Lemma 24.10.2. In the situation above, let $\mathcal{M}$ be a right graded $\mathcal{A}_ U$-module and let $\mathcal{N}$ be a left graded $\mathcal{A}$-module. Then

$j_!\mathcal{M} \otimes _\mathcal {A} \mathcal{N} = j_!(\mathcal{M} \otimes _{\mathcal{A}_ U} \mathcal{N}|_ U)$

as graded $\mathcal{O}$-modules functorially in $\mathcal{M}$ and $\mathcal{N}$.

Proof. Recall that the degree $n$ component of $j_!\mathcal{M} \otimes _\mathcal {A} \mathcal{N}$ is the cokernel of the canonical map

$\bigoplus \nolimits _{r + s + t = n} j_!\mathcal{M}^ r \otimes _\mathcal {O} \mathcal{A}^ s \otimes _\mathcal {O} \mathcal{N}^ t \longrightarrow \bigoplus \nolimits _{p + q = n} j_!\mathcal{M}^ p \otimes _\mathcal {O} \mathcal{N}^ q$

See Section 24.6. By Modules on Sites, Lemma 18.27.9 this is the same thing as the cokernel of

$\bigoplus \nolimits _{r + s + t = n} j_!(\mathcal{M}^ r \otimes _{\mathcal{O}_ U} \mathcal{A}^ s|_ U \otimes _{\mathcal{O}_ U} \mathcal{N}^ t|_ U) \longrightarrow \bigoplus \nolimits _{p + q = n} j_!(\mathcal{M}^ p \otimes _{\mathcal{O}_ U} \mathcal{N}^ q|_ U)$

and we win. An alternative proof would be to redo the Yoneda argument given in the proof of the lemma cited above. $\square$

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