Lemma 36.22.6. Consider a cartesian square

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

of quasi-compact and quasi-separated schemes. Assume $g$ and $f$ Tor independent and $S = \mathop{\mathrm{Spec}}(R)$, $S' = \mathop{\mathrm{Spec}}(R')$ affine. For $M, K \in D(\mathcal{O}_ X)$ the canonical map

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(M, K) \otimes ^\mathbf {L}_ R R' \longrightarrow R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*K) \]

in $D(R')$ is an isomorphism in the following two cases

$M \in D(\mathcal{O}_ X)$ is perfect and $K \in D_\mathit{QCoh}(X)$, or

$M \in D(\mathcal{O}_ X)$ is pseudo-coherent, $K \in D_\mathit{QCoh}^+(X)$, and $R'$ has finite tor dimension over $R$.

**Proof.**
There is a canonical map $R\mathop{\mathrm{Hom}}\nolimits _ X(M, K) \to R\mathop{\mathrm{Hom}}\nolimits _{X'}(L(g')^*M, L(g')^*K)$ in $D(\Gamma (X, \mathcal{O}_ X))$ of global hom complexes, see Cohomology, Section 20.41. Restricting scalars we can view this as a map in $D(R)$. Then we can use the adjointness of restriction and $- \otimes _ R^\mathbf {L} R'$ to get the displayed map of the lemma. Having defined the map it suffices to prove it is an isomorphism in the derived category of abelian groups.

The right hand side is equal to

\[ R\mathop{\mathrm{Hom}}\nolimits _ X(M, R(g')_*L(g')^*K) = R\mathop{\mathrm{Hom}}\nolimits _ X(M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

by Lemma 36.5.3. In both cases the complex $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K)$ is an object of $D_\mathit{QCoh}(\mathcal{O}_ X)$ by Lemma 36.10.8. There is a natural map

\[ R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K) \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'} \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

which is an isomorphism in both cases by Lemma 36.10.9. To see that this lemma applies in case (2) we note that $g'_*\mathcal{O}_{X'} = Rg'_*\mathcal{O}_{X'} = Lf^*g_*\mathcal{O}_ X$ the second equality by Lemma 36.22.5. Using Lemma 36.10.4 and Cohomology, Lemma 20.45.4 we conclude that $g'_*\mathcal{O}_{X'}$ has finite Tor dimension. Hence, in both cases by replacing $K$ by $R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (M, K)$ we reduce to proving

\[ R\Gamma (X, K) \otimes ^\mathbf {L}_ A A' \longrightarrow R\Gamma (X, K \otimes ^\mathbf {L}_{\mathcal{O}_ X} g'_*\mathcal{O}_{X'}) \]

is an isomorphism. Note that the left hand side is equal to $R\Gamma (X', L(g')^*K)$ by Lemma 36.5.3. Hence the result follows from Lemma 36.22.5.
$\square$

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