The Stacks project

75.21 Cohomology and base change, V

This section is the analogue of Derived Categories of Schemes, Section 36.26. In Section 75.20 we saw a base change theorem holds when the morphisms are tor independent. Even in the affine case there cannot be a base change theorem without such a condition, see More on Algebra, Section 15.61. In this section we analyze when one can get a base change result “one complex at a time”.

To make this work, let $S$ be a base scheme and suppose we have a commutative diagram

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

of algebraic spaces over $S$ (usually we will assume it is cartesian). Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. For a geometric point $\overline{x}'$ of $X'$ consider the geometric points $\overline{x} = g'(\overline{x}')$, $\overline{y}' = f'(\overline{x}')$, $\overline{y} = f(\overline{x}) = g(\overline{y}')$ of $X$, $Y'$, $Y$. Then we can consider the maps

\[ K_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'} \to K_{\overline{x}} \otimes _{\mathcal{O}_{X, \overline{x}}}^\mathbf {L} \mathcal{O}_{X', \overline{x}'} \to K'_{\overline{x}'} \]

where the first arrow is More on Algebra, Equation (15.61.0.1) and the second comes from $(L(g')^*K)_{\overline{x}'} = K_{\overline{x}} \otimes _{\mathcal{O}_{X, \overline{x}}}^\mathbf {L} \mathcal{O}_{X', \overline{x}'}$ and the given map $L(g')^*K \to K'$. For each $i \in \mathbf{Z}$ we obtain a $\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'}$-module structure on $H^ i(K_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'})$. Putting everything together we obtain canonical maps

75.21.0.1
\begin{equation} \label{spaces-perfect-equation-bc} H^ i(K_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'}) \otimes _{(\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'})} \mathcal{O}_{X', \overline{x}'} \longrightarrow H^ i(K'_{\overline{x}'}) \end{equation}

of $\mathcal{O}_{X', \overline{x}'}$-modules.

Lemma 75.21.1. Let $S$ be a scheme. Let

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

be a cartesian diagram of algebraic spaces over $S$. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. The following are equivalent

  1. for any $x' \in X'$ and $i \in \mathbf{Z}$ the map (75.21.0.1) is an isomorphism,

  2. for any commutative diagram

    \[ \xymatrix{ & U \ar[d] \ar[rd]^ a \\ V' \ar[r] \ar[rd]^ c & V \ar[rd]^ b & X \ar[d]^ f \\ & Y' \ar[r]^ g & Y } \]

    with $a, b, c$ étale, $U, V, V'$ schemes, and with $U' = V' \times _ V U$ the equivalent conditions of Derived Categories of Schemes, Lemma 36.26.1 hold for $(U \to X)^*K$ and $(U' \to X')^*K'$, and

  3. there is some diagram as in (2) with $U' \to X'$ surjective.

Proof. Observe that (1) is étale local on $X'$. Working through formal implications of what is known, we see that it suffices to prove condition (1) of this lemma is equivalent to condition (1) of Derived Categories of Schemes, Lemma 36.26.1 if $X, Y, Y', X'$ are representable by schemes $X_0, Y_0, Y'_0, X'_0$. Denote $f_0, g_0, g'_0, f'_0$ the morphisms between these schemes corresponding to $f, g, g', f'$. We may assume $K = \epsilon ^*K_0$ and $K' = \epsilon ^*K'_0$ for some objects $K_0 \in D_\mathit{QCoh}(\mathcal{O}_{X_0})$ and $K'_0 \in D_\mathit{QCoh}(\mathcal{O}_{X'_0})$, see Lemma 75.4.2. Moreover, the map $Lg^*K \to K'$ is the pullback of a map $L(g_0)^*K_0 \to K'_0$ with notation as in Remark 75.6.3. Recall that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x}$ (Properties of Spaces, Lemma 66.22.1) and that we have

\[ K_{\overline{x}} = K_{0, x} \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X, \overline{x}} \quad \text{and}\quad K'_{\overline{x}'} = K'_{0, x'} \otimes _{\mathcal{O}_{X', x'}}^\mathbf {L} \mathcal{O}_{X', \overline{x}'} \]

(akin to Properties of Spaces, Lemma 66.29.4). Consider the commutative diagram

\[ \xymatrix{ H^ i(K_{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'}) \otimes _{(\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'})} \mathcal{O}_{X', \overline{x}'} \ar[r] & H^ i(K'_{\overline{x}'}) \\ H^ i(K_{0, x} \otimes _{\mathcal{O}_{Y, y}}^\mathbf {L} \mathcal{O}_{Y', y'}) \otimes _{(\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y', y'})} \mathcal{O}_{X', x'} \ar[u] \ar[r] & H^ i(K'_{0, x'}) \ar[u] } \]

We have to show that the lower horizontal arrow is an isomorphism if and only if the upper horizontal arrow is an isomorphism. Since $\mathcal{O}_{X', x'} \to \mathcal{O}_{X', \overline{x}'}$ is faithfully flat (More on Algebra, Lemma 15.45.1) it suffices to show that the top arrow is the base change of the bottom arrow by this map. This follows immediately from the relationships between stalks given above for the objects on the right. For the objects on the left it suffices to show that

\begin{align*} & H^ i\left( (K_{0, x} \otimes _{\mathcal{O}_{X, x}}^\mathbf {L} \mathcal{O}_{X, \overline{x}}) \otimes _{\mathcal{O}_{Y, \overline{y}}}^\mathbf {L} \mathcal{O}_{Y', \overline{y}'}\right) \\ & = H^ i(K_{0, x} \otimes _{\mathcal{O}_{Y, y}}^\mathbf {L} \mathcal{O}_{Y', y'}) \otimes _{(\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y', y'})} (\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'}) \end{align*}

This follows from More on Algebra, Lemma 15.61.5. The flatness assumptions of this lemma hold by what was said above as well as Algebra, Lemma 10.155.12 implying that $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y, \overline{y}}$ and that $\mathcal{O}_{Y', \overline{y}'}$ is the strict henselization of $\mathcal{O}_{Y', y'} \otimes _{\mathcal{O}_{Y, y}} \mathcal{O}_{Y, \overline{y}}$. $\square$

Lemma 75.21.2. Let $S$ be a scheme. Let

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

be a cartesian diagram of algebraic spaces over $S$. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. If

  1. the equivalent conditions of Lemma 75.21.1 hold, and

  2. $f$ is quasi-compact and quasi-separated,

then the composition $Lg^*Rf_*K \to Rf'_*L(g')^*K \to Rf'_*K'$ is an isomorphism.

Proof. To check the map 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 this case, we will use the induction principle of Lemma 75.9.3 to prove that for a quasi-compact and quasi-separated algebraic space $U$ étale over $X$ the similarly constructed map $Lg^*R(U \to Y)_*K|_ U \to R(U' \to Y')_*K'|_{U'}$ is an isomorphism. Here $U' = X' \times _{g', X} U = Y' \times _{g, Y} U$.

If $U$ is a scheme (for example affine), then the result holds. Namely, then $Y, Y', U, U'$ are schemes, $K$ and $K'$ come from objects of the derived category of the underlying schemes by Lemma 75.4.2 and the condition of Derived Categories of Schemes, Lemma 36.26.1 holds for these complexes by Lemma 75.21.1. Thus (by the compatibilities explained in Remark 75.6.3) we can apply the result in the case of schemes which is Derived Categories of Schemes, Lemma 36.26.2.

The induction step. Let $(U \subset W, V \to W)$ be an elementary distinguished square with $W$ a quasi-compact and quasi-separated algebraic space étale over $X$, with $U$ quasi-compact, $V$ affine and the result holds for $U$, $V$, and $U \times _ W V$. To easy notation we replace $W$ by $X$ (this is permissible at this point). Denote $a : U \to Y$, $b : V \to Y$, and $c : U \times _ X V \to Y$ the obvious morphisms. Let $a' : U' \to Y'$, $b' : V' \to Y'$ and $c' : U' \times _{X'} V' \to Y'$ be the base changes of $a$, $b$, and $c$. Using the distinguished triangles from relative Mayer-Vietoris (Lemma 75.10.3) we obtain a commutative diagram

\[ \xymatrix{ Lg^*Rf_*K \ar[r] \ar[d] & Rf'_*K' \ar[d] \\ Lg^*Ra_*K|_ U \oplus Lg^*Rb_*K|_ V \ar[r] \ar[d] & Ra'_* K'|_{U'} \oplus Rb'_* K'|_{V'} \ar[d] \\ Lg^*Rc_*K|_{U \times _ X V} \ar[r] \ar[d] & Rc'_*K'|_{U' \times _{X'} V'} \ar[d] \\ Lg^*Rf_* K[1] \ar[r] & Rf'_* K'[1] } \]

Since the 2nd and 3rd horizontal arrows are isomorphisms so is the first (Derived Categories, Lemma 13.4.3) and the proof of the lemma is finished. $\square$

Lemma 75.21.3. Let $S$ be a scheme. Let

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

be a cartesian diagram of algebraic spaces over $S$. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. If the equivalent conditions of Lemma 75.21.1 hold, then

  1. for $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ the equivalent conditions of Lemma 75.21.1 hold for $L(g')^*(E \otimes ^\mathbf {L} K) \to L(g')^*E \otimes ^\mathbf {L} K'$,

  2. if $E$ in $D(\mathcal{O}_ X)$ is perfect the equivalent conditions of Lemma 75.21.1 hold for $L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K')$, and

  3. if $K$ is bounded below and $E$ in $D(\mathcal{O}_ X)$ pseudo-coherent the equivalent conditions of Lemma 75.21.1 hold for $L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K')$.

Proof. The statement makes sense as the complexes involved have quasi-coherent cohomology sheaves by Lemmas 75.5.5, 75.5.6, and 75.13.10 and Cohomology on Sites, Lemmas 21.45.3 and 21.47.5. Having said this, we can check the maps (75.21.0.1) are isomorphisms in case (1) by computing the source and target of (75.21.0.1) using the transitive property of tensor product, see More on Algebra, Lemma 15.59.15. The map in (2) and (3) is the composition

\[ L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, L(g')^*K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K') \]

where the first arrow is Cohomology on Sites, Remark 21.35.11 and the second arrow comes from the given map $L(g')^*K \to K'$. To prove the maps (75.21.0.1) are isomorphisms one represents $E_ x$ by a bounded complex of finite projective $\mathcal{O}_{X. x}$-modules in case (2) or by a bounded above complex of finite free modules in case (3) and computes the source and target of the arrow. Some details omitted. $\square$

Lemma 75.21.4. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Let $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$. Let $\mathcal{G}^\bullet $ be a bounded above complex of quasi-coherent $\mathcal{O}_ X$-modules flat over $Y$. Then formation of

\[ Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet ) \]

commutes with arbitrary base change (see proof for precise statement).

Proof. The statement means the following. Let $g : Y' \to Y$ be a morphism of algebraic spaces and consider the base change diagram

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

in other words $X' = Y' \times _ Y X$. The lemma asserts that

\[ Lg^*Rf_*(E \otimes ^\mathbf {L}_{\mathcal{O}_ X} \mathcal{G}^\bullet ) \longrightarrow Rf'_*(L(g')^*E \otimes ^\mathbf {L}_{\mathcal{O}_{X'}} (g')^*\mathcal{G}^\bullet ) \]

is an isomorphism. Observe that on the right hand side we do not use derived pullback on $\mathcal{G}^\bullet $. To prove this, we apply Lemmas 75.21.2 and 75.21.3 to see that it suffices to prove the canonical map

\[ L(g')^*\mathcal{G}^\bullet \to (g')^*\mathcal{G}^\bullet \]

satisfies the equivalent conditions of Lemma 75.21.1. This follows by checking the condition on stalks, where it immediately follows from the fact that $\mathcal{G}^\bullet _{\overline{x}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{Y', \overline{y}'}$ computes the derived tensor product by our assumptions on the complex $\mathcal{G}^\bullet $. $\square$

Lemma 75.21.5. Let $S$ be a scheme. Let $f : X \to Y$ be a quasi-compact and quasi-separated morphism of algebraic spaces over $S$. Let $E$ be an object of $D(\mathcal{O}_ X)$. Let $\mathcal{G}^\bullet $ be a complex of quasi-coherent $\mathcal{O}_ X$-modules. If

  1. $E$ is perfect, $\mathcal{G}^\bullet $ is a bounded above, and $\mathcal{G}^ n$ is flat over $Y$, or

  2. $E$ is pseudo-coherent, $\mathcal{G}^\bullet $ is bounded, and $\mathcal{G}^ n$ is flat over $Y$,

then formation of

\[ Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet ) \]

commutes with arbitrary base change (see proof for precise statement).

Proof. The statement means the following. Let $g : Y' \to Y$ be a morphism of algebraic spaces and consider the base change diagram

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

in other words $X' = Y' \times _ Y X$. The lemma asserts that

\[ Lg^*Rf_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, \mathcal{G}^\bullet ) \longrightarrow R(f')_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, (g')^*\mathcal{G}^\bullet ) \]

is an isomorphism. Observe that on the right hand side we do not use the derived pullback on $\mathcal{G}^\bullet $. To prove this, we apply Lemmas 75.21.2 and 75.21.3 to see that it suffices to prove the canonical map

\[ L(g')^*\mathcal{G}^\bullet \to (g')^*\mathcal{G}^\bullet \]

satisfies the equivalent conditions of Lemma 75.21.1. This was shown in the proof of Lemma 75.21.4. $\square$

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