# The Stacks Project

## Tag 017Y

Lemma 14.18.9. Let $\mathcal{A}$ be an abelian category. Let $U$ be a simplicial object of $\mathcal{A}$. Let $n \geq 0$ be an integer. The rule $$U'_m = \sum\nolimits_{\varphi : [m] \to [i], ~i\leq n} \mathop{\rm Im}(U(\varphi))$$ defines a sub simplicial object $U' \subset U$ with $U'_i = U_i$ for $i \leq n$. Moreover, $N(U'_m) = 0$ for all $m > n$.

Proof. Pick $m$, $i \leq n$ and some $\varphi : [m] \to [i]$. The image under $U(\psi)$ of $\mathop{\rm Im}(U(\varphi))$ for any $\psi : [m'] \to [m]$ is equal to the image of $U(\varphi \circ \psi)$ and $\varphi \circ \psi : [m'] \to [i]$. Hence $U'$ is a simplicial object. Pick $m > n$. We have to show $N(U'_m) = 0$. By definition of $N(U_m)$ and $N(U'_m)$ we have $N(U'_m) = U'_m \cap N(U_m)$ (intersection of subobjects). Since $U$ is split by Lemma 14.18.6, it suffices to show that $U'_m$ is contained in the sum $$\sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, ~m' < m} \mathop{\rm Im}(U(\varphi)|_{N(U_{m'})}).$$ By the splitting each $U_{m'}$ is the sum of images of $N(U_{m''})$ via $U(\psi)$ for surjective maps $\psi : [m'] \to [m'']$. Hence the displayed sum above is the same as $$\sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, ~m' < m} \mathop{\rm Im}(U(\varphi)).$$ Clearly $U'_m$ is contained in this by the simple fact that any $\varphi : [m] \to [i]$, $i \leq n$ occurring in the definition of $U'_m$ may be factored as $[m] \to [m'] \to [i]$ with $[m] \to [m']$ surjective and $m' < m$ as in the last displayed sum above. $\square$

The code snippet corresponding to this tag is a part of the file simplicial.tex and is located in lines 2082–2094 (see updates for more information).

\begin{lemma}
\label{lemma-simplicial-abelian-n-skel-sub}
Let $\mathcal{A}$ be an abelian category.
Let $U$ be a simplicial object of $\mathcal{A}$.
Let $n \geq 0$ be an integer.
The rule
$$U'_m = \sum\nolimits_{\varphi : [m] \to [i], \ i\leq n} \Im(U(\varphi))$$
defines a sub simplicial object $U' \subset U$ with $U'_i = U_i$
for $i \leq n$.
Moreover, $N(U'_m) = 0$ for all $m > n$.
\end{lemma}

\begin{proof}
Pick $m$, $i \leq n$ and some $\varphi : [m] \to [i]$.
The image under $U(\psi)$ of $\Im(U(\varphi))$
for any $\psi : [m'] \to [m]$ is
equal to the image of $U(\varphi \circ \psi)$ and
$\varphi \circ \psi : [m'] \to [i]$.
Hence $U'$ is a simplicial object.
Pick $m > n$. We have to show $N(U'_m) = 0$.
By definition of $N(U_m)$ and $N(U'_m)$ we have
$N(U'_m) = U'_m \cap N(U_m)$ (intersection of subobjects).
Since $U$ is split by Lemma \ref{lemma-splitting-abelian-category},
it suffices to show that $U'_m$ is contained in the sum
$$\sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \Im(U(\varphi)|_{N(U_{m'})}).$$
By the splitting each $U_{m'}$ is the sum of images of
$N(U_{m''})$ via $U(\psi)$ for surjective maps
$\psi : [m'] \to [m'']$. Hence the displayed sum above
is the same as
$$\sum\nolimits_{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \Im(U(\varphi)).$$
Clearly $U'_m$ is contained in this by the simple fact that
any $\varphi : [m] \to [i]$, $i \leq n$ occurring in the definition
of $U'_m$ may be factored as
$[m] \to [m'] \to [i]$ with $[m] \to [m']$ surjective
and $m' < m$ as in the last displayed sum above.
\end{proof}

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