ScalingStacks

Proof. By Lemma 13.9, it is enough to show that the strongly saturated classes $T_{\Theta_{n}}^{(b)}$ contains the generating sets $\mathrm{Segal}_{\Theta_{n}}$ and $\mathrm{Comp}_{\Theta_{n}}$. By Lemma 13.10, it suffices to show that for any $[U\to V]\in\mathrm{Segal}_{\Theta_{n}}\cup\mathrm{Comp}_{\Theta_{n}}$, any nondegenerate morphism $[H\to C_{i}]\in J_{b}$, and any nondegenerate morphism $V\to C_{i}$ of $\pre(\Theta_{n})$, we must show that

$$U^{\prime}=U\times_{C_{i}}\nu H\to V\times_{C_{i}}\nu H=V^{\prime}$$

is contained in $T_{\Theta_{n}}$. Observe the following:

• •

  If $[U\to V]\in\mathrm{Segal}_{\Theta_{n}}$ is not in the image of $\sigma_{!}$, then $U\to V$ is contained in $T_{\Theta_{n}}^{(b)}$ by Lemma 13.11.

• •

  If $[U\to V]\in\mathrm{Comp}_{\Theta_{n}}$ is not in the image of $\sigma_{!}$, then $V=C_{0}$, and the only nondegenerate map $V\to C_{i}$ occurs when $i=0$. In this case $U\to V$ is in $T_{\Theta_{n}}^{(b)}$ by [34, Proposition 6.1].

Thus we may restrict our attention to those generators $U\to V$ that lie in the image of $\sigma_{!}$. We proceed by induction. When $n=1$, the set of generators in the image of $\sigma_{!}$ is empty.

Assume that

$$T_{\Theta_{n-1}}=T_{\Theta_{n-1}}^{(a)}=T_{\Theta_{n-1}}^{(b)}=T_{\Theta_{n-1}}^{(c)},$$

and let $U\to V$ be an element of $\mathrm{Segal}_{\Theta_{n}}\cup\mathrm{Comp}_{\Theta_{n}}$ that lies in the image of $\sigma_{!}$. Now note that if $C_{i}=C_{0}$, then $U^{\prime}\to V^{\prime}$ lies in $T_{\Theta_{n}}$, again by [34, Proposition 6.1]. If $i\neq 0$, then by Lemma 13.4, the map $V\to C_{i}$ is also in the image of $\sigma_{!}$. In this case, if we have a factorization $H\to C_{0}\to C_{i}$ (which, since $H\to C_{i}$ is nondegenerate, can only happen if $H=C_{0}$), then $U^{\prime}\to V^{\prime}$ is an equivalence (as both are empty). Hence it $U\to V$ lies in $T_{\Theta_{n}}$.

This leaves the final case, where both $[U\to V]$ and $[V\to C_{i}]$ lie in the image of $\sigma_{!}$, and $[H\to C_{i}]$ is nondegenerate with $H=j({[m]};o_{1},\dots,o_{m})\neq C_{0}$, for some $m\geq 1$, $o_{i}\in\Theta_{n-1}$. The nondegenerate map $H\to C_{i}=({[1]};C_{i-1})$ is given explicitly by the following data (see also the proof of Lemma 13.3): a map $i_{k}:{[m]}\to{[1]}$ for some $1\leq k\leq m$ such that $i_{k}(i)=0$ if $i<k$ and $i_{k}(i)=1$ otherwise, together with a single (nondegenerate) map $o_{k}\to C_{i-1}$. In this case we may explicitly compute the pullback

$$U^{\prime}=U\times_{C_{i}}H\to V\times_{C_{i}}H=V^{\prime}$$

and deduce that it is contained in the class $T_{\Theta_{n}}$.

As $[U\to V]$ is in the image of $\sigma_{!}^{[1]}$, it is of the form $({[1]};U^{\prime\prime})\to({[1]};V^{\prime\prime})$ for some $[U^{\prime\prime}\to V^{\prime\prime}]$ in $\mathrm{Segal}_{\Theta_{n-1}}$. The pullback is then given explicitly as:

	$\displaystyle U^{\prime}=({[m]};$	$\displaystyle o_{1},\dots,o_{k}\times_{C_{i-1}}U^{\prime\prime},o_{k+1},\dots,o_{m})$
		$\displaystyle\to({[m]};o_{1},\dots,o_{k}\times_{C_{i-1}}V^{\prime\prime},o_{k+1},\dots,o_{m})=V^{\prime}.$

This map arises as the right-most vertical map in the following (oddly drawn) commuting square:

The left-most vertical map is a pushout of identities and (by induction) a map in $\sigma_{!}(T_{\Theta_{n-1}})$. Thus by Lemma 13.3 it is contained in $T_{\Theta_{n}}$. Both horizontal maps are contained in $T_{\Theta_{n}}$ by [34, Proposition 6.4], whence the right-most vertical map $[U^{\prime}\to V^{\prime}]$ is also contained in $T_{\Theta_{n}}$, as desired. ∎