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The short answer is that $\mathrm{im}\partial_1$ is contained in the subgroup $B$ of $C_0$ given by $\{\sum \lambda_i v_i : \sum \lambda_i = 0, \lambda_i \in \mathbb{Z}, v_i \in K_0\}$ This subgroup is proper as soon as $K_0$ is non-empty: $v \notin B$ for $v \in K_0$. The long story is that you can slightly change your simplicial complex by adding a ...


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Here you go, I will write $d_i$ instead of $\delta_i$ (because I'm lazy): $$\begin{align} \sum_{i=0}^n \sum_{j = 0}^{n+1} (-1)^{i+j} d_i d_j & = \sum_{0 \le i < j \le n+1} (-1)^{i+j} d_i d_j + \sum_{0 \le j \le i \le n} (-1)^{i+j} d_i d_j \\ & = \sum_{0 \le i < j \le n+1} (-1)^{i+j} d_{j-1} d_i + \sum_{0 \le j \le i \le n} (-1)^{i+j} d_i d_j = ...



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