Lemma 75.20.4. Let $S$ be a scheme. Let $g : Y' \to Y$ be a morphism of algebraic spaces over $S$. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Consider the base change diagram

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

If $X$ and $Y'$ are Tor independent over $Y$, then for all $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ we have $Rf'_*L(g')^*E = Lg^*Rf_*E$.

**Proof.**
For any object $E$ of $D(\mathcal{O}_ X)$ we can use Cohomology on Sites, Remark 21.19.3 to get a canonical base change map $Lg^*Rf_*E \to Rf'_*L(g')^*E$. To check this is an isomorphism we may work étale locally on $Y'$. Hence we may assume $g : Y' \to Y$ is a morphism of affine schemes. In particular, $g$ is affine and it suffices to show that

\[ Rg_*Lg^*Rf_*E \to Rg_*Rf'_*L(g')^*E = Rf_*(Rg'_* L(g')^* E) \]

is an isomorphism, see Lemma 75.6.4 (and use Lemmas 75.5.5, 75.5.6, and 75.6.1 to see that the objects $Rf'_*L(g')^*E$ and $Lg^*Rf_*E$ have quasi-coherent cohomology sheaves). Note that $g'$ is affine as well (Morphisms of Spaces, Lemma 67.20.5). By Lemma 75.6.5 the map becomes a map

\[ Rf_*E \otimes _{\mathcal{O}_ Y}^\mathbf {L} g_*\mathcal{O}_{Y'} \longrightarrow Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} g'_*\mathcal{O}_{X'}) \]

Observe that $g'_*\mathcal{O}_{X'} = f^*g_*\mathcal{O}_{Y'}$. Thus by Lemma 75.20.1 it suffices to prove that $Lf^*g_*\mathcal{O}_{Y'} = f^*g_*\mathcal{O}_{Y'}$. This follows from our assumption that $X$ and $Y'$ are Tor independent over $Y$. Namely, to check it we may work étale locally on $X$, hence we may also assume $X$ is affine. Say $X = \mathop{\mathrm{Spec}}(A)$, $Y = \mathop{\mathrm{Spec}}(R)$ and $Y' = \mathop{\mathrm{Spec}}(R')$. Our assumption implies that $A$ and $R'$ are Tor independent over $R$ (see Lemma 75.20.3 and More on Algebra, Lemma 15.61.6), i.e., $\text{Tor}_ i^ R(A, R') = 0$ for $i > 0$. In other words $A \otimes _ R^\mathbf {L} R' = A \otimes _ R R'$ which exactly means that $Lf^*g_*\mathcal{O}_{Y'} = f^*g_*\mathcal{O}_{Y'}$.
$\square$

## Comments (0)