# The Stacks Project

## Tag 017V

Lemma 14.18.6. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object in $\mathcal{A}$. Then $U$ has a splitting obtained by taking $N(U_0) = U_0$ and for $m \geq 1$ taking $$N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \mathop{\rm Ker}(d^m_i).$$ Moreover, this splitting is functorial on the category of simplicial objects of $\mathcal{A}$.

Proof. For any object $A$ of $\mathcal{A}$ we obtain a simplicial abelian group $\mathop{\rm Mor}\nolimits_\mathcal{A}(A, U)$. Each of these are canonically split by Lemma 14.18.5. Moreover, $$N(\mathop{\rm Mor}\nolimits_\mathcal{A}(A, U_m)) = \bigcap\nolimits_{i = 0}^{m - 1} \mathop{\rm Ker}(d^m_i) = \mathop{\rm Mor}\nolimits_\mathcal{A}(A, N(U_m)).$$ Hence we see that the morphism (14.18.1.1) becomes an isomorphism after applying the functor $\mathop{\rm Mor}\nolimits_\mathcal{A}(A, -)$ for any object of $\mathcal{A}$. Hence it is an isomorphism by the Yoneda lemma. $\square$

The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 2017–2028 (see updates for more information).

\begin{lemma}
\label{lemma-splitting-abelian-category}
Let $\mathcal{A}$ be an abelian category.
Let $U$ be a simplicial object in $\mathcal{A}$.
Then $U$ has a splitting obtained by taking $N(U_0) = U_0$ and
for $m \geq 1$ taking
$$N(U_m) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i).$$
Moreover, this splitting is functorial on the category of
simplicial objects of $\mathcal{A}$.
\end{lemma}

\begin{proof}
For any object $A$ of $\mathcal{A}$ we obtain
a simplicial abelian group $\Mor_\mathcal{A}(A, U)$.
Each of these are canonically split by Lemma
\ref{lemma-splitting-simplicial-groups}. Moreover,
$$N(\Mor_\mathcal{A}(A, U_m)) = \bigcap\nolimits_{i = 0}^{m - 1} \Ker(d^m_i) = \Mor_\mathcal{A}(A, N(U_m)).$$
Hence we see that the morphism (\ref{equation-splitting})
becomes an isomorphism after applying the functor
$\Mor_\mathcal{A}(A, -)$ for any object of $\mathcal{A}$.
Hence it is an isomorphism by the Yoneda lemma.
\end{proof}

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