ScalingStacks

Proof. Recall that $\nu$ commutes with limits. Let $m$ ($\leq i,j,k$) be the largest integer such that $\varphi=\sigma^{m}(g)$ and $\psi=\sigma^{m}(f)$ are both $m$-fold suspensions of maps, $g:C_{i-m}\to C_{j-m}$ and $f:C_{k-m}\to C_{j-m}$.

Suppose, without loss of generality, that $\varphi$ is not an $(m+1)$-fold suspension of a map. We thus have an $m$-suspension of the situation considered in Rk. 6.3; that is, we have a diagram of pullback squares

where as above $F$ denotes the fiber of $f:C_{k-m}\to C_{j-m}$ over the image of $g$. So let us consider each of the cases A-D of Rk. 6.3 in turn.

1. (A)

   If $F=\emptyset$, then

   $$C_{i}\times_{C_{j}}C_{k}\cong\sigma^{m}(C_{i-m}\times_{C_{0}}F)\cong\sigma^{m}(\emptyset)\cong\partial C_{m}.$$

   In this case, the morphisms of $A\subset S_{00}$ provide an iterative construction of $\nu\partial C_{m}$ as a colimit in $S_{00}^{-1}\Fun(\Upsilon_{n}^{\mathrm{op}},\set)$ of cells.

2. (B)

   Next, if $F\cong C_{0}$, then

   $$C_{i}\times_{C_{j}}C_{k}\cong\sigma^{m}(C_{i-m}\times_{C_{0}}F)\cong\sigma^{m}(C_{i-m})\cong C_{i}$$

   is already a cell.

3. (C)

   Similarly, if $F\cong C_{k-m}$, but $i=m$, then

   $$C_{i}\times_{C_{j}}C_{k}\cong\sigma^{m}(C_{0}\times_{C_{0}}F)\cong\sigma^{m}(C_{k-m})\cong C_{k}$$

   is again already a cell.

4. (D)

   Finally, let us suppose that $F\cong C_{\ell}$ with $i=m+p$ and $k=m+\ell$ for $p>0$. In this case we have,

   $$C_{i}\times_{C_{j}}C_{k}\cong C_{m+p}\times_{C_{m}}C_{m+\ell}$$

   is precisely the fiber product considered in the set $C\subset S_{00}$. One readily observes that morphisms of $B$ and $C$ provide an inductive construction of this fiber product as an iterated colimit of cells in $\mathcal{C}$.∎