ScalingStacks

Proof. Denote by $j$ any Yoneda embedding (the context will always be made clear). Let $K$ denote the simplicial set as in 12.2, which we regard as a simplicial space that is discrete in each degree. This is a pushout along an inclusion, hence this is also a (homotopy) pushout in the $\infty$-category of simplicial spaces. Now let $T$ be the strongly saturated class of morphisms of $\Fun(\mathrm{N}\Delta^{\mathrm{op}},\CSS(\Delta^{\!\times n-1}))$ generated by the three sets

One deduces immediately that a simplicial object of $\CSS(\Delta^{\!\times n-1})$ is a Segal space if and only if it is local with respect to each of the first two sets of morphisms. To show that $\CSS(\Delta^{\!\times n})$ coincides with the localization $T^{-1}\Fun(\mathrm{N}\Delta^{\mathrm{op}},\CSS(\Delta^{\!\times n-1}))$, it is enough to show that a $1$-fold Segal space $X$ is complete if and only if the natural map

is an equivalence. By the Yoneda lemma, our claim is just a restatement of [34, Proposition 10.1]. ∎

Proof. If $n=0$, there is nothing to prove. If $n$ is positive, then let us suppose that we have written $\CSS_{n-1}$ as a localization $T_{\Delta^{\!\times n-1}}^{-1}\pre(\Delta^{\!\times n-1})$ for some strongly saturated class $T_{\Delta^{\!\times n-1}}$ of small generation. Denote by

the essentially unique functor that carries pairs of the form $(j[k],j(\mathbf{m}))$ to $j([k],\mathbf{m})$ and preserves colimits separately in each variable. Now let $T_{\Delta^{\!\times n}}$ be the strongly saturated class generated by the class $T$ above along with the set

Now $\CSS(\Delta^{\!\times n})$ coincides with $T_{\Delta^{\!\times n}}^{-1}\pre(\Delta^{\!\times n})$. ∎

Proof. We will proceed by induction on $n$, the base case $n=0$ being trivial. As the functor $j[m]\boxtimes(-)$ preserves colimits and sends the generators of $T_{\Delta^{\times n-1}}$ into $T_{\Delta^{\times n}}$, we have a containment $j[m]\boxtimes T_{\Delta^{\times n-1}}\subseteq T_{\Delta^{\times n}}$. Thus by induction the canonical map,

is in $T_{\Delta^{\times n}}$. In particular when $m=0$, the map $j[m]\boxtimes\mathbf{m}\to j[m]\boxtimes\mathbf{0}$ is a composite of maps in $\mathrm{Glob}_{\Delta^{\!\times n}}$, whence the map

is in $T_{\Delta^{\times n}}$.

We will first prove the lemma for objects of the form $j[1]\boxtimes\mathbf{m}\in\Delta\times\Delta^{\times n-1}\cong\Delta^{\times n}$. One may readily check that the following is a pushout square of presheaves of sets:

Moreover, as the topmost map is an inclusion of sets and pushouts in $\pre(\Delta^{\times n})$ are computed object-wise, this is also a (homotopy) pushout square in $\pre(\Delta^{\times n})$. As we just observed, the left-most map is in the strongly saturated $T_{\Delta^{\times n}}$, whence the right-most map is also in $T_{\Delta^{\times n}}$. It follows that the composite,

is in $T_{\Delta^{\times n}}$.

To prove the general case, i.e., that the map $j[k]\boxtimes\mathbf{m}\to\delta^{*}_{n}\delta_{n}(j[k]\boxtimes\mathbf{m})$ is in $T_{\Delta^{\times n}}$, we induct on $k$. Assume the result holds when $k\leq m$. We will prove it for $k=m+1$. First consider the following commutative square:

The indicated maps are in $T_{\Delta^{\times n}}$; the topmost map is a generator and the lefttmost vertical map by induction. As $T_{\Delta^{\times n}}$ is saturated, the rightmost vertical map is in $T_{\Delta^{\times n}}$ if and only if the bottommost map is as well. Thus it suffices to prove that the natural map

is in $T_{\Delta^{\times n}}$.

The Yoneda embedding is dense for any presheaf $\infty$-category and hence the object $\delta_{n}^{*}\delta_{n}(j[m+1]\boxtimes\mathbf{m})$ may canonically be written as a colimit of representable presheaves. Let $\mathcal{D}=(\Delta^{\times n}\downarrow\delta_{n}^{*}\delta_{n}(j[m+1]\boxtimes\mathbf{m}))$ denote the overcategory consisting of pairs $(j[p]\boxtimes\mathbf{p},\phi)$ where $\phi$ is a map

Let $B:\mathcal{D}\to\pre(\Delta^{\times n})$ denote the functor which forgets the map $\phi$. We have a canonical equivalence in $\pre(\Delta^{\times n})$:

By adjunction, specifying a map $\phi:j[p]\boxtimes\mathbf{p}\to\delta_{n}^{*}\delta_{n}(j[m+1]\boxtimes\mathbf{m})$ is equivalent to a specifying a map $\phi^{\prime}:\delta_{n}(j[p]\boxtimes\mathbf{p})\to\delta_{n}(j[m+1]\boxtimes\mathbf{m})$, i.e., a map in $\pre(\Theta_{n})$:

In particular every such map includes the data of a map $\overline{\phi}:[p]\to[m+1]$. To simplify notation we will denote the object $(j[p]\boxtimes\mathbf{p},\phi)$ as $B_{\phi}$ or simply $\phi$.

Let $\mathcal{C}$ denote the full subcategory of $\mathcal{D}$ consisting of the union of the following three types of objects:

1. (a)

   those $B_{\phi}$ in which $\overline{\phi}$ factors as

   $$\overline{\phi}:[p]\to\{0,\dots,m\}\subseteq[m+1],$$

2. (b)

   those $B_{\phi}$ in which $\overline{\phi}$ factors as

   $$\overline{\phi}:[p]\to\{m,m+1\}\subseteq[m+1],\quad\text{and}$$

3. (c)

   those $B_{\phi}$ in which $(\overline{\phi})^{-1}(\{m\})=\{r\}\subseteq[p]$ consists of a singleton for some $0\leq r\leq p$.

For any object $D\in\mathcal{D}$, the under category $\mathcal{C}_{D/}$ actually has an initial object and is thus weakly contractible. Consequently (see, e.g., [28, Th. 4.1.3.1 and Proposition 4.1.1.8]), the induced morphism of (homotopy) colimits over these categories is an equivalence; in particular, it follows that the following canonical maps are equivalences in $\pre(\Delta^{\times n})$:

For each $\phi\in\mathcal{C}$, let $A_{\phi}$ denote the fiber product

This gives rise to a new functor $A:\mathcal{C}\to\pre(\Delta^{\times n})$, and as colimits in $\pre(\Delta^{\times n})$ are universal, we have natural equivalences:

Thus the desired result follows if we can demonstrate that the natural map

is in the class $T_{\Delta^{\times n}}$. This class, being saturated, is closed under colimits, and so it suffices to show that each of the maps

is in $T_{\Delta^{\times n}}$. If $B_{\phi}\in\mathcal{C}$ is of type (a) or type (b), then $A_{\phi}\simeq B_{\phi}$ is an equivalence, hence in the desired class. If $B_{\phi}=(j[p]\boxtimes\mathbf{p},\phi)$ is of type (c), so that $(\overline{\phi})^{-1}(\{m\})=\{r\}$ for $0\leq r\leq p$, then a direct calculation reveals:

As this is one of the generators of $T_{\Delta^{\times n}}$, the result follows. ∎

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 $\delta_{n}^{*}(T_{\Theta_{n}})\subseteq T_{\Delta^{\times n}}$. ∎