Lemma 48.18.1. In Situation 48.16.1 let

$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ Y' \ar[r]^ g & Y }$

be a cartesian diagram of $\textit{FTS}_ S$ with $g$ flat. Then there is an isomorphism $L(g')^* \circ f^! \to (f')^! \circ Lg^*$ on $D_\mathit{QCoh}^+(\mathcal{O}_ Y)$.

Proof. Namely, because $g$ is flat, for every choice of compactification $j : X \to \overline{X}$ of $X$ over $Y$ the scheme $\overline{X}$ is Tor independent of $Y'$. Denote $j' : X' \to \overline{X}'$ the base change of $j$ and $\overline{g}' : \overline{X}' \to \overline{X}$ the projection. We define the base change map as the composition

$L(g')^* \circ f^! = L(g')^* \circ j^* \circ a = (j')^* \circ L(\overline{g}')^* \circ a \longrightarrow (j')^* \circ a' \circ Lg^* = (f')^! \circ Lg^*$

where the middle arrow is the base change map (48.5.0.1) and $a$ and $a'$ are the right adjoints to pushforward of Lemma 48.3.1 for $\overline{X} \to Y$ and $\overline{X}' \to Y'$. This construction is independent of the choice of compactification (we will formulate a precise lemma and prove it, if we ever need this result).

To finish the proof it suffices to show that the base change map $L(g')^* \circ a \to a' \circ Lg^*$ is an isomorphism on $D_\mathit{QCoh}^+(\mathcal{O}_ Y)$. By Lemma 48.4.4 formation of $a$ and $a'$ commutes with restriction to affine opens of $Y$ and $Y'$. Thus by Remark 48.6.1 we may assume that $Y$ and $Y'$ are affine. Thus the result by Lemma 48.6.2. $\square$

Comment #5431 by CQ on

I think $g^\ast$ is already exact, since $g$ is flat. https://stacks.math.columbia.edu/tag/076W Then why $Lg^\ast$?

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