# The Stacks Project

## Tag 01E0

### 20.8. Locality of cohomology

The following lemma says there is no ambiguity in defining the cohomology of a sheaf $\mathcal{F}$ over an open.

Lemma 20.8.1. Let $X$ be a ringed space. Let $U \subset X$ be an open subspace.

1. If $\mathcal{I}$ is an injective $\mathcal{O}_X$-module then $\mathcal{I}|_U$ is an injective $\mathcal{O}_U$-module.
2. 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 12.25.1.

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 $$\tag{20.8.1.1} H^n(V, \mathcal{F}) \longrightarrow H^n(U, \mathcal{F}), \quad \xi \longmapsto \xi|_U$$ 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 $$\Gamma(V, \mathcal{I}^\bullet) \longrightarrow \Gamma(U, \mathcal{I}^\bullet)$$ 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 20.12.4.

Lemma 20.8.2. 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 $$H^n(U, \mathcal{F}) = \frac{\text{Ker}(\mathcal{I}^n(U) \to \mathcal{I}^{n + 1}(U))} {\text{Im}(\mathcal{I}^{n - 1}(U) \to \mathcal{I}^n(U))}.$$ 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 $\text{Im}(\mathcal{I}^{n - 1} \to \mathcal{I}^n) = \text{Ker}(\mathcal{I}^n \to \mathcal{I}^{n + 1})$ 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 20.8.3. 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 $$V \longmapsto H^i(f^{-1}(V), \mathcal{F})$$ with restriction mappings as in Equation (20.8.1.1). There is a similar statement for $R^if_*$ applied to a bounded below complex $\mathcal{F}^\bullet$.

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 $$f_*\mathcal{I}^0 \to f_*\mathcal{I}^1 \to f_*\mathcal{I}^2 \to \ldots$$ By definition of the abelian category structure on $\mathcal{O}_Y$-modules this cohomology sheaf is the sheaf associated to the presheaf $$V \longmapsto \frac{\text{Ker}(f_*\mathcal{I}^i(V) \to f_*\mathcal{I}^{i + 1}(V))} {\text{Im}(f_*\mathcal{I}^{i - 1}(V) \to f_*\mathcal{I}^i(V))}$$ and this is obviously equal to $$\frac{\text{Ker}(\mathcal{I}^i(f^{-1}(V)) \to \mathcal{I}^{i + 1}(f^{-1}(V)))} {\text{Im}(\mathcal{I}^{i - 1}(f^{-1}(V)) \to \mathcal{I}^i(f^{-1}(V)))}$$ which is equal to $H^i(f^{-1}(V), \mathcal{F})$ and we win. $\square$

Lemma 20.8.4. 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 $$R^pg_*(\mathcal{F}|_{f^{-1}(V)}) = (R^pf_*\mathcal{F})|_V$$ 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.

Proof. First proof. Apply Lemmas 20.8.3 and 20.8.1 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 20.8.5. Here is a different approach to the proofs of Lemmas 20.8.2 and 20.8.3 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.

1. First prove Lemma 20.12.4 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 12.11.
2. A restatement of Lemma 20.8.2 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 $$(\underline{H}^p(\mathcal{F}))^\# = (R^pi_X\mathcal{F})^\# = R^p(\# \circ i_X)(\mathcal{F}) = 0$$ because $\# \circ i_X$ is the identity functor.
3. 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 20.8.3 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 $$R^p f_* \mathcal{F} = R^p(\# \circ i_Y \circ f_*)\mathcal{F} = (R^p (i_Y \circ f_*)\mathcal{F})^\#$$ as desired.

The code snippet corresponding to this tag is a part of the file cohomology.tex and is located in lines 490–705 (see updates for more information).

\section{Locality of cohomology}
\label{section-locality}

\noindent
The following lemma says there is no ambiguity in defining the cohomology
of a sheaf $\mathcal{F}$ over an open.

\begin{lemma}
\label{lemma-cohomology-of-open}
Let $X$ be a ringed space.
Let $U \subset X$ be an open subspace.
\begin{enumerate}
\item If $\mathcal{I}$ is an injective $\mathcal{O}_X$-module
then $\mathcal{I}|_U$ is an injective $\mathcal{O}_U$-module.
\item For any sheaf of $\mathcal{O}_X$-modules $\mathcal{F}$ we have
$H^p(U, \mathcal{F}) = H^p(U, \mathcal{F}|_U)$.
\end{enumerate}
\end{lemma}

\begin{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 \ref{sheaves-lemma-j-shriek-modules}.
Moreover, $j_!$ is exact. Hence (1) follows from

\medskip\noindent
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).
\end{proof}

\noindent
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 {\it restriction mapping}

\label{equation-restriction-mapping}
H^n(V, \mathcal{F})
\longrightarrow
\xi \longmapsto \xi|_U

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
$$\Gamma(V, \mathcal{I}^\bullet) \longrightarrow \Gamma(U, \mathcal{I}^\bullet)$$
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 \ref{lemma-include}.

\begin{lemma}
\label{lemma-kill-cohomology-class-on-covering}
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 $i \in I$.
\end{lemma}

\begin{proof}
Let $\mathcal{F} \to \mathcal{I}^\bullet$ be an injective resolution.
Then
$$H^n(U, \mathcal{F}) = \frac{\Ker(\mathcal{I}^n(U) \to \mathcal{I}^{n + 1}(U))} {\Im(\mathcal{I}^{n - 1}(U) \to \mathcal{I}^n(U))}.$$
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
$\Im(\mathcal{I}^{n - 1} \to \mathcal{I}^n) = \Ker(\mathcal{I}^n \to \mathcal{I}^{n + 1})$ 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.
\end{proof}

\begin{lemma}
\label{lemma-describe-higher-direct-images}
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
$$V \longmapsto H^i(f^{-1}(V), \mathcal{F})$$
with restriction mappings as in Equation (\ref{equation-restriction-mapping}).
There is a similar statement for $R^if_*$ applied to a
bounded below complex $\mathcal{F}^\bullet$.
\end{lemma}

\begin{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
$$f_*\mathcal{I}^0 \to f_*\mathcal{I}^1 \to f_*\mathcal{I}^2 \to \ldots$$
By definition of the abelian category structure on $\mathcal{O}_Y$-modules
this cohomology sheaf is the sheaf associated to the presheaf
$$V \longmapsto \frac{\Ker(f_*\mathcal{I}^i(V) \to f_*\mathcal{I}^{i + 1}(V))} {\Im(f_*\mathcal{I}^{i - 1}(V) \to f_*\mathcal{I}^i(V))}$$
and this is obviously equal to
$$\frac{\Ker(\mathcal{I}^i(f^{-1}(V)) \to \mathcal{I}^{i + 1}(f^{-1}(V)))} {\Im(\mathcal{I}^{i - 1}(f^{-1}(V)) \to \mathcal{I}^i(f^{-1}(V)))}$$
which is equal to $H^i(f^{-1}(V), \mathcal{F})$
and we win.
\end{proof}

\begin{lemma}
\label{lemma-localize-higher-direct-images}
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
$$R^pg_*(\mathcal{F}|_{f^{-1}(V)}) = (R^pf_*\mathcal{F})|_V$$
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.
\end{lemma}

\begin{proof}
First proof. Apply Lemmas \ref{lemma-describe-higher-direct-images}
and \ref{lemma-cohomology-of-open} 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.
\end{proof}

\begin{remark}
\label{remark-daniel}
Here is a different approach to the proofs of
Lemmas \ref{lemma-kill-cohomology-class-on-covering} and
\ref{lemma-describe-higher-direct-images} 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.
\begin{enumerate}
\item First prove Lemma \ref{lemma-include} 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 \ref{homology-section-cohomological-delta-functor}.
\item A restatement of Lemma \ref{lemma-kill-cohomology-class-on-covering}
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
$$(\underline{H}^p(\mathcal{F}))^\# = (R^pi_X\mathcal{F})^\# = R^p(\# \circ i_X)(\mathcal{F}) = 0$$
because $\# \circ i_X$ is the identity functor.
\item 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 \ref{lemma-describe-higher-direct-images}
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
$$R^p f_* \mathcal{F} = R^p(\# \circ i_Y \circ f_*)\mathcal{F} = (R^p (i_Y \circ f_*)\mathcal{F})^\#$$
as desired.
\end{enumerate}
\end{remark}

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