Lemma 48.17.2. In Situation 48.16.1 let
be a commutative diagram of $\textit{FTS}_ S$ where $j$ and $j'$ are open immersions. Then $j^* \circ f^! = g^! \circ (j')^*$ as functors $D^+_\mathit{QCoh}(\mathcal{O}_ Y) \to D^+(\mathcal{O}_ U)$.
Lemma 48.17.2. In Situation 48.16.1 let
be a commutative diagram of $\textit{FTS}_ S$ where $j$ and $j'$ are open immersions. Then $j^* \circ f^! = g^! \circ (j')^*$ as functors $D^+_\mathit{QCoh}(\mathcal{O}_ Y) \to D^+(\mathcal{O}_ U)$.
Proof. Let $h = f \circ j = j' \circ g$. By Lemma 48.16.3 we have $h^! = j^! \circ f^! = g^! \circ (j')^!$. By Lemma 48.17.1 we have $j^! = j^*$ and $(j')^! = (j')^*$. $\square$
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