ScalingStacks

Proof. That $N$ preserves products is clear.

To show that $N(D^{C})\rightarrow(ND)^{NC}$ is an isomorphism, we must show that for each $m,n\geq 0$ this map induces a one-to-one correspondence between functors $[m]\times I[n]\rightarrow D^{C}$ and maps $F(m)\times\Delta[n]\rightarrow(ND)^{NC}$. By (3.8) it will suffice to show this for the case $m=n=0$; that is, to show that functors $C\rightarrow D$ are in one-to-one correspondence with maps $NC\rightarrow ND$, or in other words, that $N\colon{\operatorname{\mathcal{C}at}}\rightarrow s{\operatorname{\mathcal{S}}}$ is a full embedding of categories.

To see that $N$ is a full embedding, note that any map $NC\rightarrow ND$ is determined by how it acts on the $0$th and $1$st spaces of $NC$. The result follows from a straightforward argument using (3.2) and the fact that $(d_{1},d_{0})$ is a simplicial covering map such that $d_{1}s_{0}=1=d_{0}s_{0}$ for both $NC$ and $ND$.

It is immediate that naturally isomorphic functors induce simplicially homotopic maps of simplicial spaces since $N(C^{I[1]})\approx(NC)^{\Delta[1]}$ by (3.8), and thus an equivalence of categories induces a weak equivalence of simplicial spaces. To prove the converse, note that (3.9) will show that $(ND)^{NC}\approx N(D^{C})$ is Reedy fibrant, and in particular $\Map_{{\operatorname{\mathcal{S}}}}(NC,ND)\approx(ND^{NC})_{0}$ is a Kan complex. Therefore, if $Nf\colon NC\rightarrow ND$ is a weak equivalence of simplicial spaces it must be a simplicial homotopy equivalence. Furthermore, the homotopy inverse is a $0$-simplex of $N(C^{D})_{0}$ and the simplicial homotopies are $1$-simplices of $N(D^{C})_{0}$ and $N(C^{D})_{0}$; by what we have already shown these correspond precisely to a functor $g\colon D\rightarrow C$ and natural isomorphisms $fg\sim 1_{D}$ and $gf\sim 1_{C}$, as desired. ∎