$\Newextarrow{\xhookrightarrow}{10,10}{0x21AA}$
Theorem 8.5.8.4. Let $n \geq 3$ be an integer. The inclusion map $\operatorname{N}_{\leq n}( \operatorname{Idem}) \hookrightarrow \operatorname{N}_{\bullet }( \operatorname{Idem})$ admits a factorization
\[ \operatorname{N}_{\leq n}(\operatorname{Idem}) \xrightarrow { \iota } \operatorname{\mathcal{E}}\xrightarrow { U } \operatorname{N}_{\bullet }( \operatorname{Idem}), \]
where $\operatorname{\mathcal{E}}$ is an $\infty $-category, $\iota $ is an inner anodyne morphism which is bijective on simplices of dimension $< n$, and the functor $U$ admits a right inverse $V: \operatorname{N}_{\bullet }( \operatorname{Idem}) \rightarrow \operatorname{\mathcal{E}}$.
Proof of Theorem 8.5.8.4.
Fix an integer $n \geq 0$. Using Example 8.5.8.12, we can choose an $\infty $-category $\operatorname{\mathcal{E}}$ and an inner anodyne morphism $\iota : \operatorname{N}_{\leq n}(\operatorname{Idem}) \hookrightarrow \operatorname{\mathcal{E}}$ which is bijective on simplices of dimension $< n$. Since $\iota $ is inner anodyne, there is a unique functor $U: \operatorname{\mathcal{E}}\rightarrow \operatorname{N}_{\bullet }( \operatorname{Idem})$ for which $U \circ \iota $ is coincides with the inclusion map $\operatorname{N}_{\leq n}( \operatorname{Idem}) \hookrightarrow \operatorname{N}_{\bullet }( \operatorname{Idem})$. If $n \geq 3$, then Proposition 8.5.8.13 guarantees that $\iota (\widetilde{e})$ is an idempotent endomorphism in $\operatorname{\mathcal{E}}$: that is, there exists a functor $V: \operatorname{N}_{\bullet }( \operatorname{Idem}) \rightarrow \operatorname{\mathcal{E}}$ satisfying $V(\widetilde{e}) = \iota (\widetilde{e})$. To complete the proof, it will suffice to show that the composition $\operatorname{N}_{\bullet }( \operatorname{Idem}) \xrightarrow {V} \operatorname{\mathcal{E}}\xrightarrow {U} \operatorname{N}_{\bullet }( \operatorname{Idem})$ is the identity functor. This follows from the universal property of Remark 8.5.2.8 (together with Proposition 1.3.3.1), it since the functor $U \circ V$ carries the morphism $e$ to itself.
$\square$