ScalingStacks

Proof of (8.11). Using (8.15) we can reinterpret (8.14) as stating that there is a weak equivalence

$$\class({\operatorname{\mathcal{S}}}^{I})\xrightarrow{\sim}\Map_{s{\operatorname{\mathcal{S}}}}(\discnerve I,N^{f}({\operatorname{\mathcal{S}}})).$$

Substituting $[m]\times I$ for $I$ in the above for all $m\geq 0$ leads to a Reedy weak equivalence

$$N({\operatorname{\mathcal{S}}}^{I})\xrightarrow{\sim}N^{f}({\operatorname{\mathcal{S}}})^{\discnerve I},$$

which is the special case of (8.10) with ${\operatorname{\mathbf{M}}}={\operatorname{\mathcal{S}}}$. To obtain the case of ${\operatorname{\mathbf{M}}}={\operatorname{\mathcal{S}}}^{J}$, note that by what we have just shown the maps in

$$N({\operatorname{\mathcal{S}}}^{I\times J})\xrightarrow{\sim}N^{f}({\operatorname{\mathcal{S}}})^{\discnerve(I\times J)}\approx N^{f}({\operatorname{\mathcal{S}}})^{\discnerve I\times\discnerve J}\xleftarrow{\sim}N^{f}({\operatorname{\mathcal{S}}}^{J})^{\discnerve I}$$

must be Reedy weak equivalences. ∎