I'm trying to understand the following lemma from Weiebl.
https://i.stack.imgur.com/PVKiS.jpg (found from here: https://people.math.rochester.edu/faculty/doug//otherpapers/weibel-hom.pdf)
I will list the necessary definitions.
Let $G$ be a simplicail abelian group. Let $N_n(G) = \{x \in G_n \colon x \partial_i = 1 \text{ for all } i \neq n \}$. Clearly, $N_n(G) = \cap_{i = 0}^{n-1} \text{ker}(\partial_i G_n \to G_{n-1})$.
$N(G)$ is a sub chain comeplex of $C(G)$, the unnormalized complex. Let $D(G)$ be the subcomplex of $C(G)$ generated by image of $\sigma_i \colon G_{n-1} \to G_n$.
$N(G)$ is also a chain complex with differential $\partial_n \colon N(G)_n \to N(G)_{n-1}$, which picks out the $n$-th face that is not $1$.
The theorem asserts that $C(G) = N(G) \oplus D(G)$.
The part from the proof I don't understand is showing $N(G)$ and $D(G)$ have trivial intersection. To show this, they assume $y \in D_n(G)$, so $y = \sum \sigma_j(x_j)$, $x_j \in C_{n-1}(G)$. Then we assume $y \in N_n(G)$ too to get a contradiction. To get this, they assume $i$ is the smallest integer such that $\sigma_i (x_i) \neq 1$. Then they say $\partial_i(y) = x_i$, but this is the equality that I do not understand. I know that $\partial_i \sigma_i x_i = x_i$, but how does rest of the elements cancel?
Thanks for your help!