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 $\check H^ i(\mathcal{U}, \mathcal{F})$.

## 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

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

by the formula

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

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

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

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.
$\square$

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