The Stacks project

95.18 The Čech complex

To compute the cohomology of a sheaf on an algebraic stack we compare it to the cohomology of the sheaf restricted to coverings of the given algebraic stack.

Throughout this section the situation will be as follows. We are given a $1$-morphism of categories fibred in groupoids
\begin{equation} \label{stacks-sheaves-equation-covering} \vcenter { \xymatrix{ \mathcal{U} \ar[rr]_ f \ar[rd]_ q & & \mathcal{X} \ar[ld]^ p \\ & (\mathit{Sch}/S)_{fppf} } } \end{equation}

We are going to think about $\mathcal{U}$ as a “covering” of $\mathcal{X}$. Hence we want to consider the simplicial object

\[ \xymatrix{ \mathcal{U} \times _\mathcal {X} \mathcal{U} \times _\mathcal {X} \mathcal{U} \ar@<1ex>[r] \ar@<0ex>[r] \ar@<-1ex>[r] & \mathcal{U} \times _\mathcal {X} \mathcal{U} \ar@<0.5ex>[r] \ar@<-0.5ex>[r] & \mathcal{U} } \]

in the category of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. However, since this is a $(2, 1)$-category and not a category, we should say explicitly what we mean. Namely, we let $\mathcal{U}_ n$ be the category with objects $(u_0, \ldots , u_ n, x, \alpha _0, \ldots , \alpha _ n)$ where $\alpha _ i : f(u_ i) \to x$ is an isomorphism in $\mathcal{X}$. We denote $f_ n : \mathcal{U}_ n \to \mathcal{X}$ the $1$-morphism which assigns to $(u_0, \ldots , u_ n, x, \alpha _0, \ldots , \alpha _ n)$ the object $x$. Note that $\mathcal{U}_0 = \mathcal{U}$ and $f_0 = f$. Given a map $\varphi : [m] \to [n]$ we consider the $1$-morphism $\mathcal{U}_\varphi : \mathcal{U}_ n \longrightarrow \mathcal{U}_ n$ given by

\[ (u_0, \ldots , u_ n, x, \alpha _0, \ldots , \alpha _ n) \longmapsto (u_{\varphi (0)}, \ldots , u_{\varphi (m)}, x, \alpha _{\varphi (0)}, \ldots , \alpha _{\varphi (m)}) \]

on objects. All of these $1$-morphisms compose correctly on the nose (no $2$-morphisms required) and all of these $1$-morphisms are $1$-morphisms over $\mathcal{X}$. We denote $\mathcal{U}_\bullet $ this simplicial object. If $\mathcal{F}$ is a presheaf of sets on $\mathcal{X}$, then we obtain a cosimplicial set

\[ \xymatrix{ \Gamma (\mathcal{U}_0, f_0^{-1}\mathcal{F}) \ar@<0.5ex>[r] \ar@<-0.5ex>[r] & \Gamma (\mathcal{U}_1, f_1^{-1}\mathcal{F}) \ar@<1ex>[r] \ar@<0ex>[r] \ar@<-1ex>[r] & \Gamma (\mathcal{U}_2, f_2^{-1}\mathcal{F}) } \]

Here the arrows are the pullback maps along the given morphisms of the simplicial object. If $\mathcal{F}$ is a presheaf of abelian groups, this is a cosimplicial abelian group.

Let $\mathcal{U} \to \mathcal{X}$ be as above and let $\mathcal{F}$ be an abelian presheaf on $\mathcal{X}$. The Čech complex associated to the situation is denoted $\check{\mathcal{C}}^\bullet (\mathcal{U} \to \mathcal{X}, \mathcal{F})$. It is the cochain complex associated to the cosimplicial abelian group above, see Simplicial, Section 14.25. It has terms

\[ \check{\mathcal{C}}^ n(\mathcal{U} \to \mathcal{X}, \mathcal{F}) = \Gamma (\mathcal{U}_ n, f_ n^{-1}\mathcal{F}). \]

The boundary maps are the maps

\[ d^ n = \sum \nolimits _{i = 0}^{n + 1} (-1)^ i \delta ^{n + 1}_ i : \Gamma (\mathcal{U}_ n, f_ n^{-1}\mathcal{F}) \longrightarrow \Gamma (\mathcal{U}_{n + 1}, f_{n + 1}^{-1}\mathcal{F}) \]

where $\delta ^{n + 1}_ i$ corresponds to the map $[n] \to [n + 1]$ omitting the index $i$. Note that the map $\Gamma (\mathcal{X}, \mathcal{F}) \to \Gamma (\mathcal{U}_0, f_0^{-1}\mathcal{F}_0)$ is in the kernel of the differential $d^0$. Hence we define the extended Čech complex to be the complex

\[ \ldots \to 0 \to \Gamma (\mathcal{X}, \mathcal{F}) \to \Gamma (\mathcal{U}_0, f_0^{-1}\mathcal{F}_0) \to \Gamma (\mathcal{U}_1, f_1^{-1}\mathcal{F}_1) \to \ldots \]

with $\Gamma (\mathcal{X}, \mathcal{F})$ placed in degree $-1$. The extended Čech complex is acyclic if and only if the canonical map

\[ \Gamma (\mathcal{X}, \mathcal{F})[0] \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{U} \to \mathcal{X}, \mathcal{F}) \]

is a quasi-isomorphism of complexes.

Lemma 95.18.1. Generalities on Čech complexes.

  1. If

    \[ \xymatrix{ \mathcal{V} \ar[d]_ g \ar[r]_ h & \mathcal{U} \ar[d]^ f \\ \mathcal{Y} \ar[r]^ e & \mathcal{X} } \]

    is $2$-commutative diagram of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$, then there is a morphism of Čech complexes

    \[ \check{\mathcal{C}}^\bullet (\mathcal{U} \to \mathcal{X}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}^\bullet (\mathcal{V} \to \mathcal{Y}, e^{-1}\mathcal{F}) \]
  2. if $h$ and $e$ are equivalences, then the map of (1) is an isomorphism,

  3. if $f, f' : \mathcal{U} \to \mathcal{X}$ are $2$-isomorphic, then the associated Čech complexes are isomorphic,

Proof. In the situation of (1) let $t : f \circ h \to e \circ g$ be a $2$-morphism. The map on complexes is given in degree $n$ by pullback along the $1$-morphisms $\mathcal{V}_ n \to \mathcal{U}_ n$ given by the rule

\[ (v_0, \ldots , v_ n, y, \beta _0, \ldots , \beta _ n) \longmapsto (h(v_0), \ldots , h(v_ n), e(y), e(\beta _0) \circ t_{v_0}, \ldots , e(\beta _ n) \circ t_{v_ n}). \]

For (2), note that pullback on global sections is an isomorphism for any presheaf of sets when the pullback is along an equivalence of categories. Part (3) follows on combining (1) and (2). $\square$

Lemma 95.18.2. If there exists a $1$-morphism $s : \mathcal{X} \to \mathcal{U}$ such that $f \circ s$ is $2$-isomorphic to $\text{id}_\mathcal {X}$ then the extended Čech complex is homotopic to zero.

Proof. Set $\mathcal{U}' = \mathcal{U} \times _\mathcal {X} \mathcal{X}$ equal to the fibre product as described in Categories, Lemma 4.32.3. Set $f' : \mathcal{U}' \to \mathcal{X}$ equal to the second projection. Then $\mathcal{U} \to \mathcal{U}'$, $u \mapsto (u, f(x), 1)$ is an equivalence over $\mathcal{X}$, hence we may replace $(\mathcal{U}, f)$ by $(\mathcal{U}', f')$ by Lemma 95.18.1. The advantage of this is that now $f'$ has a section $s'$ such that $f' \circ s' = \text{id}_\mathcal {X}$ on the nose. Namely, if $t : s \circ f \to \text{id}_\mathcal {X}$ is a $2$-isomorphism then we can set $s'(x) = (s(x), x, t_ x)$. Thus we may assume that $f \circ s = \text{id}_\mathcal {X}$.

In the case that $f \circ s = \text{id}_\mathcal {X}$ the result follows from general principles. We give the homotopy explicitly. Namely, for $n \geq 0$ define $s_ n : \mathcal{U}_ n \to \mathcal{U}_{n + 1}$ to be the $1$-morphism defined by the rule on objects

\[ (u_0, \ldots , u_ n, x, \alpha _0, \ldots , \alpha _ n) \longmapsto (u_0, \ldots , u_ n, s(x), x, \alpha _0, \ldots , \alpha _ n, \text{id}_ x). \]


\[ h^{n + 1} : \Gamma (\mathcal{U}_{n + 1}, f_{n + 1}^{-1}\mathcal{F}) \longrightarrow \Gamma (\mathcal{U}_ n, f_ n^{-1}\mathcal{F}) \]

as pullback along $s_ n$. We also set $s_{-1} = s$ and $h^0 : \Gamma (\mathcal{U}_0, f_0^{-1}\mathcal{F}) \to \Gamma (\mathcal{X}, \mathcal{F})$ equal to pullback along $s_{-1}$. Then the family of maps $\{ h^ n\} _{n \geq 0}$ is a homotopy between $1$ and $0$ on the extended Čech complex. $\square$

Comments (2)

Comment #4287 by Félix Baril Boudreau on

I suspect that there is a small typo: the -morphism induced by the map of posets should be

given by

as otherwise it would be independent of .

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 06X3. Beware of the difference between the letter 'O' and the digit '0'.