The Stacks project

20.9 The Čech complex and Čech cohomology

Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering, see Topology, Basic notion (13). As is customary we denote $U_{i_0\ldots i_ p} = U_{i_0} \cap \ldots \cap U_{i_ p}$ for the $(p + 1)$-fold intersection of members of $\mathcal{U}$. Let $\mathcal{F}$ be an abelian presheaf on $X$. Set

\[ \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F}) = \prod \nolimits _{(i_0, \ldots , i_ p) \in I^{p + 1}} \mathcal{F}(U_{i_0\ldots i_ p}). \]

This is an abelian group. For $s \in \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F})$ we denote $s_{i_0\ldots i_ p}$ its value in $\mathcal{F}(U_{i_0\ldots i_ p})$. Note that if $s \in \check{\mathcal{C}}^1(\mathcal{U}, \mathcal{F})$ and $i, j \in I$ then $s_{ij}$ and $s_{ji}$ are both elements of $\mathcal{F}(U_ i \cap U_ j)$ but there is no imposed relation between $s_{ij}$ and $s_{ji}$. In other words, we are not working with alternating cochains (these will be defined in Section 20.23). We define

\[ d : \check{\mathcal{C}}^ p(\mathcal{U}, \mathcal{F}) \longrightarrow \check{\mathcal{C}}^{p + 1}(\mathcal{U}, \mathcal{F}) \]

by the formula
\begin{equation} \label{cohomology-equation-d-cech} d(s)_{i_0\ldots i_{p + 1}} = \sum \nolimits _{j = 0}^{p + 1} (-1)^ j s_{i_0\ldots \hat i_ j \ldots i_{p + 1}}|_{U_{i_0\ldots i_{p + 1}}} \end{equation}

It is straightforward to see that $d \circ d = 0$. In other words $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is a complex.

Definition 20.9.1. Let $X$ be a topological space. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. Let $\mathcal{F}$ be an abelian presheaf on $X$. The complex $\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F})$ is the Čech complex associated to $\mathcal{F}$ and the open covering $\mathcal{U}$. Its cohomology groups $H^ i(\check{\mathcal{C}}^\bullet (\mathcal{U}, \mathcal{F}))$ are called the Čech cohomology groups associated to $\mathcal{F}$ and the covering $\mathcal{U}$. They are denoted $\check H^ i(\mathcal{U}, \mathcal{F})$.

Lemma 20.9.2. Let $X$ be a topological space. Let $\mathcal{F}$ be an abelian presheaf on $X$. The following are equivalent

  1. $\mathcal{F}$ is an abelian sheaf and

  2. for every open covering $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ the natural map

    \[ \mathcal{F}(U) \to \check{H}^0(\mathcal{U}, \mathcal{F}) \]

    is bijective.

Proof. This is true since the sheaf condition is exactly that $\mathcal{F}(U) \to \check{H}^0(\mathcal{U}, \mathcal{F})$ is bijective for every open covering. $\square$

Lemma 20.9.3. Let $X$ be a topological space. Let $\mathcal{F}$ be an abelian presheaf on $X$. Let $\mathcal{U} : U = \bigcup _{i \in I} U_ i$ be an open covering. If $U_ i = U$ for some $i \in I$, then the extended Čech complex

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

obtained by putting $\mathcal{F}(U)$ in degree $-1$ with differential given by the canonical map of $\mathcal{F}(U)$ into $\check{\mathcal{C}}^0(\mathcal{U}, \mathcal{F})$ is homotopy equivalent to $0$.

Proof. Fix an element $i \in I$ with $U = U_ i$. Observe that $U_{i_0 \ldots i_ p} = U_{i_0 \ldots \hat i_ j \ldots i_ p}$ if $i_ j = i$. Let us define a homotopy

\[ h : \prod \nolimits _{i_0 \ldots i_{p + 1}} \mathcal{F}(U_{i_0 \ldots i_{p + 1}}) \longrightarrow \prod \nolimits _{i_0 \ldots i_ p} \mathcal{F}(U_{i_0 \ldots i_ p}) \]

by the rule

\[ h(s)_{i_0 \ldots i_ p} = s_{i i_0 \ldots i_ p} \]

In other words, $h : \prod _{i_0} \mathcal{F}(U_{i_0}) \to \mathcal{F}(U)$ is projection onto the factor $\mathcal{F}(U_ i) = \mathcal{F}(U)$ and in general the map $h$ equals the projection onto the factors $\mathcal{F}(U_{i i_1 \ldots i_{p + 1}}) = \mathcal{F}(U_{i_1 \ldots i_{p + 1}})$. We compute

\begin{align*} (dh + hd)(s)_{i_0 \ldots i_ p} & = \sum \nolimits _{j = 0}^ p (-1)^ j h(s)_{i_0 \ldots \hat i_ j \ldots i_ p} + d(s)_{i i_0 \ldots i_ p}\\ & = \sum \nolimits _{j = 0}^ p (-1)^ j s_{i i_0 \ldots \hat i_ j \ldots i_ p} + s_{i_0 \ldots i_ p} + \sum \nolimits _{j = 0}^ p (-1)^{j + 1} s_{i i_0 \ldots \hat i_ j \ldots i_ p} \\ & = s_{i_0 \ldots i_ p} \end{align*}

This proves the identity map is homotopic to zero as desired. $\square$

Comments (2)

Comment #783 by Anfang Zhou on

Typo. In the second paragraph, it should be .

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