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
as graded $\mathcal{O}$-modules functorially in $\mathcal{M}$ and $\mathcal{N}$.
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
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
See Section 24.6. By Modules on Sites, Lemma 18.27.9 this is the same thing as the cokernel of
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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