The Stacks project

Proof. Without any assumptions there is a map $Rf_*(E) \otimes _{\mathcal{O}_ Y}^\mathbf {L} K \to Rf_*(E \otimes _{\mathcal{O}_ X}^\mathbf {L} Lf^*K)$. Namely, it is the adjoint to the canonical map

coming from the map $Lf^*Rf_*E \to E$. See Cohomology on Sites, Lemmas 21.18.4 and 21.19.1. To check it is an isomorphism we may work étale locally on $Y$. Hence we reduce to the case that $Y$ is an affine scheme.

Suppose that $K = \bigoplus K_ i$ is a direct sum of some complexes $K_ i \in D_\mathit{QCoh}(\mathcal{O}_ Y)$. If the statement holds for each $K_ i$, then it holds for $K$. Namely, the functors $Lf^*$ and $\otimes ^\mathbf {L}$ preserve direct sums by construction and $Rf_*$ commutes with direct sums (for complexes with quasi-coherent cohomology sheaves) by Lemma 75.6.2. Moreover, suppose that $K \to L \to M \to K[1]$ is a distinguished triangle in $D_\mathit{QCoh}(Y)$. Then if the statement of the lemma holds for two of $K, L, M$, then it holds for the third (as the functors involved are exact functors of triangulated categories).

Assume $Y$ affine, say $Y = \mathop{\mathrm{Spec}}(A)$. The functor $\widetilde{\ } : D(A) \to D_\mathit{QCoh}(\mathcal{O}_ Y)$ is an equivalence by Lemma 75.4.2 and Derived Categories of Schemes, Lemma 36.3.5. Let $T$ be the property for $K \in D(A)$ that the statement of the lemma holds for $\widetilde{K}$. The discussion above and More on Algebra, Remark 15.59.11 shows that it suffices to prove $T$ holds for $A[k]$. This finishes the proof, as the statement of the lemma is clear for shifts of the structure sheaf. $\square$

Proof. For an étale morphism $\varphi : U \to X$ of algebraic spaces and geometric point $\overline{u}$ the map of local rings $\mathcal{O}_{X, \varphi (\overline{u})} \to \mathcal{O}_{U, \overline{u}}$ is an isomorphism. Hence the equivalence of (1) and (2) follows. So does the implication (1) $\Rightarrow $ (3). Assume (3) and pick a diagram of geometric points as in Definition 75.20.2. The assumptions imply that we can first lift $\overline{b}$ to a geometric point $\overline{w}$ of $W$, then lift the geometric point $(\overline{x}, \overline{b})$ to a geometric point $\overline{u}$ of $U$, and finally lift the geometric point $(\overline{y}, \overline{b})$ to a geometric point $\overline{v}$ of $V$. Use Properties of Spaces, Lemma 66.19.4 to find the lifts. Using the remark on local rings above we conclude that the condition of the definition is satisfied for the given diagram.

Having made these initial points, it is clear that (4) comes down to the statement that Definition 75.20.2 agrees with Derived Categories of Schemes, Definition 36.22.2 when $X$, $Y$, and $B$ are schemes.

Let $\overline{x}, \overline{b}, \overline{y}$ be as in Definition 75.20.2 lying over the points $x, y, b$. Recall that $\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{X, x}^{sh}$ (Properties of Spaces, Lemma 66.22.1) and similarly for the other two. By Algebra, Lemma 10.155.12 we see that $\mathcal{O}_{X, \overline{x}}$ is a strict henselization of $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{B, \overline{b}}$. In particular, the ring map

is flat (More on Algebra, Lemma 15.45.1). By More on Algebra, Lemma 15.61.3 we see that

Hence it follows that if $X$ and $Y$ are Tor independent over $B$ as schemes, then $X$ and $Y$ are Tor independent as algebraic spaces over $B$.

For the converse, we may assume $X$, $Y$, and $B$ are affine. Observe that the ring map

is flat by the observations given above. Moreover, the image of the map on spectra includes all primes $\mathfrak s \subset \mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{Y, y}$ lying over $\mathfrak m_ x$ and $\mathfrak m_ y$. Hence from this and the displayed formula of Tor's above we see that if $X$ and $Y$ are Tor independent over $B$ as algebraic spaces, then

for all $i > 0$ and all $\mathfrak s$ as above. By More on Algebra, Lemma 15.61.6 applied to the ring maps $\Gamma (B, \mathcal{O}_ B) \to \Gamma (X, \mathcal{O}_ X)$ and $\Gamma (B, \mathcal{O}_ B) \to \Gamma (X, \mathcal{O}_ X)$ this implies that $X$ and $Y$ are Tor independent over $B$. $\square$

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

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

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 $Lf^*g_*\mathcal{O}_{Y'} = f^*g_*\mathcal{O}_{Y'}$. $\square$

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 on Sites, Section 21.36. 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

by Lemma 75.6.5. 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 75.13.10. There is a natural map

which is an isomorphism in both cases Lemma 75.13.11. 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 75.20.4. Using Derived Categories of Schemes, Lemma 36.10.4, Lemma 75.13.3, and Cohomology on Sites, Lemma 21.46.5 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

is an isomorphism. Note that the left hand side is equal to $R\Gamma (X', L(g')^*K)$ by Lemma 75.6.5. Hence the result follows from Lemma 75.20.4. $\square$

Proof. We can compute tor dimension at stalks, see Cohomology on Sites, Lemma 21.46.10 and Properties of Spaces, Theorem 66.19.12. If $\overline{x}'$ is a geometric point of $X'$ with image $\overline{x}$ in $X$, then

Let $\overline{y}'$ in $Y'$ and $\overline{y}$ in $Y$ be the image of $\overline{x}'$ and $\overline{x}$. Since $X$ and $Y'$ are tor independent over $Y$, we can apply More on Algebra, Lemma 15.61.2 to see that the right hand side of the displayed formula is equal to $E_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'}$ in $D(\mathcal{O}_{Y', \overline{y}'})$. Thus (1) follows from More on Algebra, Lemma 15.66.13. To see (2) observe that flatness of $\mathcal{G}$ is equivalent to the condition that $\mathcal{G}[0]$ has tor amplitude in $[0, 0]$. Applying (1) we conclude. $\square$