Lemma 62.13.3. The formation of $\alpha \circ \beta $ is compatible with base change.
Proof. Let $g : S' \to S$ be a morphism of schemes. Denote $X' \to Y'$ the base change of $X \to Y$ by $g$. Denote $\alpha '$ the base change of $\alpha $ with respect to $Y' \to Y$. Denote $\beta '$ the base change of $\beta $ with respect to $S' \to S$. The assertion means that $\alpha ' \circ \beta '$ is the base change of $\alpha \circ \beta $ by $g : S' \to S$.
Let $s' \in S'$ be a point with image $s \in S$. Then
\[ (\alpha ' \circ \beta ')_{s'} = (Y'_{s'} \to Y')^*\alpha ' \cap \beta '_{s'} \]
We observe that
\[ (Y'_{s'} \to Y')^*\alpha ' = (Y'_{s'} \to Y')^*(Y' \to Y)^*\alpha = (Y'_{s'} \to Y_ s)^*(Y_ s \to Y)^*\alpha \]
and that $\beta '_{s'}$ is the base change of $\beta _ s$ by $s' = \mathop{\mathrm{Spec}}(\kappa (s')) \to \mathop{\mathrm{Spec}}(\kappa (s)) = s$. Hence the result follows from Lemma 62.11.3 applied to $(Y_ s \to Y)^*\alpha $, $\beta _ s$, $X_ s \to Y_ s \to s$, and base change by $s' \to s$. $\square$
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