Lemma 48.4.1. In diagram (48.4.0.1) assume that $g$ is flat or more generally that $f$ and $g$ are Tor independent. Then $a \circ Rg_* \leftarrow Rg'_* \circ a'$ is an isomorphism.

**Proof.**
In this case the base change map $Lg^* \circ Rf_* K \longrightarrow Rf'_* \circ L(g')^*K$ is an isomorphism for every $K$ in $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Derived Categories of Schemes, Lemma 36.22.5. Thus the corresponding transformation between adjoint functors is an isomorphism as well.
$\square$

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