We continue the discussion started in Section 20.21.
Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $Z \subset X$ be a closed subset. In this situation we can consider the functor $\textit{Mod}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X(X))$ given by $\mathcal{F} \mapsto \Gamma _ Z(X, \mathcal{F})$. See Modules, Definition 17.5.1 and Modules, Lemma 17.5.2. Using K-injective resolutions, see Section 20.28, we obtain the right derived functor
Given an object $K$ in $D(\mathcal{O}_ X)$ we denote $H^ q_ Z(X, K) = H^ q(R\Gamma _ Z(X, K))$ the cohomology module with support in $Z$. We will see later (Lemma 20.34.8) that this agrees with the construction in Section 20.21.
For an $\mathcal{O}_ X$-module $\mathcal{F}$ we can consider the subsheaf of sections with support in $Z$, denoted $\mathcal{H}_ Z(\mathcal{F})$, defined by the rule
As discussed in Modules, Remark 17.13.5 we may view $\mathcal{H}_ Z(\mathcal{F})$ as an $\mathcal{O}_ X|_ Z$-module on $Z$ and we obtain a functor
This functor is left exact, but in general not exact. Exactly as above we obtain a right derived functor
We set $\mathcal{H}^ q_ Z(K) = H^ q(R\mathcal{H}_ Z(K))$ so that $\mathcal{H}^0_ Z(\mathcal{F}) = \mathcal{H}_ Z(\mathcal{F})$ for any sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$.
Lemma 20.34.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset.
$R\mathcal{H}_ Z : D(\mathcal{O}_ X) \to D(\mathcal{O}_ X|_ Z)$ is right adjoint to $i_* : D(\mathcal{O}_ X|_ Z) \to D(\mathcal{O}_ X)$.
For $K$ in $D(\mathcal{O}_ X|_ Z)$ we have $R\mathcal{H}_ Z(i_*K) = K$.
Let $\mathcal{G}$ be a sheaf of $\mathcal{O}_ X|_ Z$-modules on $Z$. Then $\mathcal{H}^ p_ Z(i_*\mathcal{G}) = 0$ for $p > 0$.
Proof. The functor $i_*$ is exact, so $i_* = Ri_* = Li_*$. Hence part (1) of the lemma follows from Modules, Lemma 17.13.6 and Derived Categories, Lemma 13.30.3. Let $K$ be as in (2). We can represent $K$ by a K-injective complex $\mathcal{I}^\bullet $ of $\mathcal{O}_ X|_ Z$-modules. By Lemma 20.32.9 the complex $i_*\mathcal{I}^\bullet $, which represents $i_*K$, is a K-injective complex of $\mathcal{O}_ X$-modules. Thus $R\mathcal{H}_ Z(i_*K)$ is computed by $\mathcal{H}_ Z(i_*\mathcal{I}^\bullet ) = \mathcal{I}^\bullet $ which proves (2). Part (3) is a special case of (2). $\square$
Let $(X, \mathcal{O}_ X)$ be a ringed space and let $Z \subset X$ be a closed subset. The category of $\mathcal{O}_ X$-modules whose support is contained in $Z$ is a Serre subcategory of the category of all $\mathcal{O}_ X$-modules, see Homology, Definition 12.10.1 and Modules, Lemma 17.5.2. We denote $D_ Z(\mathcal{O}_ X)$ the strictly full saturated triangulated subcategory of $D(\mathcal{O}_ X)$ consisting of complexes whose cohomology sheaves are supported on $Z$, see Derived Categories, Section 13.17.
Lemma 20.34.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset.
For $K$ in $D(\mathcal{O}_ X|_ Z)$ we have $i_*K$ in $D_ Z(\mathcal{O}_ X)$.
The functor $i_* : D(\mathcal{O}_ X|_ Z) \to D_ Z(\mathcal{O}_ X)$ is an equivalence with quasi-inverse $i^{-1}|_{D_ Z(\mathcal{O}_ X)} = R\mathcal{H}_ Z|_{D_ Z(\mathcal{O}_ X)}$.
The functor $i_* \circ R\mathcal{H}_ Z : D(\mathcal{O}_ X) \to D_ Z(\mathcal{O}_ X)$ is right adjoint to the inclusion functor $D_ Z(\mathcal{O}_ X) \to D(\mathcal{O}_ X)$.
Proof. Part (1) is immediate from the definitions. Part (3) is a formal consequence of part (2) and Lemma 20.34.1. In the rest of the proof we prove part (2).
Let us think of $i$ as the morphism of ringed spaces $i : (Z, \mathcal{O}_ X|_ Z) \to (X, \mathcal{O}_ X)$. Recall that $i^*$ and $i_*$ is an adjoint pair of functors. Since $i$ is a closed immersion, $i_*$ is exact. Since $i^{-1}\mathcal{O}_ X = \mathcal{O}_ X|_ Z$ is the structure sheaf of $(Z, \mathcal{O}_ X|_ Z)$ we see that $i^* = i^{-1}$ is exact and we see that that $i^*i_* = i^{-1}i_*$ is isomorphic to the identify functor. See Modules, Lemmas 17.3.3 and 17.6.1. Thus $i_* : D(\mathcal{O}_ X|_ Z) \to D_ Z(\mathcal{O}_ X)$ is fully faithful and $i^{-1}$ determines a left inverse. On the other hand, suppose that $K$ is an object of $D_ Z(\mathcal{O}_ X)$ and consider the adjunction map $K \to i_*i^{-1}K$. Using exactness of $i_*$ and $i^{-1}$ this induces the adjunction maps $H^ n(K) \to i_*i^{-1}H^ n(K)$ on cohomology sheaves. Since these cohomology sheaves are supported on $Z$ we see these adjunction maps are isomorphisms and we conclude that $i_* : D(\mathcal{O}_ X|_ Z) \to D_ Z(\mathcal{O}_ X)$ is an equivalence.
To finish the proof it suffices to show that $R\mathcal{H}_ Z(K) = i^{-1}K$ if $K$ is an object of $D_ Z(\mathcal{O}_ X)$. To do this we can use that $K = i_*i^{-1}K$ as we've just proved this is the case. Then Lemma 20.34.1 tells us what we want. $\square$
Lemma 20.34.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset. If $\mathcal{I}^\bullet $ is a K-injective complex of $\mathcal{O}_ X$-modules, then $\mathcal{H}_ Z(\mathcal{I}^\bullet )$ is K-injective complex of $\mathcal{O}_ X|_ Z$-modules.
Proof. Since $i_* : \textit{Mod}(\mathcal{O}_ X|_ Z) \to \textit{Mod}(\mathcal{O}_ X)$ is exact and left adjoint to $\mathcal{H}_ Z$ (Modules, Lemma 17.13.6) this follows from Derived Categories, Lemma 13.31.9. $\square$
Lemma 20.34.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset. Then $R\Gamma (Z, - ) \circ R\mathcal{H}_ Z = R\Gamma _ Z(X, - )$ as functors $D(\mathcal{O}_ X) \to D(\mathcal{O}_ X(X))$.
Proof. Follows from the construction of right derived functors using K-injective resolutions, Lemma 20.34.3, and the fact that $\Gamma _ Z(X, -) = \Gamma (Z, -) \circ \mathcal{H}_ Z$. $\square$
Lemma 20.34.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset. Let $U = X \setminus Z$. There is a distinguished triangle in $D(\mathcal{O}_ X(X))$ functorial for $K$ in $D(\mathcal{O}_ X)$.
\[ R\Gamma _ Z(X, K) \to R\Gamma (X, K) \to R\Gamma (U, K) \to R\Gamma _ Z(X, K)[1] \]
Proof. Choose a K-injective complex $\mathcal{I}^\bullet $ all of whose terms are injective $\mathcal{O}_ X$-modules representing $K$. See Section 20.28. Recall that $\mathcal{I}^\bullet |_ U$ is a K-injective complex of $\mathcal{O}_ U$-modules, see Lemma 20.32.1. Hence each of the derived functors in the distinguished triangle is gotten by applying the underlying functor to $\mathcal{I}^\bullet $. Hence we find that it suffices to prove that for an injective $\mathcal{O}_ X$-module $\mathcal{I}$ we have a short exact sequence
This follows from Lemma 20.8.1 and the definitions. $\square$
Lemma 20.34.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset. Denote $j : U = X \setminus Z \to X$ the inclusion of the complement. There is a distinguished triangle in $D(\mathcal{O}_ X)$ functorial for $K$ in $D(\mathcal{O}_ X)$.
\[ i_*R\mathcal{H}_ Z(K) \to K \to Rj_*(K|_ U) \to i_*R\mathcal{H}_ Z(K)[1] \]
Proof. Choose a K-injective complex $\mathcal{I}^\bullet $ all of whose terms are injective $\mathcal{O}_ X$-modules representing $K$. See Section 20.28. Recall that $\mathcal{I}^\bullet |_ U$ is a K-injective complex of $\mathcal{O}_ U$-modules, see Lemma 20.32.1. Hence each of the derived functors in the distinguished triangle is gotten by applying the underlying functor to $\mathcal{I}^\bullet $. Hence it suffices to prove that for an injective $\mathcal{O}_ X$-module $\mathcal{I}$ we have a short exact sequence
This follows from Lemma 20.8.1 and the definitions. $\square$
Lemma 20.34.7. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $Z \subset X$ be a closed subset. Let $j : U \to X$ be the inclusion of an open subset with $U \cap Z = \emptyset $. Then $R\mathcal{H}_ Z(Rj_*K) = 0$ for all $K$ in $D(\mathcal{O}_ U)$.
Proof. Choose a K-injective complex $\mathcal{I}^\bullet $ of $\mathcal{O}_ U$-modules representing $K$. Then $j_*\mathcal{I}^\bullet $ represents $Rj_*K$. By Lemma 20.32.9 the complex $j_*\mathcal{I}^\bullet $ is a K-injective complex of $\mathcal{O}_ X$-modules. Hence $\mathcal{H}_ Z(j_*\mathcal{I}^\bullet )$ represents $R\mathcal{H}_ Z(Rj_*K)$. Thus it suffices to show that $\mathcal{H}_ Z(j_*\mathcal{G}) = 0$ for any abelian sheaf $\mathcal{G}$ on $U$. Thus we have to show that a section $s$ of $j_*\mathcal{G}$ over some open $W$ which is supported on $W \cap Z$ is zero. The support condition means that $s|_{W \setminus W \cap Z} = 0$. Since $j_*\mathcal{G}(W) = \mathcal{G}(U \cap W) = j_*\mathcal{G}(W \setminus W \cap Z)$ this implies that $s$ is zero as desired. $\square$
Lemma 20.34.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $Z \subset X$ be a closed subset. Let $K$ be an object of $D(\mathcal{O}_ X)$ and denote $K_{ab}$ its image in $D(\underline{\mathbf{Z}}_ X)$.
There is a canonical map $R\Gamma _ Z(X, K) \to R\Gamma _ Z(X, K_{ab})$ which is an isomorphism in $D(\textit{Ab})$.
There is a canonical map $R\mathcal{H}_ Z(K) \to R\mathcal{H}_ Z(K_{ab})$ which is an isomorphism in $D(\underline{\mathbf{Z}}_ Z)$.
Proof. Proof of (1). The map is constructed as follows. Choose a K-injective complex of $\mathcal{O}_ X$-modules $\mathcal{I}^\bullet $ representing $K$. Choose a quasi-isomorpism $\mathcal{I}^\bullet \to \mathcal{J}^\bullet $ where $\mathcal{J}^\bullet $ is a K-injective complex of abelian groups. Then the map in (1) is given by
determined by the fact that $\Gamma _ Z$ is a functor on abelian sheaves. An easy check shows that the resulting map combined with the canonical maps of Lemma 20.32.7 fit into a morphism of distinguished triangles
of Lemma 20.34.5. Since two of the three arrows are isomorphisms by the lemma cited, we conclude by Derived Categories, Lemma 13.4.3.
The proof of (2) is omitted. Hint: use the same argument with Lemma 20.34.6 for the distinguished triangle. $\square$
Remark 20.34.9. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i : Z \to X$ be the inclusion of a closed subset. Given $K$ and $M$ in $D(\mathcal{O}_ X)$ there is a canonical map in $D(\mathcal{O}_ X|_ Z)$. Here $K|_ Z = i^{-1}K$ is the restriction of $K$ to $Z$ viewed as an object of $D(\mathcal{O}_ X|_ Z)$. By adjointness of $i_*$ and $R\mathcal{H}_ Z$ of Lemma 20.34.1 to construct this map it suffices to produce a canonical map To construct this map, we choose a K-injective complex $\mathcal{I}^\bullet $ of $\mathcal{O}_ X$-modules representing $M$ and a K-flat complex $\mathcal{K}^\bullet $ of $\mathcal{O}_ X$-modules representing $K$. Observe that $\mathcal{K}^\bullet |_ Z$ is a K-flat complex of $\mathcal{O}_ X|_ Z$-modules representing $K|_ Z$, see Lemma 20.26.8. Hence we need to produce a map of complexes of $\mathcal{O}_ X$-modules. For this it suffices to produce maps Looking at stalks (for example), we see that the left hand side of this formula is equal to $\mathcal{K}^ a \otimes _{\mathcal{O}_ X} i_*\mathcal{H}_ Z(\mathcal{I}^ b)$ and we can use the inclusion $\mathcal{H}_ Z(\mathcal{I}^ b) \to \mathcal{I}^ b$ to get our map.
\[ K|_ Z \otimes _{\mathcal{O}_ X|_ Z}^\mathbf {L} R\mathcal{H}_ Z(M) \longrightarrow R\mathcal{H}_ Z(K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \]
\[ i_*\left(K|_ Z \otimes _{\mathcal{O}_ X|_ Z}^\mathbf {L} R\mathcal{H}_ Z(M)\right) \longrightarrow K \otimes _{\mathcal{O}_ X}^\mathbf {L} M \]
\[ i_*\text{Tot}\left( \mathcal{K}^\bullet |_ Z \otimes _{\mathcal{O}_ X|_ Z} \mathcal{H}_ Z(\mathcal{I}^\bullet )\right) \longrightarrow \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{I}^\bullet ) \]
\[ i_*(\mathcal{K}^ a|_ Z \otimes _{\mathcal{O}_ X|_ Z} \mathcal{H}_ Z(\mathcal{I}^ b)) \longrightarrow \mathcal{K}^ a \otimes _{\mathcal{O}_ X} \mathcal{I}^ b \]
Remark 20.34.10. With notation as in Remark 20.34.9 we obtain a canonical cup product Here the equal signs are given by Lemma 20.34.4, the first arrow is restriction to $Z$, the second arrow is the cup product (Section 20.31), and the third arrow is the map from Remark 20.34.9.
\begin{align*} H^ a(X, K) \times H^ b_ Z(X, M) & = H^ a(X, K) \times H^ b(Z, R\mathcal{H}_ Z(M)) \\ & \to H^ a(Z, K|_ Z) \times H^ b(Z, R\mathcal{H}_ Z(M)) \\ & \to H^{a + b}(Z, K|_ Z \otimes _{\mathcal{O}_ X|_ Z}^\mathbf {L} R\mathcal{H}_ Z(M)) \\ & \to H^{a + b}(Z, R\mathcal{H}_ Z(K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)) \\ & = H^{a + b}_ Z(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \end{align*}
Lemma 20.34.11. With notation as in Remark 20.34.9 the diagram commutes where the top horizontal arrow is the cup product of Remark 20.34.10.
\[ \xymatrix{ H^ i(X, K) \times H^ j_ Z(X, M) \ar[r] \ar[d] & H^{i + j}_ Z(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \ar[d] \\ H^ i(X, K) \times H^ j(X, M) \ar[r] & H^{i + j}(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) } \]
Proof. Omitted. $\square$
Remark 20.34.12. Let $f : (X', \mathcal{O}_{X'}) \to (X, \mathcal{O}_ X)$ be a morphism of ringed spaces. Let $Z \subset X$ be a closed subset and $Z' = f^{-1}(Z)$. Denote $f|_{Z'} : (Z', \mathcal{O}_{X'}|_{Z'}) \to (Z, \mathcal{O}_ X|Z)$ be the induced morphism of ringed spaces. For any $K$ in $D(\mathcal{O}_ X)$ there is a canonical map in $D(\mathcal{O}_{X'}|_{Z'})$. Denote $i : Z \to X$ and $i' : Z' \to X'$ the inclusion maps. By Lemma 20.34.2 part (2) applied to $i'$ it is the same thing to give a map in $D_{Z'}(\mathcal{O}_{X'})$. The map of functors $Lf^* \circ i_* \to i'_* \circ L(f|_{Z'})^*$ of Remark 20.28.3 is an isomorphism in this case (follows by checking what happens on stalks using that $i_*$ and $i'_*$ are exact and that $\mathcal{O}_{Z, z} = \mathcal{O}_{X, z}$ and similarly for $Z'$). Hence it suffices to construct a the top horizonal arrow in the following diagram The complex $Lf^* i_* R\mathcal{H}_ Z(K)$ is supported on $Z'$. The south-east arrow comes from the adjunction mapping $i_*R\mathcal{H}_ Z(K) \to K$ (Lemma 20.34.1). Since the adjunction mapping $i'_* R\mathcal{H}_{Z'}(Lf^*K) \to Lf^*K$ is universal by Lemma 20.34.2 part (3), we find that the south-east arrow factors uniquely over the south-west arrow and we obtain the desired arrow.
\[ L(f|_{Z'})^*R\mathcal{H}_ Z(K) \longrightarrow R\mathcal{H}_{Z'}(Lf^*K) \]
\[ i'_* L(f|_{Z'})^* R\mathcal{H}_ Z(K) \longrightarrow i'_*R\mathcal{H}_{Z'}(Lf^*K) \]
\[ \xymatrix{ Lf^* i_* R\mathcal{H}_ Z(K) \ar[rr] \ar[rd] & & i'_* R\mathcal{H}_{Z'}(Lf^*K) \ar[ld] \\ & Lf^*K } \]
Lemma 20.34.13. With notation and assumptions as in Remark 20.34.12 the diagram commutes. Here the top horizontal arrow comes from the identifications $H^ p_ Z(X, K) = H^ p(Z, R\mathcal{H}_ Z(K))$ and $H^ p_{Z'}(X', Lf^*K) = H^ p(Z', R\mathcal{H}_{Z'}(K'))$, the pullback map $H^ p(Z, R\mathcal{H}_ Z(K)) \to H^ p(Z', L(f|_{Z'})^*R\mathcal{H}_ Z(K))$, and the map constructed in Remark 20.34.12.
\[ \xymatrix{ H^ p_ Z(X, K) \ar[r] \ar[d] & H^ p_{Z'}(X, Lf^*K) \ar[d] \\ H^ p(X, K) \ar[r] & H^ p(X', Lf^*K) } \]
Proof. Omitted. Hints: Using that $H^ p(Z, R\mathcal{H}_ Z(K)) = H^ p(X, i_*R\mathcal{H}_ Z(K))$ and similarly for $R\mathcal{H}_{Z'}(Lf^*K)$ this follows from the functoriality of the pullback maps and the commutative diagram used to define the map of Remark 20.34.12. $\square$
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