Lemma 62.13.4. Let $f : X \to Y$ and $Y \to S$ be morphisms of schemes, both locally of finite type. Let $r, e \geq 0$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module of finite type, with $\dim (\text{Supp}(\mathcal{F}_ y)) \leq r$ for all $y \in Y$. Let $\mathcal{G}$ be a quasi-coherent $\mathcal{O}_ Y$-module of finite type, with $\dim (\text{Supp}(\mathcal{G}_ s)) \leq e$ for all $s \in S$. If $\alpha = [\mathcal{F}/X/Y]_ r$ and $\beta = [\mathcal{G}/Y/S]_ e$ (Example 62.5.2) and $\mathcal{F}$ is flat over $Y$, then $\alpha \circ \beta = [\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}/X/S]_{r + e}$.
Proof. First we observe that $\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G}$ is a quasi-coherent $\mathcal{O}_ X$-module of finite type. Let $s \in S$. Observe that
\[ (\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G})_ s = \mathcal{F}_ s \otimes _{\mathcal{O}_{X_ s}} f_ s^*\mathcal{G}_ s \]
by right exactness of tensor products. Moreover $\mathcal{F}_ s$ is flat over $Y_ s$ as a base change of a flat module. Thus the equality $(\alpha \circ \beta )_ s = [(\mathcal{F} \otimes _{\mathcal{O}_ X} f^*\mathcal{G})_ s]_{r + e}$ follows from Lemma 62.11.4. $\square$
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