Lemma 14.18.8. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object in $\mathcal{A}$. Let $N(U_ m)$ as in Lemma 14.18.6 above. Then $d^ m_ m(N(U_ m)) \subset N(U_{m - 1})$.

**Proof.**
For $j = 0, \ldots , m - 2$ we have $d^{m - 1}_ j d^ m_ m = d^{m - 1}_{m - 1} d^ m_ j$ by the relations in Remark 14.3.3. The result follows.
$\square$

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