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Unbounded cohomological descent for hypercoverings

Thanks to Bhargav Bhatt
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aisejohan committed Oct 10, 2018
1 parent 3752d1a commit 620effaa97812b6d1178de720939fb3ba8170618
Showing with 285 additions and 15 deletions.
  1. +1 −0 homology.tex
  2. +85 −0 more-algebra.tex
  3. +36 −0 sites-modules.tex
  4. +163 −15 spaces-simplicial.tex
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@@ -6001,6 +6001,7 @@ \section{Double complexes of abelian groups}
\section{Injectives}
\label{section-injectives}
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@@ -24314,6 +24314,91 @@ \section{Miscellany}
\section{Tricks with double complexes}
\label{section-tricks}
\noindent
This section continues the discussion in
Homology, Section \ref{homology-section-double-complexes-abelian-groups}.
\begin{lemma}
\label{lemma-prod-qis-gives-qis}
Let
$$
(A_0^\bullet \to A_1^\bullet \to A_2^\bullet \to \ldots)
\longrightarrow
(B_0^\bullet \to B_1^\bullet \to B_2^\bullet \to \ldots)
$$
be a map between two complexes of complexes of abelian groups.
Set $A^{p, q} = A_p^q$, $B^{p, q} = B_p^q$ to obtain double complexes.
Let $\text{Tot}_\pi(A^{\bullet, \bullet})$
and $\text{Tot}_\pi(B^{\bullet, \bullet})$ be the
product total complexes associated to the double complexes.
If each $A_p^\bullet \to B_p^\bullet$ is a
quasi-isomorphism, then
$\text{Tot}_\pi(A^{\bullet, \bullet}) \to \text{Tot}_\pi(B^{\bullet, \bullet})$
is a quasi-isomorphism.
\end{lemma}
\begin{proof}
Recall that $\text{Tot}_\pi(A^{\bullet, \bullet})$ in degree
$n$ is given by $\prod_{p + q = n} A^{p, q} = \prod_{p + 1 = n} A^q_p$.
Let $C_p^\bullet$ be the cone on the map $A_p^\bullet \to B_p^\bullet$,
see Derived, Section \ref{derived-section-cones}.
By the functoriality of the cone construction we obtain a
complex of complexes
$$
C_0^\bullet \to C_1^\bullet \to C_2^\bullet \to \ldots
$$
Then we see $\text{Tot}_\pi(C^{\bullet, \bullet})$ in degree $n$
is given by
$$
\prod_{p + q = n} C^{p, q} = \prod_{p + q = n} C^q_p =
\prod_{p + q = n} (B^q_p \oplus A^{q + 1}_p) =
\prod_{p + q = n} B^q_p \oplus \prod_{p + q = n} A^{q + 1}_p
$$
We conclude that $\text{Tot}_\pi(C^{\bullet, \bullet})$
is the cone of the map
$\text{Tot}_\pi(A^{\bullet, \bullet}) \to \text{Tot}_\pi(B^{\bullet, \bullet})$
(We omit the verification that the differentials agree.)
Thus it suffices to show $\text{Tot}_\pi(A^{\bullet, \bullet})$ is
acyclic if each $A_p^\bullet$ is acyclic.
\medskip\noindent
Denote $f_p : A_p^\bullet \to A_{p + 1}^\bullet$ the given maps
of complexes. Recall that the differential on
$\text{Tot}_\pi(A^{\bullet, \bullet})$ is given by
$$
\prod\nolimits_{p + q = n} (f^q_p + (-1)^p\text{d}^q_{A_p^\bullet})
$$
on elements in degree $n$.
Let $\xi \in H^0(\text{Tot}_\pi(A^{\bullet, \bullet}))$ be a cohomology
class. We will show $\xi$ is zero; the same argument works in other
degrees. Represent $\xi$ as the class of an cocycle
$x = (x_p) \in \prod A^{p, -p}$.
Since $\text{d}(x) = 0$ we find that
$\text{d}_{A_0^\bullet}(x_0) = 0$.
Thus there exists a $y_{-1} \in A^{0, -1}$ with
$\text{d}_{A_0^\bullet}(y_{-1}) = x_0$.
Then we see that $\text{d}_{A_1^\bullet}(x_1 + f_0(y_{-1})) = 0$.
Thus we can find a $y_{-2} \in A^{1, -2}$ such that
$-\text{d}_{A_1^\bullet}(y_{-2}) = x_1 + f_0(y_{-1})$.
By induction we can find
$y_{-p - 1} \in A^{p, -p - 1}$ such that
$$
(-1)^p\text{d}_{A_p^\bullet}(y_{-p - 1}) = x_p + f_{p - 1}(y_{-p})
$$
This implies that $\text{d}(y) = x$ where $y = (y_{-p - 1})$.
\end{proof}
\section{Weakly \'etale ring maps}
\label{section-weakly-etale}
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@@ -3849,6 +3849,42 @@ \section{Internal Hom}
The lemma follows from this and the Yoneda lemma.
\end{proof}
\begin{remark}
\label{remark-j-shriek-tensor}
Let $\mathcal{C}$ be a site. Let $\mathcal{F}$ be a sheaf of
sets on $\mathcal{C}$ and consider the localization morphism
$j : \Sh(\mathcal{C})/\mathcal{F} \to \Sh(\mathcal{C})$.
See Sites, Section \ref{sites-definition-localize-topos}.
We claim that (a) $j_!\mathbf{Z} = \mathbf{Z}_\mathcal{F}^\#$
and (b) $j_!(j^{-1}\mathcal{H}) = j_!\mathbf{Z} \otimes_\mathbf{Z} \mathcal{H}$
for any abelian sheaf $\mathcal{H}$ on $\mathcal{C}$.
Let $\mathcal{G}$ be an abelian on $\mathcal{C}$.
Part (a) follows from the Yoneda lemma because
$$
\Hom(j_!\mathbf{Z}, \mathcal{G}) =
\Hom(\mathbf{Z}, j^{-1}\mathcal{G}) =
\Hom(\mathbf{Z}_\mathcal{F}^\#, \mathcal{G})
$$
where the second equality holds because both sides of
the equality evaluate to the set of maps from $\mathcal{F} \to \mathcal{G}$
viewed as an abelian group. For (b) we use the Yoneda lemma and
\begin{align*}
\Hom(j_!(j^{-1}\mathcal{H}), \mathcal{G})
& =
\Hom(j^{-1}\mathcal{H}, j^{-1}\mathcal{G}) \\
& =
\Hom(\mathbf{Z}, \SheafHom(j^{-1}\mathcal{H}, j^{-1}\mathcal{G})) \\
& =
\Hom(\mathbf{Z}, j^{-1}\SheafHom(\mathcal{H}, \mathcal{G})) \\
& =
\Hom(j_!\mathbf{Z}, \SheafHom(\mathcal{H}, \mathcal{G})) \\
& =
\Hom(j_!\mathbf{Z} \otimes_\mathbf{Z} \mathcal{H}, \mathcal{G})
\end{align*}
Here we use adjunction, the fact that taking $\SheafHom$ commutes
with localization, and Lemma \ref{lemma-internal-hom-adjoint-tensor}.
\end{remark}
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