Lemma 13.5 (0MK9) · ScalingStacks

[0MK9]

Lemma 13.5. Let $b_{1},\dots,b_{p}$ be elements of $\pre(\Theta_{n-1})$ , and let $0\leq r\leq s\leq p$ . Let $A$ and $B$ be defined as follows:

	$\displaystyle A$	$\displaystyle=\sigma_{!}^{\{0,\dots,s\}}(b_{1},\dots,b_{s})\cup^{\sigma^{\{r,\dots,s\}}_{!}(b_{r+1},\dots,b_{s})}\sigma_{!}^{\{r,\dots,p\}}(b_{r+1},\dots,b_{p}),\text{ and}$
	$\displaystyle B$	$\displaystyle=\sigma_{!}^{[p]}(b_{1},\dots,b_{p}).$

Then the natural map $A\to B$ is in $T_{\Theta_{n}}$ .

[0MKA]

Proof. First note that if each of the $b_{i}$ were a representable presheaf, then the map $A\to B$ may be written as a pushout of the generating morphism $\mathcal{T}_{n,\infty}$ , hence is manifestly an element of $T_{\Theta_{n}}$ (we leave this as an exercise). The general case, however, reduces to this case as every presheaf is (canonically) a colimit of representables, the functors $\sigma^{[\ell]}_{!}$ commute with these colimits separately in each variable, and $T_{\Theta_{n}}$ , being a saturated class, is closed under colimits. ∎