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At the page 74 of HTT, there is the following theorem

Let $S$ be a simplicial set, $\mathcal{C}$ a simplicial category, and $\phi: \mathfrak{C}[S] \rightarrow \mathcal{C}^{op}$ a simplicial functor. The straightening and unstraigntening functors determine a Quillen adjunction $$ St_{\phi} : (Set_{\Delta})_{/S} \leftrightarrows Set_{\Delta}^{\mathcal{C}} :Un_{\phi}$$ where $(Set_{\Delta})_{/S}$ is endowed with the contravariant model structure and $Set_{\Delta}^{\mathcal{C}}$ with the projective model structure. [...]

In then says that the proof is easy, but I can't manage to show that $St_{\phi}$ sends cofibrations to projective cofibrations. I thought that since the the class of morphisms which are sent to projective cofibrations is weakly saturated it is enough to show the result for all inclusions $\partial \Delta^n \subseteq \Delta^n$.

I did not have much success for the simplicial category $\mathcal{C}$ and the map $\phi$ could be anything and I have a hard time dealing with it.

Furthermore there is something else which troubles me: the model structure on the $Set^{\mathcal{C}}_{\Delta}$ makes no use of the simplicial enrichement on both $\mathcal{C}$ and $sSet$ so I was wondering if I was not missing something by believing that the that model structure on $Set^{\mathcal{C}}_{\Delta}$ is really the projective model structure coming from the Kan model structure on $sSet$ and not the one coming somehow from an other model stucture using the simplicial enrichement.

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    $\begingroup$ Cross-posted: math.stackexchange.com/questions/3013094 $\endgroup$ – Watson Nov 28 '18 at 10:36
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    $\begingroup$ Maybe I'm misunderstanding you, but are you aware that $Set_\Delta^{\mathcal{C}}$ is the category simplicial functors, i.e. the functors that preserve the simplicial enrichment? The projective model structure in question is that of section A.3.3 $\endgroup$ – Denis Nardin Nov 28 '18 at 11:16
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First, notice that if $X\hookrightarrow Y$ is an injective map over $S$, then the map $M_{X,\phi} \to M_{Y,\phi}$ is a cofibration of simplicial categories. To see this, notice that it is a pushout of the map $M_{X,id}\to M_{Y,id},$ which can easily be verified to be a cofibration, as it is the image of the injective map $$X^\triangleright\coprod_X S \hookrightarrow Y^\triangleright \coprod_Y S$$ under the functor $\mathfrak{C},$ which preserves cofibrations.

Then we see that if $C\to D$ is a cofibration of simplicial categories, then given any $x\in C$ such that $C(-,x)$ is projectively cofibrant, the induced natural transformation $C(-, x)\to D(f-,fx)$ is a projective cofibration by A.3.3.9.ii.

The last thing that remains to be proved is that $M_{\phi,X}(-,p),$ where $p$ denotes the image of the cone point, is projectively cofibrant. The previous revision to this answer did not answer this question, and I am leaving it open until I have more time later or until someone adds an answer.

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  • $\begingroup$ I just realized there's still a minor gap here: It's not clear from what I wrote why $M_{X,\phi}(-,p)$ where $p$ is the image of the cone point is projectively cofibrant, which you need to apply A.3.3.9.ii. I'll figure it out later if still nobody has answered. $\endgroup$ – Harry Gindi Nov 28 '18 at 12:57
  • $\begingroup$ I think the projective cofibrancy can be shown to follow in the special case of the above argument by taking $X=\emptyset$ and $Y=X$, since in this case the straightening of the empty simplicial set over $S$ can be computed by hand to be the constant functor hitting the empty simplicial set, but I have to work it out. $\endgroup$ – Harry Gindi Nov 28 '18 at 13:10
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I think what Lurie might have meant when he wrote "It is easy to see that $St_{\phi}$ preserves cofibrations" in the proof of Theorem 2.2.1.2, is that it is easy to see it if you take into account the compatibility of straightening with left Kan extensions as described just above in Proposition 2.2.1.1. Indeed, combining (1) and (2) of the latter proposition we see that given any generating cofibration $\partial \Delta^n \to \Delta^n \to S$ in $({\rm Set}_{\Delta})_{/S}$, in order to show that $St_{\phi}(\partial \Delta^n \to \Delta^n)$ is a projective cofibration in ${\rm Set}_{\Delta}^{{\cal C}}$ it is enough to show that $St_{Id}(\partial \Delta^n \to \Delta^n)$ is a projective cofibration in ${\rm Set}_{\Delta}^{\mathfrak{C}[\Delta^n]^{op}}$. The map $St_{Id}(\partial \Delta^n \to \Delta^n)$ can then be described very explicitly, and one can check that it is a pushout of a map of the form $(i_0)_!\partial ((\Delta^1)^n) \to (i_0)_!(\Delta^1)^n$, where $(i_0)_!$ denotes the left Kan extension functor along the terminal object inclusion $\{0\} \subseteq \mathfrak{C}[\Delta^n]^{op}$ and $\partial ((\Delta^1)^n) \to (\Delta^1)^n$ is the inclusion of the boundary of the $n$-cube inside the full $n$-cube.

Edit:

In response to the comment, here are some more details on the computation. First note that by definition the map $St_{Id}(\partial \Delta^n)(0) \to St_{Id}(\Delta^n)(0)$ can be identified with the inclusion $$ (*)\quad {\rm Map}_{\mathfrak{C}[\partial \Delta^{n+1}]}(0,n+1) \subseteq {\rm Map}_{\mathfrak{C}[\Delta^{n+1}]}(0,n+1) = {\rm N}({\rm P}(\{1,...,n\})) \cong (\Delta^1)^n ,$$ where ${\rm P}(\{1,...,n\})$ is the poset of subsets of $\{1,...,n\}$ and ${\rm N}(-)$ is the nerve. Now for $i \in \{1,...,n\}$, image of $\mathfrak{C}[\Delta^{\{0,...,\hat{i},...,n+1\}}] \to \mathfrak{C}[\Delta^{n+1}]$ on the mapping space from $0$ to $n+1$ is exactly the face of ${\rm N}({\rm P}(\{1,...,n\}))$ corresponding to the subposet spanned by those subsets which do not contain $i$. On the other hand, the face of ${\rm N}({\rm P}(\{1,...,n\}))$ corresponding to the subposet spanned by those subsets which do contain $i$ is exactly the image of $${\rm Map}_{\mathfrak{C}[\Delta^{\{0,...,i\}}]}(0,i) \times {\rm Map}_{\mathfrak{C}[\Delta^{\{i,...,n+1\}}]}(i,n+1) \subseteq {\rm Map}_{\mathfrak{C}[\partial \Delta^{n+1}]}(0,n+1)$$ in ${\rm Map}_{\mathfrak{C}[\Delta^{n+1}]}(0,n+1)$. This shows that the image of (*) contains all the boundary of the cube. It is also not difficult to check that this image is contained in the boundary of the cube, and hence coincides with it.

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    $\begingroup$ I wrote up a proof of 2.2.1.1 that now appears on the nLab page ncatlab.org/nlab/show/(infinity%2C1)-Grothendieck+construction (search for my name). It's not so obvious, but together with your answer, it provides the full answer! $\endgroup$ – Harry Gindi Nov 28 '18 at 17:10
  • $\begingroup$ First and foremost thanks to both of you for your answers! Then when I want to compute $St_{Id}(\partial \rightarrow \Delta^n)$ I have a small issue actually computing $St_{Id}(\partial \Delta^n)(0)$. It seems to me that it should some simplicial subset of the cube $(\Delta^1)^{n-1}$ (something like the cube with the interior and a face removed?) but I can't produce a concrete proof. $\endgroup$ – Oscar P. Dec 8 '18 at 17:13
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    $\begingroup$ It is actually the entire boundary of the cube. I added an edit with more details on this in the answer above. $\endgroup$ – Yonatan Harpaz Dec 9 '18 at 9:38
  • $\begingroup$ @ValerianMontessuit please accept this answer, by the way, as it is definitive. $\endgroup$ – Harry Gindi Dec 9 '18 at 9:58
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    $\begingroup$ @ValérianMontessuit, It's not just the cone you need to take in the definition of $St_{\Delta^n}(\partial \Delta^n)$, it's the pushout $(\partial \Delta^n)^{\triangleright} \coprod_{\partial \Delta^n} \Delta^n = \partial \Delta^{n+1}$. $\endgroup$ – Yonatan Harpaz Dec 9 '18 at 17:57

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