The Stacks project

48.10 Right adjoint of pushforward for closed immersions and base change

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 there is a canonical base change map

48.10.0.1
\begin{equation} \label{duality-equation-base-change-exact-support} Lg^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Z'}, Lf^*K) \end{equation}

in $D(\mathcal{O}_{Z'})$ functorial for $K$ in $D(\mathcal{O}_ X)$. Namely, by adjointness of Lemma 48.9.2 such an arrow is the same thing as a map

\[ Ri'_*Lg^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \longrightarrow Lf^*K \]

in $D(\mathcal{O}_{X'})$. By tor independence we have $Ri'_* \circ Lg^* = Lf^* \circ Ri_*$ (see Derived Categories of Schemes, Lemma 36.22.9). Thus this is the same thing as a map

\[ Lf^*Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \longrightarrow Lf^*K \]

For this we can use $Lf^*(can)$ where $can : Ri_* R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \to K$ is the counit of the adjunction.

Lemma 48.10.1. In the situation above, the map (48.10.0.1) is an isomorphism if and only if the base change map

\[ Lf^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ X}(\mathcal{O}_ Z, K) \longrightarrow R\mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_{X'}}(\mathcal{O}_{Z'}, Lf^*K) \]

of Cohomology, Remark 20.39.13 is an isomorphism.

Proof. The statement makes sense because $\mathcal{O}_{Z'} = Lf^*\mathcal{O}_ Z$ by the assumed tor independence. Since $i'_*$ is exact and faithful we see that it suffices to show the map (48.10.0.1) is an isomorphism after applying $Ri'_*$. Since $Ri'_* \circ Lg^* = Lf^* \circ Ri_*$ by the assumed tor indepence and Derived Categories of Schemes, Lemma 36.22.9 we obtain a map

\[ Lf^*Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \longrightarrow Ri'_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_{Z'}, Lf^*K) \]

whose source and target are as in the statement of the lemma by Lemma 48.9.3. We omit the verification that this is the same map as the one constructed in Cohomology, Remark 20.39.13. $\square$

Lemma 48.10.2. In the situation above, assume $f$ is flat and $i$ pseudo-coherent. Then (48.10.0.1) is an isomorphism for $K$ in $D^+_\mathit{QCoh}(\mathcal{O}_ X)$.

Proof. First proof. To prove this map is an isomorphism, we may work locally. Hence we may assume $X$, $X'$, $Z$, $Z'$ are affine, say corresponding to the rings $A$, $A'$, $B$, $B'$. Then $B$ and $A'$ are tor independent over $A$. By Lemma 48.10.1 it suffices to check that

\[ R\mathop{\mathrm{Hom}}\nolimits _ A(B, K) \otimes _ A^\mathbf {L} A' = R\mathop{\mathrm{Hom}}\nolimits _{A'}(B', K \otimes _ A^\mathbf {L} A') \]

in $D(A')$ for all $K \in D^+(A)$. Here we use Derived Categories of Schemes, Lemma 36.10.8 and the fact that $B$, resp. $B'$ is pseudo-coherent as an $A$-module, resp. $A'$-module to compare derived hom on the level of rings and schemes. The displayed equality follows from More on Algebra, Lemma 15.98.3 part (3). See also the discussion in Dualizing Complexes, Section 47.14.

Second proof1. Let $z' \in Z'$ with image $z \in Z$. First show that (48.10.0.1) on stalks at $z'$ induces the map

\[ R\mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_{Z, z}, K_ z) \otimes _{\mathcal{O}_{Z, x}}^\mathbf {L} \mathcal{O}_{Z', z'} \longrightarrow R\mathop{\mathrm{Hom}}\nolimits (\mathcal{O}_{Z', z'}, K_ z \otimes _{\mathcal{O}_{X, z}}^\mathbf {L} \mathcal{O}_{X', z'}) \]

from Dualizing Complexes, Equation (47.14.0.1). Namely, the constructions of these maps are identical. Then apply Dualizing Complexes, Lemma 47.14.2. $\square$

Lemma 48.10.3. Let $i : Z \to X$ be a pseudo-coherent closed immersion of schemes. Let $M \in D_\mathit{QCoh}(\mathcal{O}_ X)$ locally have tor-amplitude in $[a, \infty )$. Let $K \in D_\mathit{QCoh}^+(\mathcal{O}_ X)$. Then there is a canonical isomorphism

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

in $D(\mathcal{O}_ Z)$.

Proof. A map from LHS to RHS is the same thing as a map

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

by Lemmas 48.9.2 and 48.9.3. For this map we take the counit $Ri_*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{O}_ Z, K) \to K$ tensored with $\text{id}_ M$. To see this map is an isomorphism under the hypotheses given, translate into algebra using Lemma 48.9.5 and then for example use More on Algebra, Lemma 15.98.3 part (3). Instead of using Lemma 48.9.5 you can look at stalks as in the second proof of Lemma 48.10.2. $\square$

[1] This proof shows it suffices to assume $K$ is in $D^+(\mathcal{O}_ X)$.

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