ScalingStacks

The map $s_{0}\colon(W^{F(1)})_{0}\rightarrow(W^{F(1)})_{1}$ is obtained by taking limits of the rows in the diagram:

By hypothesis, $s_{0}\colon W_{0}\rightarrow W_{1}$ is a homotopy monomorphism. Thus the maps $s_{0},s_{1}\colon W_{1}\rightarrow W_{2}$ are homotopy monomorphisms, since they are weakly equivalent to $W_{1}\times_{W_{0}}s_{0}\colon W_{1}\times_{W_{0}}W_{0}\rightarrow W_{1}\times_{W_{0}}W_{1}$ and $s_{0}\times_{W_{0}}W_{1}\colon W_{0}\times_{W_{0}}W_{1}\rightarrow W_{1}\times_{W_{0}}W_{1}$. It follows that $s_{0}\colon(W^{F(1)})_{0}\rightarrow(W^{F(1)})_{1}$ is a homotopy monomorphism.

Thus both $s_{0}\colon(W^{F(1)})_{0}\rightarrow(W^{F(1)})_{1}$ and $(W^{F(1)})_{\hoequiv}\rightarrow(W^{F(1)})_{1}$ are homotopy monomorphisms. So to prove the proposition it suffices to show that both these maps hit the same components. As we already know that $(W^{F(1)})_{0}\rightarrow(W^{F(1)})_{1}$ factors through a map $(W^{F(1)})_{0}\rightarrow(W^{F(1)})_{\hoequiv}$, it suffices to show that this last map is surjective on $\pi_{0}$.

Using the part of the proof already completed and (12.4), one observes that a point $x\in(W^{F(1)})_{\hoequiv}$ lies in a component hit by $(W^{F(1)})_{0}\rightarrow(W^{F(1)})_{\hoequiv}$ if and only if the images $fx,gx\in(W^{F(0)})_{1}\approx W_{1}$ are homotopy equivalences in $W$, where $f,g\colon W^{F(1)}\rightarrow W^{F(0)}$ are the maps induced by the two inclusions $d^{0},d^{1}\colon F(0)\rightarrow F(1)$. But if $x\in(W^{F(1)})_{1}$ is a homotopy equivalence of $W^{F(1)}$ then certainly its images under $f$ and $g$ are homotopy equivalences. Thus the result is proved.