# Kerodon

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Proposition 2.5.5.6. Let $A_{\bullet }$ be a simplicial abelian group. For every positive integer $n$, the boundary operator $\partial : \mathrm{C}_{n}(A) \rightarrow \mathrm{C}_{n-1}(A)$ carries the subgroup $\mathrm{D}_{n}(A)$ into $\mathrm{D}_{n-1}(A)$. Consequently, we can regard $\mathrm{D}_{\ast }(A)$ as a subcomplex of the Moore complex $\mathrm{C}_{\ast }(A)$.

Proof. Choose an element $\sigma \in \mathrm{D}_{n}(A)$; we wish to show that $\partial (\sigma )$ belongs to $\mathrm{D}_{n-1}(A)$. Without loss of generality, we may assume that $\sigma = s_{i}(\tau )$ for some $0 \leq i \leq n-1$ and some $\tau \in A_{n-1}$. We now compute

\begin{eqnarray*} \partial (\sigma ) & = & \sum _{ j =0 }^{n} (-1)^{j} d_ j(\sigma ) \\ & = & (\sum _{j=0}^{i-1} (-1)^{j} d_ j s_ i \tau ) + (-1)^{i} d_ i s_ i \tau + (-1)^{i+1} d_{i+1} s_ i \tau + (\sum _{j = i+2}^{n} (-1)^{j} d_ j s_ i \tau ) \\ & = & (\sum _{j < i} (-1)^{j} s_{i-1} d_ j \tau ) + (-1)^{i} \tau + (-1)^{i+1} \tau + (\sum _{j = i+2}^{n} (-1)^{j} s_{i} d_{j-1}\tau ) \\ & \in & \operatorname{im}( s_{i-1} ) + \operatorname{im}( s_ i ) \\ & \subseteq & \mathrm{D}_{n-1}(A). \end{eqnarray*}
$\square$