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While studying the book Higher Topos Theory I have encountered some difficulty with Lemma 3.3.4.1, which says that the pullback along a cartesian fibration of a map q such that $q^{op}$ is cofinal is a cartesian equivalence (i.e. a weak equivalence in the model structure on marked simplicial sets) once we declare the marked 1-simplices in both the domain and codomain to be the cartesian edges with respect to the appropriate cartesian fibrations. But this seems to imply that such a q is a weak categorical equivalence, i.e. a weak equivalence in the Joyal model structure, at least when it's between fibrant objects.

This essentially because we can consider the pullback along the identity map on the codomain of q, which surely is a cartesian fibration, and then use the fact that the cartesian model structure on marked simplicial sets is Quillen equivalent to the Joyal one.

Now, this fact is clearly false, since we can pick one of the two maps $\Delta^0 \to \Delta^1$ which is not a categorical equivalence.

Am I missing something or is there a problem in the proof? I think, if ever, the problem might lie in the commutative square that appears in the proof, and in the statement that the upper horizontal arrow is a Joyal equivalence.

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The lemma is true as stated. The Cartesian equivalence in question does not imply an equivalence in the Joyal model structure after forgetting the markings; rather, it implies an equivalence after formally inverting the markings in question. In more detail:

  • The forgetful functor $\mathrm{Set}^+_{\Delta} \to \mathrm{Set}_{\Delta}$ is a (right) Quillen equivalence with the Cartesian model structure on the left and the Joyal model structure on the right. But one must take care that the derived functor is computed on fibrant objects. Fibrant objects on the source are $\infty$-categories $\mathcal{C}$ marked by their equivalences. So, given a pair $(\mathcal{C}, \mathcal{E})$, the corresponding $\infty$-category under this equivalence is really a model for the localization $\mathcal{C}[\mathcal{E}^{-1}]$.
  • For the case in question, $X$ is marked by the $p$-cartesian morphisms, and $X'$ by the $p'$-cartesian morphisms. So the fact that $X' \to X$ is a Cartesian equivalence means that the functor becomes an equivalence after inverting those morphisms.
  • The example with the identity map is actually a great example: if $X = S$ then every morphism is $p$-cartesian. So the lemma now says: if $S' \to S$ is initial, then it induces an equivalence after inverting all the morphisms. Another way of saying that is that it induces an equivalence on 'classifying spaces', or that it induces an equivalence in the Quillen model structure- and that's true! Initial and final maps are homotopy equivalences.

The intuition is supposed to be that, if $p: X \to S$ is Cartesian, $X$ is like an oplax colimit of the diagram $S^{op} \to \mathsf{Cat}_{\infty}$ (this has been made precise by Gepner-Haugseng-Nikolaus). The difference between an oplax colimit and a colimit is that some comparison morphisms are just morphisms instead of equivalences, so you fix that by inverting them. So (as is shown in different language in this section of HTT) the localization $X[\mathcal{E}^{-1}]$ is meant to model the colimit of an $S^{op}$ shaped diagram. But colimits may be computed after restricting along a final map $S^{'op} \to S^{op}$, whence the lemma. (Obviously this reasoning is circular, since the point of the section is to justify the claim that $X[\mathcal{E}^{-1}]$ models the colimit, but hopefully the lemma is more believable now.)

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  • $\begingroup$ The lemma states "$X'^\natural \rightarrow X^\natural$ is a cartesian equivalence (in $\mathbf{Set}_{\Delta}^+$)". By definition, $Y^\natural$ makes sense if $p \colon Y \to S$ is fibrant in $\mathbf{Set}_\Delta^+/S$ (i.e. it is a cartesian fibration). If I understand correctly, we are to interpret that sentence of the lemma by considering the image of $X' \to X$ by the left Quillen functor $t_!\colon \mathbf{Set}_\Delta^+/S \to \mathbf{Set}_\Delta^+$, then take a fibrant replacement of this (a model for the localisation, as you say) and this will be a categorical equivalence. Is this correct? $\endgroup$ – Andrea Gagna Jan 28 '18 at 17:57
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    $\begingroup$ Almost: the equivalences in $\mathrm{Set}^{+}_{\Delta}$ are the Cartesian equivalences (not 'categorical equivalences'), and every object in the source of $t_!$ is cofibrant so you don't need to do any replacement to the morphism $X' \to X$ to get an equivalence. It's only when you do the right Quillen functor down to $\mathrm{Set}_{\Delta}$ that you have to replace to get an equivalence, because not every object is fibrant in $\mathrm{Set}^+_{\Delta}$ for the Cartesian model structure. $\endgroup$ – Dylan Wilson Jan 28 '18 at 19:44

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