ScalingStacks

Proof. We will show that the triple $(\Delta^{\times n},T_{\Delta^{\times n}},d)$ satisfies conditions (R.1-4) of Th. 11.2. Condition (R.4) clearly holds. Condition (R.3) is the statement of Lemma 14.5.

For condition (R.2) we must show that $i_{!}(\delta_{n})_{!}(T_{\Delta^{\times n}})\subseteq S$. By Lemma 13.14 it is sufficient to show that $(\delta_{n})_{!}(T_{\Delta^{\times n}})\subseteq T_{\Theta_{n}}$, and as $(\delta_{n})_{!}$ preserves colimits it is sufficient to check this on the generating classes $\mathrm{Segal}_{\Delta^{\!\times n}}$, $\mathrm{Glob}_{\Delta^{\!\times n}}$, and $\mathrm{Comp}_{\Delta^{\!\times n}}$. In each case this is clear: the set $\mathrm{Glob}_{\Delta^{\!\times n}}$ maps under $(\delta_{n})_{!}$ to equivalences in $\pre(\Theta_{n})$, the set $\mathrm{Comp}_{\Delta^{\!\Theta_{n}}}$ is constructed as the image of $\mathrm{Comp}_{\Delta^{\!\times n}}$ under $(\delta_{n})_{!}$, and the image of $\mathrm{Segal}_{\Delta^{\!\times n}}$ under $(\delta_{n})_{!}$ is a subset of $\mathrm{Segal}_{\Delta^{\!\Theta_{n}}}$.

For the final condition (R.1) we must show that $\delta_{n}^{*}i^{*}(S)\subseteq T_{\Delta^{\times n}}$. By Th. 13.13, it suffices to show that $\delta_{n}^{*}(T_{\Theta_{n}})\subseteq T_{\Delta^{\times n}}$. As $\delta_{n}^{*}$ preserves colimits, it is sufficient to prove this for the generating class of $T_{\Theta_{n}}$. As we previously mentioned, the set $\mathrm{Comp}_{\Delta^{\!\Theta_{n}}}$ consists of elements in the image of $(\delta_{n})_{!}$. By Proposition 12.4 the remaining generators of $T_{\Theta_{n}}$ are retracts of maps in the image of $(\delta_{n})_{!}$. Thus $T_{\Theta_{n}}$ is contained in the strongly saturated class generated from $(\delta_{n})_{!}(T_{\Delta^{\times n}})$. Hence $\delta_{n}^{*}(T_{\Theta_{n}})$ is contained in the strongly saturated class generated by $\delta_{n}^{*}(\delta_{n})_{!}(T_{\Delta^{\times n}})$. Using Lemma 14.5 one readily deduces that the generators of $T_{\Delta^{\times n}}$ are mapped, via $\delta_{n}^{*}(\delta_{n})_{!}$ back into $T_{\Delta^{\times n}}$. As the composite functor $\delta_{n}^{*}(\delta_{n})_{!}$ preserves colimits, this implies that the saturated class generated by $\delta_{n}^{*}(\delta_{n})_{!}(T_{\Delta^{\times n}})$ is contained in $T_{\Delta^{\times n}}$, whence $\delta_{n}^{*}(T_{\Theta_{n}})\subseteq T_{\Delta^{\times n}}$. ∎