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

$U'_ m = \sum \nolimits _{\varphi : [m] \to [i], \ i\leq n} \mathop{\mathrm{Im}}(U(\varphi ))$

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 $\mathop{\mathrm{Im}}(U(\varphi ))$ 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

$\sum \nolimits _{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \mathop{\mathrm{Im}}(U(\varphi )|_{N(U_{m'})}).$

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

$\sum \nolimits _{\varphi : [m] \to [m']\text{ surjective}, \ m' < m} \mathop{\mathrm{Im}}(U(\varphi )).$

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. $\square$

There are also:

• 6 comment(s) on Section 14.18: Splitting simplicial objects

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).