Finding a cochain with specific coboundary Let $K$ be the complex whose space is the real line and whose vertices are the integers. Let $\sigma_n = [n,n+1]$. Show that $H^1(K)=0$ by finding a specific cochain whose coboundary is $\Sigma^\infty_{i=-\infty}a_i \sigma_i^*$.
Would the specific cochain here be the formal sum of all the vertices? Or would that just be a cocycle but not a coboundary? I'm new to this, any insight helps!
 A: The first step should be to compute what the coboundary of the dual of a single vertex is. Vertices are integers, so for notational convenience I let $v_n$ be the vertex corresponding to $n \in \mathbb{Z}$.
The dual element is $v_n^* \in C^0(K)$. What's $\delta(v_n^*)$? You need to look at all $1$-simplices (edges) of $K$ and determine when their boundary contains $n$. You know all the edges of $K$: they're all of the form $\sigma_k = [k, k+1]$. The boundary of $\sigma_k$ is $\partial \sigma_k = v_{k+1} - v_k$. So clearly, only $\partial\sigma_{n-1} = v_n - v_{n-1}$ and $\partial\sigma_n = v_{n+1} - v_n$ can contain $v_n$ in their boundary. It follows, from the definition of the differential on cochains, that:
$$\delta(v_n^*) = \sigma_{n-1}^* - \sigma_n^*.$$
Armed with this relation, it becomes a simpler matter of algebra. You are looking for a $0$-cochain,
$$X = \sum_{n \in \mathbb{Z}} b_n v_n^*,$$
such that $\delta(X) = \sum_{n \in \mathbb{Z}} a_n \sigma_n^*$. A simple computation shows that:
$$\delta(X) = \sum_n b_n \delta(v_n^*) = \sum_n b_n (\sigma_{n-1}^* - \sigma_n^*).$$
It then remains to identify the coefficients in $\delta(X) = \sum_n a_n \sigma_n^*$. You see that you need to solve the system of equations:
$$\{ a_n = b_{n+1} - b_n \mid n \in \mathbb{Z} \}.$$
There are several ways to solve this system (the solution is not unique). You basically need to fix one of the coefficients, say $b_0 = 0$. Then by writing down the first few equations, you find $b_1 = a_0$, $b_2 = a_0 + a_1$, $b_3 = a_0 + a_1 + a_2$ and so on, and in the reverse direction $b_{-1} = -a_{-1}$, $b_{-2} = -a_{-1} - a_{-2}$, $b_{-3} = -a_{-1} - a_{-2} - a_{-3}$ and so on.
