Lemma 20.22.1. Let $i : Z \to X$ be the inclusion of a closed subset. Let $\mathcal{I}$ be an injective abelian sheaf on $X$. Then $\mathcal{H}_ Z(\mathcal{I})$ is an injective abelian sheaf on $Z$.

## 20.22 Cohomology with support in a closed

Let $X$ be a topological space and let $Z \subset X$ be a closed subset. Let $\mathcal{F}$ be an abelian sheaf on $X$. We let

be the sections with support in $Z$ (Modules, Definition 17.5.1). This is a left exact functor which is not exact in general. Hence we obtain a derived functor

and cohomology groups with support in $Z$ defined by $H^ q_ Z(X, \mathcal{F}) = R^ q\Gamma _ Z(X, \mathcal{F})$.

Let $\mathcal{I}$ be an injective abelian sheaf on $X$. Let $U = X \setminus Z$. Then the restriction map $\mathcal{I}(X) \to \mathcal{I}(U)$ is surjective (Lemma 20.9.1) with kernel $\Gamma _ Z(X, \mathcal{I})$. It immediately follows that for $K \in D(X)$ there is a distinguished triangle

in $D(\textit{Ab})$. As a consequence we obtain a long exact cohomology sequence

for any $K$ in $D(X)$.

For an abelian sheaf $\mathcal{F}$ on $X$ we can consider the *subsheaf of sections with support in $Z$*, denoted $\mathcal{H}_ Z(\mathcal{F})$, defined by the rule

Using the equivalence of Modules, Lemma 17.6.1 we may view $\mathcal{H}_ Z(\mathcal{F})$ as an abelian sheaf on $Z$ (see also Modules, Lemmas 17.6.2 and 17.6.3). Thus we obtain a functor

which is left exact, but in general not exact.

**Proof.**
Observe that for any abelian sheaf $\mathcal{G}$ on $Z$ we have

because after all any section of $i_*\mathcal{G}$ has support in $Z$. Since $i_*$ is exact (Modules, Lemma 17.6.1) and $\mathcal{I}$ injective on $X$ we conclude that $\mathcal{H}_ Z(\mathcal{I})$ is injective on $Z$. $\square$

Denote

the derived functor. We set $\mathcal{H}^ q_ Z(\mathcal{F}) = R^ q\mathcal{H}_ Z(\mathcal{F})$ so that $\mathcal{H}^0_ Z(\mathcal{F}) = \mathcal{H}_ Z(\mathcal{F})$. By the lemma above we have a Grothendieck spectral sequence

Lemma 20.22.2. Let $i : Z \to X$ be the inclusion of a closed subset. Let $\mathcal{G}$ be an injective abelian sheaf on $Z$. Then $\mathcal{H}^ p_ Z(i_*\mathcal{G}) = 0$ for $p > 0$.

**Proof.**
This is true because the functor $i_*$ is exact and transforms injective abelian sheaves into injective abelian sheaves by Lemma 20.12.11.
$\square$

Let $X$ be a topological space and let $Z \subset X$ be a closed subset. We denote $D_ Z(X)$ the strictly full saturated triangulated subcategory of $D(X)$ consisting of complexes whose cohomology sheaves are supported on $Z$.

Lemma 20.22.3. Let $i : Z \to X$ be the inclusion of a closed subset of a topological space $X$. The map $Ri_* = i_* : D(Z) \to D(X)$ induces an equivalence $D(Z) \to D_ Z(X)$ with quasi-inverse

**Proof.**
Recall that $i^{-1}$ and $i_*$ is an adjoint pair of exact functors such that $i^{-1}i_*$ is isomorphic to the identify functor on abelian sheaves. See Modules, Lemmas 17.3.3 and 17.6.1. Thus $i_* : D(Z) \to D_ Z(X)$ is fully faithful and $i^{-1}$ determines a left inverse. On the other hand, suppose that $K$ is an object of $D_ Z(X)$ 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 $D(Z) \to D_ Z(X)$ is an equivalence.

To finish the proof we have to show that $R\mathcal{H}_ Z(K) = i^{-1}K$ if $K$ is an object of $D_ Z(X)$. To do this we can use that $K = i_*i^{-1}K$ as we've just proved this is the case. Then we can choose a K-injective representative $\mathcal{I}^\bullet $ for $i^{-1}K$. Since $i_*$ is the right adjoint to the exact functor $i^{-1}$, the complex $i_*\mathcal{I}^\bullet $ is K-injective (Derived Categories, Lemma 13.29.9). We see that $R\mathcal{H}_ Z(K)$ is computed by $\mathcal{H}_ Z(i_*\mathcal{I}^\bullet ) = \mathcal{I}^\bullet $ as desired. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (1)

Comment #2123 by Arun Debray on