ScalingStacks

Proof of statements (1) and (3). For each simplicial map $\delta\colon[n]\rightarrow[m]\in\boldsymbol{\Delta}$ there is a diagram

By (13.8) the maps $W^{E(m)}\rightarrow W^{E(n)}$ are Dwyer-Kan equivalences, so for each set of objects $x_{0},\dots,x_{k}\in{\operatorname{ob}}W$ the morphism

$$\map_{W^{E(m)}}(x_{0},\dots,x_{k})\rightarrow\map_{W^{E(n)}}(\delta x_{0},\dots,\delta x_{k})$$

between the fibers of the vertical maps in the above diagram is a weak equivalence. Thus, the square is a homotopy pullback, with fibers which are weakly equivalent to the products of mapping spaces.

Thus, the induced map of realizations $\diag^{\prime}(W^{E(-)})_{k}\rightarrow\diag^{\prime}(W^{E(-)})_{0}^{\times k}$ has its homotopy fibers weakly equivalent to a $k$-fold product of mapping spaces, and thus we have shown that $\widehat{W}$ is a Segal space, and that $\map_{W}(x,y)\rightarrow\map_{\widehat{W}}(i(x),i(y))$ are weak equivalences for all $x,y\in{\operatorname{ob}}W$.

By construction the map $\pi_{0}W_{0}\rightarrow\pi_{0}\widehat{W}_{0}$ is surjective; it follows that $\ho W\rightarrow\ho\widehat{W}$ is surjective on isomorphism classes of objects. Therefore we have shown that $W\rightarrow\widehat{W}$ is a Dwyer-Kan equivalence, proving statement (3).

It remains to show that $\widehat{W}$ is a complete Segal space. Consider the square

induced by a map $j\colon E(n)\rightarrow E(m)$. Since $W^{E(m)}\rightarrow W^{E(n)}$ is a categorical equivalence by (13.5) and thus a Dwyer-Kan equivalence by (13.8), we may conclude that the induced map $\hoequiv_{W^{E(m)}}(x,y)\rightarrow\hoequiv_{W^{E(n)}}(j(x),j(y))$ is a weak equivalence for each pair $(x,y)\in(W^{E(m)})_{0}\times(W^{E(m)})_{0}$. Thus the above square is a homotopy pullback, and so the induced map $\diag^{\prime}(W^{E(-)})_{\hoequiv}\rightarrow\diag^{\prime}(W^{E(-)})_{0}^{\times 2}$ has its homotopy fibers weakly equivalent to the spaces $\hoequiv_{W}(x,y)$. That is,

$$(\widehat{W})_{\hoequiv}\approx\diag([m]\mapsto(W^{E(m)})_{\hoequiv}).$$

Since $(W^{E(m)})_{\hoequiv}\approx(W^{E(m)\times E(1)})_{0}$ by (6.2), the above really says that there is an equivalence $(\widehat{W})_{\hoequiv}\approx(\widehat{W^{E(1)}})_{0}$. Now (14.3) shows that since $W^{E(1)}$ is categorically equivalent to $W$, we have that $(\widehat{W})_{\hoequiv}\approx(\widehat{W^{E(1)}})_{0}\approx(\widehat{W})_{0}$; in other words, $\widehat{W}$ is a complete Segal space. This proves statement (1), and completes the proof. ∎