ScalingStacks

Proof. Both statements follow by induction. First note that $\Theta_{n}$ itself is closed under retracts [9, Proposition 3.14]. In the base case, the category $\Theta_{1}=\Delta$ consists of precisely the grids, and the sets of morphisms agree $\mathrm{Segal}_{\Theta_{1}}=\mathrm{Segal}_{\Delta}$. Now assume, by induction, that every object of $o\in\Theta_{n-1}$ is the retract of a grid $\delta_{n}(\mathbf{m}^{o})$, for some object $\mathbf{m}^{o}=[m_{1}^{o}]\times\cdots\times[m_{n-1}^{o}]\in\Delta^{\times n}$. In fact, given any finite collection of objects $\{o_{i}\in\Theta_{n-1}\}$ they may be obtained as the retract of a single grid. This grid may be obtained as the image of $\mathbf{k}=[k_{1}]\times\cdots\times[k_{n-1}]$, where $k_{j}$ is the maximum of the collection $\{m_{j}^{o_{i}}\}$. It now follows easily that the object $([n];o_{1},\dots,o_{i})\in\Theta_{n}$ is a retract of the grid coming from the object $[n]\times\mathbf{k}$.

To prove the second statement we note that there are two types of maps in $\mathrm{Segal}_{\Theta_{n}}$, those in $\sigma_{!}(\mathrm{Segal}_{\Theta_{n-1}})$ and the maps

$$j({\{0,\dots,k\}};o_{1},\dots,o_{k})\cup^{j({\{k\}})}j({\{k,\dots,m\}};o_{k+1},\dots,o_{m})\to j({[m]};o_{1},\dots,o_{m})$$

for $0\leq k\leq m$ and $o_{i}\in\Theta_{n-1}$. This later map is a retract of the image under $(\delta_{n})_{!}$ of the map

$$\left(j{\{0,1,\dots,k\}}\cup^{j{\{k\}}}j{\{k,\dots,m\}}\to j[m]\right)\boxtimes j(\mathbf{m})$$

which is a map in $\mathrm{Segal}_{\Delta^{\times n}}$. Here $\mathbf{m}$ is such that $({\{0,\dots,k\}};o_{1},\dots,o_{k})$ is the retract of the grid corresponding to $[k]\times\mathbf{m}$.

The former class of morphisms in $\mathrm{Segal}_{\Theta_{n}}$, those in $\sigma_{!}(\mathrm{Segal}_{\Theta_{n-1}})$, are also retracts on elements in $\mathrm{Segal}_{\Delta^{\times n}}$. Specifically, if $\sigma_{!}(f)\in\sigma_{!}(\mathrm{Segal}_{\Theta_{n-1}})$, then by induction $f$ is the retract of $(\delta_{n-1})_{!}(g)$ for some $g\in\mathrm{Segal}_{\Delta^{\times n-1}}$. One may then readily check that $\sigma(f)$ is the retract of $j[1]\boxtimes g\in\mathrm{Segal}_{\Delta^{\times n}}$. ∎