# The Stacks Project

## Tag 017X

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. $\square$

The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 2067–2073 (see updates for more information).

\begin{lemma}
\label{lemma-N-d-in-N}
Let $\mathcal{A}$ be an abelian category.
Let $U$ be a simplicial object in $\mathcal{A}$.
Let $N(U_m)$ as in Lemma \ref{lemma-splitting-abelian-category} above.
Then $d^m_m(N(U_m)) \subset N(U_{m - 1})$.
\end{lemma}

\begin{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 \ref{remark-relations}.
The result follows.
\end{proof}

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