## 20.34 Cohomology with support in a closed, II

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 $\textit{Mod}(\mathcal{O}_ X) \to \text{Mod}(\mathcal{O}_ X(X))$ 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

$R\Gamma _ Z(X, - ) : D(\mathcal{O}_ X) \to D(\mathcal{O}_ X(X))$

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

$\mathcal{H}_ Z(\mathcal{F})(U) = \{ s \in \mathcal{F}(U) \mid \text{Supp}(s) \subset U \cap Z\} = \Gamma _{Z \cap U}(U, \mathcal{F}|_ U)$

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

$\textit{Mod}(\mathcal{O}_ X) \longrightarrow \textit{Mod}(\mathcal{O}_ X|_ Z), \quad \mathcal{F} \longmapsto \mathcal{H}_ Z(\mathcal{F}) \text{ viewed as an }\mathcal{O}_ X|_ Z\text{-module on }Z$

This functor is left exact, but in general not exact. Exactly as above we obtain a right derived functor

$R\mathcal{H}_ Z : D(\mathcal{O}_ X) \longrightarrow D(\mathcal{O}_ X|_ Z)$

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.

1. $R\mathcal{H}_ Z : D(\mathcal{O}_ X) \to D(\mathcal{O}_ X|_ Z)$ is right adjoint to $i_* : D(\mathcal{O}_ X|_ Z) \to D(\mathcal{O}_ X)$.

2. For $K$ in $D(\mathcal{O}_ X|_ Z)$ we have $R\mathcal{H}_ Z(i_*K) = K$.

3. 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.

1. For $K$ in $D(\mathcal{O}_ X|_ Z)$ we have $i_*K$ in $D_ Z(\mathcal{O}_ X)$.

2. The functor $i_* : D(\mathcal{O}_ X|_ Z) \to D_ Z(\mathcal{O}_ X)$ is an equivalence with quasi-inverse $i^{-1}|_{D_ Z(\mathcal{O}_ X)} = R\mathcal{H}_ Z|_{D_ Z(\mathcal{O}_ X)}$.

3. 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

$R\Gamma _ Z(X, K) \to R\Gamma (X, K) \to R\Gamma (U, K) \to R\Gamma _ Z(X, K)[1]$

in $D(\mathcal{O}_ X(X))$ functorial for $K$ in $D(\mathcal{O}_ X)$.

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

$0 \to \Gamma _ Z(X, \mathcal{I}) \to \Gamma (X, \mathcal{I}) \to \Gamma (U, \mathcal{I}) \to 0$

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

$i_*R\mathcal{H}_ Z(K) \to K \to Rj_*(K|_ U) \to i_*R\mathcal{H}_ Z(K)[1]$

in $D(\mathcal{O}_ X)$ functorial for $K$ in $D(\mathcal{O}_ X)$.

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

$0 \to i_*\mathcal{H}_ Z(\mathcal{I}) \to \mathcal{I} \to j_*(\mathcal{I}|_ U) \to 0$

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 $D(\underline{\mathbf{Z}}_ X)$.

1. There is a canonical map $R\Gamma _ Z(X, K) \to R\Gamma _ Z(X, K_{ab})$ which is an isomorphism in $D(\textit{Ab})$.

2. 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

$\Gamma _ Z(X, \mathcal{I}^\bullet ) \to \Gamma _ Z(X, \mathcal{J}^\bullet )$

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

$\xymatrix{ R\Gamma _ Z(X, K) \ar[r] \ar[d] & R\Gamma (X, K) \ar[r] \ar[d] & R\Gamma (U, K) \ar[d] \\ R\Gamma _ Z(X, K_{ab}) \ar[r] & R\Gamma (X, K_{ab}) \ar[r] & R\Gamma (U, K_{ab}) }$

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

$K|_ Z \otimes _{\mathcal{O}_ X|_ Z}^\mathbf {L} R\mathcal{H}_ Z(M) \longrightarrow R\mathcal{H}_ Z(K \otimes _{\mathcal{O}_ X}^\mathbf {L} M)$

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

$i_*\left(K|_ Z \otimes _{\mathcal{O}_ X|_ Z}^\mathbf {L} R\mathcal{H}_ Z(M)\right) \longrightarrow K \otimes _{\mathcal{O}_ X}^\mathbf {L} M$

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

$i_*\text{Tot}\left( \mathcal{K}^\bullet |_ Z \otimes _{\mathcal{O}_ X|_ Z} \mathcal{H}_ Z(\mathcal{I}^\bullet )\right) \longrightarrow \text{Tot}(\mathcal{K}^\bullet \otimes _{\mathcal{O}_ X} \mathcal{I}^\bullet )$

of $\mathcal{O}_ X$-modules. For this it suffices to produce maps

$i_*(\mathcal{K}^ a|_ Z \otimes _{\mathcal{O}_ X|_ Z} \mathcal{H}_ Z(\mathcal{I}^ b)) \longrightarrow \mathcal{K}^ a \otimes _{\mathcal{O}_ X} \mathcal{I}^ b$

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.

Remark 20.34.10. With notation as in Remark 20.34.9 we obtain a canonical cup product

\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*}

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.

Lemma 20.34.11. With notation as in Remark 20.34.9 the diagram

$\xymatrix{ H^ i(X, K) \times H^ j_ Z(X, M) \ar[r] \ar[d] & H^{i + j}_ Z(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) \ar[d] \\ H^ i(X, K) \times H^ j(X, M) \ar[r] & H^{i + j}(X, K \otimes _{\mathcal{O}_ X}^\mathbf {L} M) }$

commutes where the top horizontal arrow is the cup product of Remark 20.34.10.

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

$L(f|_{Z'})^*R\mathcal{H}_ Z(K) \longrightarrow R\mathcal{H}_{Z'}(Lf^*K)$

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

$i'_* L(f|_{Z'})^* R\mathcal{H}_ Z(K) \longrightarrow i'_*R\mathcal{H}_{Z'}(Lf^*K)$

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

$\xymatrix{ Lf^* i_* R\mathcal{H}_ Z(K) \ar[rr] \ar[rd] & & i'_* R\mathcal{H}_{Z'}(Lf^*K) \ar[ld] \\ & Lf^*K }$

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.

Lemma 20.34.13. With notation and assumptions as in Remark 20.34.12 the diagram

$\xymatrix{ H^ p_ Z(X, K) \ar[r] \ar[d] & H^ p_{Z'}(X, Lf^*K) \ar[d] \\ H^ p(X, K) \ar[r] & H^ p(X', Lf^*K) }$

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.

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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