The Stacks project

Corrected tiny typos.

Tags: added new tags to the project

	modified:   tags/tags

Hi, I somehow felt that there should be a cleaner proof of locality of
cohomology, where 'cleaner' means avoiding any appeal to injectives
which I find a bit unappealing... So this is how I would rewrite the
proofs of that and two nearby lemmas.
Best,
Daniel

P.S.: I am realizing now that I am not following the directions to
contributors... sorry!

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Put Lemma 13.10.3 somewhere at the beginning of 13.5.

Proof of Lemma [[13.10.3]]

Given a short exact sequence of sheaves, the long exact sequences in
cohomology associated to their restrictions to each $U\subset X$ open
can be viewed as a long exact sequence in the category
$\textit{PMod}(\mathcal{O}_X)$.
This shows that the functors $i$ and $(\underline{H}^n)_{ n\geq 1}$
together form a $\delta$-functor, which is universal by Lemma 9.8.4
since $\underline{H}^n(\mathcal{I})=0$ whenever $\mathcal{I}$ is
injective and $n>0$.
The conclusion then follows from Lemma 9.8.5.

Proof of Lemma 13.5.2

Let $\sharp: \textit{PMod}(\mathcal{O}_X) \rightarrow
\textit{Mod}(\mathcal{O}_X)$ denote the sheafification functor.
We have to prove that $\sharp\underline{H}^n(\mathcal{F})=0$ for all
$p>0$.
By Lemma [[13.10.3]] we have $\underline{H}^n(\mathcal{F})=R^n
i(\mathcal{F})$, where $i\textit{Mod}(\mathcal{O}_X) \rightarrow
\textit{PMod}(\mathcal{O}_X)$ is the inclusion.
Now since the functor $\sharp$ is exact we have $\sharp R^n
i(\mathcal{F})=R^n(\sharp \circ i)(\mathcal{F})$ and this is $0$ for
$n>0$ since $\sharp\circ i$ is the identity functor.

Proof of Lemma 13.5.2

The same argument used in the proof of Lemma [[13.10.3]] above shows
that the displayed presheaves are the right derived functors of the
composition $i\circ f_*: \textit{Mod}(\mathcal{O}_X) \rightarrow
\textit{PMod}(\mathcal{O}_Y)$, where $i:\textit{Mod}(\mathcal{O}_Y)
\rightarrow \textit{PMod}(\mathcal{O}_Y)$ is the inclusion.
Then we have to show that $R^p f_* \mathcal{F}=\sharp R^p (i\circ
f_*)\mathcal{F}$.
Since the sheafification functor $\sharp$ is exact and $\sharp\circ i$
is the identity we  have $R^p f_* \mathcal{F} = R^p(\sharp\circ i
\circ f_*)\mathcal{F} =\sharp R^p (i\circ f_*)\mathcal{F}$, as
desired.

Proof of Lemma 14.4.1

Same as for 13.5.2

	modified:   CONTRIBUTORS
	modified:   cohomology.tex

Signed-off-by: Aise Johan de Jong -- Strider <aise.johan.de.jong@gmail.com>