# stacks/stacks-project

Unbounded cohomological descent for hypercoverings

Thanks to Bhargav Bhatt
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
 @@ -6001,6 +6001,7 @@ \section{Double complexes of abelian groups} \section{Injectives} \label{section-injectives}
 @@ -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}
 @@ -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}