The following lemma says there is no ambiguity in defining the cohomology of a sheaf $\mathcal{F}$ over an open.
Lemma 20.7.1. Let $X$ be a ringed space. Let $U \subset X$ be an open subspace.
If $\mathcal{I}$ is an injective $\mathcal{O}_ X$-module then $\mathcal{I}|_ U$ is an injective $\mathcal{O}_ U$-module.
For any sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$ we have $H^ p(U, \mathcal{F}) = H^ p(U, \mathcal{F}|_ U)$.
Proof. Denote $j : U \to X$ the open immersion. Recall that the functor $j^{-1}$ of restriction to $U$ is a right adjoint to the functor $j_!$ of extension by $0$, see Sheaves, Lemma 6.31.8. Moreover, $j_!$ is exact. Hence (1) follows from Homology, Lemma 12.29.1.
By definition $H^ p(U, \mathcal{F}) = H^ p(\Gamma (U, \mathcal{I}^\bullet ))$ where $\mathcal{F} \to \mathcal{I}^\bullet $ is an injective resolution in $\textit{Mod}(\mathcal{O}_ X)$. By the above we see that $\mathcal{F}|_ U \to \mathcal{I}^\bullet |_ U$ is an injective resolution in $\textit{Mod}(\mathcal{O}_ U)$. Hence $H^ p(U, \mathcal{F}|_ U)$ is equal to $H^ p(\Gamma (U, \mathcal{I}^\bullet |_ U))$. Of course $\Gamma (U, \mathcal{F}) = \Gamma (U, \mathcal{F}|_ U)$ for any sheaf $\mathcal{F}$ on $X$. Hence the equality in (2). $\square$
Let $X$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $U \subset V \subset X$ be open subsets. Then there is a canonical restriction mapping
functorial in $\mathcal{F}$. Namely, choose any injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $. The restriction mappings of the sheaves $\mathcal{I}^ p$ give a morphism of complexes
The LHS is a complex representing $R\Gamma (V, \mathcal{F})$ and the RHS is a complex representing $R\Gamma (U, \mathcal{F})$. We get the map on cohomology groups by applying the functor $H^ n$. As indicated we will use the notation $\xi \mapsto \xi |_ U$ to denote this map. Thus the rule $U \mapsto H^ n(U, \mathcal{F})$ is a presheaf of $\mathcal{O}_ X$-modules. This presheaf is customarily denoted $\underline{H}^ n(\mathcal{F})$. We will give another interpretation of this presheaf in Lemma 20.11.4.
Lemma 20.7.2. Let $X$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. Let $U \subset X$ be an open subspace. Let $n > 0$ and let $\xi \in H^ n(U, \mathcal{F})$. Then there exists an open covering $U = \bigcup _{i\in I} U_ i$ such that $\xi |_{U_ i} = 0$ for all $i \in I$.
Proof. Let $\mathcal{F} \to \mathcal{I}^\bullet $ be an injective resolution. Then
Pick an element $\tilde\xi \in \mathcal{I}^ n(U)$ representing the cohomology class in the presentation above. Since $\mathcal{I}^\bullet $ is an injective resolution of $\mathcal{F}$ and $n > 0$ we see that the complex $\mathcal{I}^\bullet $ is exact in degree $n$. Hence $\mathop{\mathrm{Im}}(\mathcal{I}^{n - 1} \to \mathcal{I}^ n) = \mathop{\mathrm{Ker}}(\mathcal{I}^ n \to \mathcal{I}^{n + 1})$ as sheaves. Since $\tilde\xi $ is a section of the kernel sheaf over $U$ we conclude there exists an open covering $U = \bigcup _{i \in I} U_ i$ such that $\tilde\xi |_{U_ i}$ is the image under $d$ of a section $\xi _ i \in \mathcal{I}^{n - 1}(U_ i)$. By our definition of the restriction $\xi |_{U_ i}$ as corresponding to the class of $\tilde\xi |_{U_ i}$ we conclude. $\square$
Lemma 20.7.3. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be a $\mathcal{O}_ X$-module. The sheaves $R^ if_*\mathcal{F}$ are the sheaves associated to the presheaves with restriction mappings as in Equation (20.7.1.1). There is a similar statement for $R^ if_*$ applied to a bounded below complex $\mathcal{F}^\bullet $.
\[ V \longmapsto H^ i(f^{-1}(V), \mathcal{F}) \]
Proof. Let $\mathcal{F} \to \mathcal{I}^\bullet $ be an injective resolution. Then $R^ if_*\mathcal{F}$ is by definition the $i$th cohomology sheaf of the complex
By definition of the abelian category structure on $\mathcal{O}_ Y$-modules this cohomology sheaf is the sheaf associated to the presheaf
and this is obviously equal to
which is equal to $H^ i(f^{-1}(V), \mathcal{F})$ and we win. $\square$
Lemma 20.7.4. Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. Let $V \subset Y$ be an open subspace. Denote $g : f^{-1}(V) \to V$ the restriction of $f$. Then we have There is a similar statement for the derived image $Rf_*\mathcal{F}^\bullet $ where $\mathcal{F}^\bullet $ is a bounded below complex of $\mathcal{O}_ X$-modules.
\[ R^ pg_*(\mathcal{F}|_{f^{-1}(V)}) = (R^ pf_*\mathcal{F})|_ V \]
Proof. First proof. Apply Lemmas 20.7.3 and 20.7.1 to see the displayed equality. Second proof. Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet $ and use that $\mathcal{F}|_{f^{-1}(V)} \to \mathcal{I}^\bullet |_{f^{-1}(V)}$ is an injective resolution also. $\square$
Remark 20.7.5. Here is a different approach to the proofs of Lemmas 20.7.2 and 20.7.3 above. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $i_ X : \textit{Mod}(\mathcal{O}_ X) \to \textit{PMod}(\mathcal{O}_ X)$ be the inclusion functor and let $\# $ be the sheafification functor. Recall that $i_ X$ is left exact and $\# $ is exact.
First prove Lemma 20.11.4 below which says that the right derived functors of $i_ X$ are given by $R^ pi_ X\mathcal{F} = \underline{H}^ p(\mathcal{F})$. Here is another proof: The equality is clear for $p = 0$. Both $(R^ pi_ X)_{p \geq 0}$ and $(\underline{H}^ p)_{p \geq 0}$ are delta functors vanishing on injectives, hence both are universal, hence they are isomorphic. See Homology, Section 12.12.
A restatement of Lemma 20.7.2 is that $(\underline{H}^ p(\mathcal{F}))^\# = 0$, $p > 0$ for any sheaf of $\mathcal{O}_ X$-modules $\mathcal{F}$. To see this is true, use that ${}^\# $ is exact so
because $\# \circ i_ X$ is the identity functor.
Let $f : X \to Y$ be a morphism of ringed spaces. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module. The presheaf $V \mapsto H^ p(f^{-1}V, \mathcal{F})$ is equal to $R^ p (i_ Y \circ f_*)\mathcal{F}$. You can prove this by noticing that both give universal delta functors as in the argument of (1) above. Hence Lemma 20.7.3 says that $R^ p f_* \mathcal{F}= (R^ p (i_ Y \circ f_*)\mathcal{F})^\# $. Again using that $\# $ is exact a that $\# \circ i_ Y$ is the identity functor we see that
as desired.
\[ (\underline{H}^ p(\mathcal{F}))^\# = (R^ pi_ X\mathcal{F})^\# = R^ p(\# \circ i_ X)(\mathcal{F}) = 0 \]
\[ R^ p f_* \mathcal{F} = R^ p(\# \circ i_ Y \circ f_*)\mathcal{F} = (R^ p (i_ Y \circ f_*)\mathcal{F})^\# \]
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