ScalingStacks

Proof. In this proof, weak equivalence will mean weak equivalence in the Segal space model category structure. The composite map $ji$ is a weak equivalence by construction, so it suffices to show that $i$ is also a weak equivalence. Given any $\alpha^{i_{1}}F(k_{1}),\alpha^{i_{2}}F(k_{2})\subset F(n)$, we see that the intersection $\alpha^{i_{1}}F(k_{1})\cap\alpha^{i_{2}}F(k_{2})$ is either empty, or is equal to $\alpha^{i_{3}}F(k_{3})$ for some $i_{3}$ and $k_{3}$. Thus $G$ can be written as a colimit over a partially ordered set of subcomplexes of the form $\alpha^{i}F(k)$. Since $G(n)\cap\alpha^{i}F(k)=\alpha^{i}G(k)$, we see that $G(n)$ is obtained as a colimit over the same indexing category of subobjects of the form $\alpha^{i}G(k)$. Since by hypothesis the map $\Map_{s{\operatorname{\mathcal{S}}}}(\alpha^{i}F(k),W)\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(\alpha^{i}G(k),W)$ is a weak equivalence for any Segal space $W$, we conclude that $\Map_{s{\operatorname{\mathcal{S}}}}(G,W)\rightarrow\Map_{s{\operatorname{\mathcal{S}}}}(G(n),W)$ is also a weak equivalence for any Segal space $W$, and hence $i$ is a weak equivalence in the Segal space model category, as desired. ∎