ScalingStacks

Proof. Let $\gamma^{i}\colon[n+1]\rightarrow[1]\times[n]$ denote the map defined by

$$\gamma^{i}(j)=\begin{cases}(0,j)&\text{if $j\leq i$,}\\ (1,j-1)&\text{if $j>i$.}\end{cases}$$

Likewise, let $\delta^{i}\colon[n]\rightarrow[1]\times[n]$ denote the map defined by

$$\delta^{i}(j)=\begin{cases}(0,j)&\text{if $j\leq i$,}\\ (1,j)&\text{if $j>i$.}\end{cases}$$

Then one can write $F(1)\times F(n)$ as a colimit of the diagram

$$\gamma^{0}F(n+1)\leftarrow\delta^{0}F(n)\rightarrow\gamma^{1}F(n+1)\leftarrow\delta^{1}F(n)\rightarrow\dots\rightarrow\gamma^{n}F(n+1)$$

(10.4)

of subobjects. (This is analogous to the decomposition of the simplicial set $\Delta[1]\times\Delta[n]$ into a union of $(n+1)$ copies of $\Delta[n+1]$, attached along faces.) A straightforward computation shows that the maps $\gamma^{i}F(n+1)\cap(F(1)\times G(n))\rightarrow\gamma^{i}F(n+1)$ and $\delta^{i}F(n)\cap(F(1)\times G(n))\rightarrow\delta^{i}F(n)$ are covers, and hence by (10.1) are weak equivalences. Thus the result follows by comparing diagram (10.4) with the diagram obtained by intersecting each object of (10.4) with $F(1)\times G(n)$. ∎