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
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
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
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. $\square$
Your email address will not be published. Required fields are marked.
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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: