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. $\square$

There are also:

• 2 comment(s) on Section 20.9: The Čech complex and Čech cohomology

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).