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aisejohan committed Oct 3, 2019
1 parent ff09225 commit 48e1102588fe51cba5623f4ff147188ad7ca88aa
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  1. +50 −0 cohomology.tex
@@ -4888,6 +4888,56 @@ \section{{\v C}ech cohomology of complexes}
Immediate from the spectral sequence of Lemma \ref{lemma-cech-complex-complex}.

Let $(X, \mathcal{O}_X)$ be a ringed space. Let
$\mathcal{U} : X = \bigcup_{i \in I} U_i$ be
an open covering. Let $\mathcal{F}^\bullet$ be a bounded below complex
of $\mathcal{O}_X$-modules. Let $b$ be an integer.
We claim there is a commutative diagram
\text{Tot}(\check{\mathcal{C}}^\bullet(\mathcal{U}, \mathcal{F}^\bullet))[b]
\ar[r] \ar[d]_\gamma &
R\Gamma(X, \mathcal{F}^\bullet)[b] \ar[d] \\
\text{Tot}(\check{\mathcal{C}}^\bullet(\mathcal{U}, \mathcal{F}^\bullet[b]))
\ar[r] &
R\Gamma(X, \mathcal{F}^\bullet[b])
in the derived category where the map $\gamma$ is the map on complexes
constructed in Homology, Remark \ref{homology-remark-shift-double-complex}.
This makes sense because the double complex
$\check{\mathcal{C}}^\bullet(\mathcal{U}, \mathcal{F}^\bullet[b])$
is clearly the same as the double complex
$\check{\mathcal{C}}^\bullet(\mathcal{U}, \mathcal{F}^\bullet)[0, b]$
introduced in Homology, Remark \ref{homology-remark-shift-double-complex}.
To check that the diagram commutes, we may choose an injective resolution
$\mathcal{F}^\bullet \to \mathcal{I}^\bullet$ as in the proof of
Lemma \ref{lemma-cech-complex-complex}. Chasing diagrams, we see that
it suffices to check the diagram commutes when we replace $\mathcal{F}^\bullet$
by $\mathcal{I}^\bullet$. Then we consider the extended diagram
\Gamma(X, \mathcal{I}^\bullet)[b] \ar[r] \ar[d] &
\text{Tot}(\check{\mathcal{C}}^\bullet(\mathcal{U}, \mathcal{I}^\bullet))[b]
\ar[r] \ar[d]_\gamma &
R\Gamma(X, \mathcal{I}^\bullet)[b] \ar[d] \\
\Gamma(X, \mathcal{I}^\bullet[b]) \ar[r] &
\text{Tot}(\check{\mathcal{C}}^\bullet(\mathcal{U}, \mathcal{I}^\bullet[b]))
\ar[r] &
R\Gamma(X, \mathcal{I}^\bullet[b])
where the left horizontal arrows are (\ref{equation-global-sections-to-cech}).
Since in this case the horizonal arrows are isomorphisms in the derived
category (see proof of Lemma \ref{lemma-cech-complex-complex}) it
suffices to show that the left square commutes. This is true because
the map $\gamma$ uses the sign $1$ on the summands
$\check{\mathcal{C}}^0(\mathcal{U}, \mathcal{I}^{q + b})$, see
formula in Homology, Remark \ref{homology-remark-shift-double-complex}.

Let $X$ be a topological space, let $\mathcal{U} : X = \bigcup_{i \in I} U_i$
be an open covering, and let $\mathcal{F}^\bullet$ be a bounded below

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