ScalingStacks

Proof. We must show that the map

$${\tt E}_{/{\tt f}}\rightarrow{\tt E}_{/{\tt f}(1)}\underset{{\tt B}_{/{\tt p}({\tt f}(1))}}{\times}{\tt B}_{/{\tt p}({\tt f})}$$

in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$ lies in $({\bf W}\cap{\bf F})_{\textup{Joyal}}$.

First of all, using the characterization ${\bf F}_{\textup{Joyal}}=\textup{rlp}(({\bf W}\cap{\bf C})_{\textup{Joyal}})$, it is easy to see

• •

  that the object ${\tt E}_{/{\tt f}(1)}\in s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$ is fibrant and

• •

  that the map ${\tt B}_{/{\tt p}({\tt f})}\rightarrow{\tt B}_{/{\tt p}({\tt f}(1))}$ is a fibration in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$.

Hence, it follows from the Reedy trick that this fiber product is in fact a homotopy pullback in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$. Thus, this map in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$ presents the map

$$\mathcal{E}_{/f}\rightarrow\mathcal{E}_{/f(1)}\underset{\mathcal{B}_{/p(f(1))}}{\times}\mathcal{B}_{/p(f)},$$

in ${{\mathcal{C}\textup{at}}_{\infty}}$, which can be canonically identified with the map

$$\mathcal{E}_{/f(0)}\rightarrow\mathcal{E}_{f(1)}\underset{\mathcal{B}_{/p(f(1))}}{\times}\mathcal{B}_{/p(f(0))}$$

in ${{\mathcal{C}\textup{at}}_{\infty}}$ and therefore lies in ${\bf W}_{\textup{Joyal}}$ by assumption.

To see that it also lies in ${\bf F}_{\textup{Joyal}}$, we argue as follows. We claim that there is a Quillen adjunction

$$\alpha:\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}\rightleftarrows(s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\Delta^{1}/}:\beta,$$

where

• •

  we equip $[1]$ with the Reedy category structure determined by the degree function $0\mapsto 0$ and $1\mapsto 1$,

• •

  we define

  $$\beta(\Delta^{1}\xrightarrow{{\tt y}}{\tt Y})=({\tt Y}_{/{\tt y}}\to{\tt Y}_{/{\tt y}(1)}),$$

  and

• •

  we define

  $$\alpha({\tt Z}\rightarrow{\tt W})=\left(\Delta^{1}\rightarrow\left({\tt Z}\star\Delta^{1}\coprod_{{\tt Z}\star\Delta^{\{1\}}}{\tt W}\star\Delta^{\{1\}}\right)\right).$$

It is not hard to see that indeed $\alpha\dashv\beta$, so it suffices to show that $\alpha$ is a left Quillen functor. For this, given a map

in $\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}$ (reading the square vertically), observe that for this to be a (resp.​ acyclic) cofibration is precisely to require that the two relative latching maps ${\tt Z}_{1}\rightarrow{\tt Z}_{2}$ and ${\tt W}_{1}\coprod_{{\tt Z}_{1}}{\tt Z}_{2}\rightarrow{\tt W}_{2}$ are (resp.​ acyclic) cofibrations. For simplicity, let us write the composite

$$\textup{Fun}([1],s{\mathcal{S}\textup{et}})\xrightarrow{\alpha}s{\mathcal{S}\textup{et}}_{\Delta^{1}/}\rightarrow s{\mathcal{S}\textup{et}}$$

of our left adjoint with the evident forgetful functor simply as $\textup{Fun}([1],s{\mathcal{S}\textup{et}})\xrightarrow{\alpha^{\prime}}s{\mathcal{S}\textup{et}}$. Now, assuming our map ${\tt g}_{1}\rightarrow{\tt g}_{2}$ is a cofibration in $\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}$, then its image $\alpha^{\prime}({\tt g}_{1})\rightarrow\alpha^{\prime}({\tt g}_{2})$ fits into the diagram

in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$ of Figure 3, in which

• •

  the front and back faces are pushouts by definition;

• •

  the quadrilateral contained in the top face is a pushout since in the composite

  $$s{\mathcal{S}\textup{et}}\xrightarrow{-\star\Delta^{\{1\}}}s{\mathcal{S}\textup{et}}_{\Delta^{\{1\}}/}\rightarrow s{\mathcal{S}\textup{et}}$$

  where the second functor is forgetful,

  • –

    the first functor commutes with colimits by [Lur09, Remark 1.2.8.2] and

  • –

    the second functor commutes with pushouts since the walking span $\textup{N}^{-1}(\Lambda^{2}_{0})\in{\mathcal{C}\textup{at}}$ has an initial object

  (although really we have only rewritten this pushout to improve readability), and the dotted arrow is then the induced map;

• •

  the left face is a pushout by inspection;

• •

  all maps labeled as cofibrations are such

  • –

    by inspection,

  • –

    by the assumption that ${\tt g}_{1}\rightarrow{\tt g}_{2}$ is a cofibration in $\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}$,

  • –

    because ${\bf C}_{\textup{Joyal}}\subset s{\mathcal{S}\textup{et}}$ is closed under pushouts, or

  • –

    because ${\bf C}_{\textup{Joyal}}\subset s{\mathcal{S}\textup{et}}$ is closed under composition;

  and

• •

  the maps labeled with the symbol $\stackrel{{\scriptstyle?}}{{\approx}}$ are weak equivalences in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$ if ${\tt g}_{1}\rightarrow{\tt g}_{2}$ is additionally a weak equivalence in $\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}$

  • –

    by the assumption that ${\tt g}_{1}\rightarrow{\tt g}_{2}$ is an acyclic cofibration in $\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}$,

  • –

    because $({\bf W}\cap{\bf C})_{\textup{Joyal}}\subset s{\mathcal{S}\textup{et}}$ is closed under pushouts, or

  • –

    because $({\bf W}\cap{\bf C})_{\textup{Joyal}}\subset s{\mathcal{S}\textup{et}}$ is closed under composition.

Now, because the left and back faces are both pushouts, then the composite rectangle which they form is also a pushout. But this is the same as the composite rectangle formed by the front and right faces. As the front face is a pushout, it follows that the right face is also a pushout as well. Thus, the functor

$$\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}\xrightarrow{\alpha^{\prime}}s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$$

preserves both cofibrations and acyclic cofibrations, since these are each closed under pushout in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$. But the cofibrations and acyclic cofibrations in $(s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\Delta^{1}/}$ are created by the forgetful functor $s{\mathcal{S}\textup{et}}_{\Delta^{1}/}\rightarrow s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$, and so the functor

$$\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}\xrightarrow{\alpha}(s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\Delta^{1}/}$$

is indeed a left Quillen functor, as claimed.

We now return to our given composite

$$\Delta^{1}\xrightarrow{{\tt f}}{\tt E}\stackrel{{\scriptstyle{\tt p}}}{{\twoheadrightarrow}}{\tt B}$$

in $s{\mathcal{S}\textup{et}}_{\textup{Joyal}}$. This can be considered as defining a fibration in $(s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\Delta^{1}/}$, and hence applying our right Quillen functor

$$(s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\Delta^{1}/}\xrightarrow{\beta}\textup{Fun}([1],s{\mathcal{S}\textup{et}}_{\textup{Joyal}})_{\textup{Reedy}}$$

yields another fibration. In particular, the resulting relative matching map

$${\tt E}_{/{\tt f}}\rightarrow{\tt E}_{/{\tt f}(1)}\underset{{\tt B}_{/{\tt p}({\tt f}(1))}}{\times}{\tt B}_{/{\tt p}({\tt f})}$$

at the object $1\in[1]$ must lie in ${\bf F}_{\textup{Joyal}}\subset s{\mathcal{S}\textup{et}}$, as desired. ∎