Lemma 36.22.9. Consider a cartesian diagram of schemes

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

where $i$ is a closed immersion. If $Z$ and $X'$ are tor independent over $X$, then $Ri'_* \circ Lg^* = Lf^* \circ Ri_*$ as functors $D(\mathcal{O}_ Z) \to D(\mathcal{O}_{X'})$.

Proof. Note that the lemma is supposed to hold for all $K \in D(\mathcal{O}_ Z)$. Observe that $i_*$ and $i'_*$ are exact functors and hence $Ri_*$ and $Ri'_*$ are computed by applying $i_*$ and $i'_*$ to any representatives. Thus the base change map

$Lf^*(Ri_*(K)) \longrightarrow Ri'_*(Lg^*(K))$

on stalks at a point $z' \in Z'$ with image $z \in Z$ is given by

$K_ z \otimes _{\mathcal{O}_{X, z}}^\mathbf {L} \mathcal{O}_{X', z'} \longrightarrow K_ z \otimes _{\mathcal{O}_{Z, z}}^\mathbf {L} \mathcal{O}_{Z', z'}$

This map is an isomorphism by More on Algebra, Lemma 15.61.2 and the assumed tor independence. $\square$

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