ScalingStacks

Proof of (8.3). Let $U=N({\operatorname{\mathbf{M}}})$, so that $U_{n}=\nerve\we({\operatorname{\mathbf{M}}}^{[n]})$ and $U_{n}\rightarrow V_{n}$ is a weak equivalence of spaces. For each $n\geq 0$ there is a map $\pi_{n}\colon U_{n}\rightarrow U_{0}^{n+1}$ which “remembers” only objects. The remarks above together with (8.7) show that for each $(n+1)$-tuple of objects $(X_{0},\dots,X_{n})$ in ${\operatorname{\mathbf{M}}}$ the homotopy fiber of $\pi_{n}$ over the point corresponding to $(X_{0},\dots,X_{n})$ is in a natural way weakly equivalent to a product

$$\map_{{\operatorname{\mathbf{M}}}}(X_{n-1}^{\prime},X_{n}^{\prime})\times\dots\times\map_{{\operatorname{\mathbf{M}}}}(X_{0}^{\prime},X_{1}^{\prime}),$$

where $X_{i}^{\prime}$ is a fibrant-and-cofibrant object of ${\operatorname{\mathbf{M}}}$ which is weakly equivalent to $X_{i}$.

Note that it is an immediate consequence of the above that $V$ is a Segal space. Since $\pi_{0}U_{0}$ is just the set of weak homotopy types in ${\operatorname{\mathbf{M}}}$, and since $\ho{\operatorname{\mathbf{M}}}(X,Y)\approx\pi_{0}\map_{{\operatorname{\mathbf{M}}}}(X^{\prime},Y^{\prime})$ where $X^{\prime}$ and $Y^{\prime}$ are fibrant-and-cofibrant replacements of $X$ and $Y$ respectively, we see that $\ho{\operatorname{\mathbf{M}}}\approx\ho V$.

Let $U_{\hoequiv}\subset U_{1}$ denote the subspace of $U_{1}$ which corresponds to the subspace $V_{\hoequiv}\subset V_{1}$. By the equivalence of homotopy categories above, we see that $U_{\hoequiv}$ consists of precisely the components of $U_{1}$ whose points go to isomorphisms in $\ho{\operatorname{\mathbf{M}}}$. Since ${\operatorname{\mathbf{M}}}$ is a closed model category, this means that the $0$-simplices of $U_{\hoequiv}$ are precisely the objects of ${\operatorname{\mathbf{M}}}^{[1]}$ which are weak equivalences, so $U_{\hoequiv}=\nerve\we((\we{\operatorname{\mathbf{M}}})^{[1]})$. There is an adjoint functor pair $F\colon{\operatorname{\mathbf{M}}}^{[1]}\leftrightarrows{\operatorname{\mathbf{M}}}\;{:}\,G$ in which the right adjoint takes $G(X)=\id_{X}$, and the left adjoint takes $F(X\rightarrow Y)=X$; this pair restricts to an adjoint pair $\we((\we{\operatorname{\mathbf{M}}})^{[1]})\leftrightarrows\we{\operatorname{\mathbf{M}}}$ and thus induces a weak equivalence $U_{\hoequiv}\approx U_{0}$ of the nerves. Thus $V$ is a complete Segal space. ∎