\kappa (t')} \kappa (\xi ))$. This is possible by Divisors on Spaces, Lemma 69.4.7. As the closure $Z_1$ of $\{ \xi _1\}$ in $|X_{T_1}|$ maps into the closure of $\{ \xi \}$ in $|X_ T|$ we conclude that the image of $Z_1$ in $|T_1|$ cannot contain $t_1$. Hence $(T_1 \to Y_1, t'_1 \leadsto t_1, \xi _1)$ is an impurity of $\mathcal{F}_1$ above $Y_1$. $\square$

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