Timeline for On HTT's Lemma 3.3.4.1
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2018 at 0:25 | vote | accept | Edoardo Lanari | ||
| Jan 28, 2018 at 19:44 | comment | added | Dylan Wilson | 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. | |
| Jan 28, 2018 at 17:57 | comment | added | Andrea Gagna | 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? | |
| Jan 28, 2018 at 13:04 | history | answered | Dylan Wilson | CC BY-SA 3.0 |