Theorem 8.1 in textbook Differential Forms in Algebraic Topology (Bott and Tu)

Question

In theorem 8.1 the following statement is shown: \begin{equation} H_D(C^*(\mathfrak U,\Omega^*))\simeq H^*_\text{DR}(M), \end{equation} where the lefthand side is cohomology of a double complex and the righthand side is the de Rham cohomology.

In this proof the map \begin{equation} r: \Omega^*(M) \rightarrow \Omega^*(U)\oplus\Omega^*(V)\subset C^*(\mathfrak U,\Omega^*), \end{equation} is considered and the commutativity of the following is used.

\begin{equation} \begin{array}{ccc} \Omega^*(M)& \overset{r}\longrightarrow& C^*(\mathfrak U,\Omega^*)\\ d\Big\uparrow& & \Big\uparrow D\\ \Omega^*(M)& \underset{r}\rightarrow& C^*(\mathfrak U,\Omega^*) \end{array} \end{equation}

The book says it is commutative because \begin{align*} Dr &= (\delta + (-1)^pd)r & [\because\text{it is the definition of $D$ and $p=0$}]\\ &= dr = rd. \end{align*}

I don't understand why operator $\delta$ vanishes in the second line?

p.s. The double complex $C^*(\mathfrak U,\Omega^*)$ is defined, for open cover $\mathfrak U = \{U,V\}$, as \begin{equation} C^0(\mathfrak U,\Omega^q) = \Omega^q(U)\oplus\Omega^q(V),\;\; C^1(\mathfrak U,\Omega^q) = \Omega^q(U\cap V). \end{equation}

Answer 1

Since the open cover has only two elements, we know that $\delta$ vanishes when $p>0$. But, regardless, $\delta\circ r = 0$ by definition, since $(r\omega)|_{U\cap V} = (r\omega)|_{U\cap V}$ for any $\omega\in\Omega^*(M)$.

EDIT: I had to go look at the book. I see that they haven't been explicit yet about defining the coboundary map on Čech cochains. They do in the paragraphs preceding this theorem say that $\delta$ is the difference map going horizontally. That is, if $\sigma = (\sigma_i)\in C^0(\mathfrak U,\Omega^q)$ is a $0$-cochain (i.e., an assignment of a smooth $q$-form to each open set $U_i$), then $\delta(\sigma)\in C^1(\mathfrak U,\Omega^q)$ is the $1$-cochain that assigns to the open set $U_i\cap U_j$ the $q$-form $\sigma_i\big|_{U_i\cap U_j} - \sigma_j\big|_{U_i\cap U_j}$. Thus, $\sigma$ is a $0$-cocycle if and only if $\sigma_i$ and $\sigma_j$ agree on the overlap $U_i\cap U_j$. This means that a $0$-cocycle gives a global form on $M$, and conversely.

Answer 2

I don't think you need to actually go into later discussions of Cech cohomology to get this result. The simpler answer is that the vanishing of $\delta r$ follows immediately from the exactness of the sequence:

$$ 0\rightarrow \Omega^*(U\cup V) \overset{r}\longrightarrow \Omega^*(U)\oplus \Omega^*(V) \overset{\delta}\longrightarrow \Omega^*(U\cap V) \rightarrow 0$$

If I recall correctly, the authors discuss this sequence a page or two before Theorem 8.1, and they prove the exactness of the above sequence (in a manner similar to Ted's answer) in Chapter 5.