Timeline for A combinatorial approximation functor sSet->qCat
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 27, 2023 at 14:48 | comment | added | Dmitri Pavlov | @lvl: Kan's Ex^∞ does not preserve all limits (it does not preserve infinite products, for example), and for this reason neither does this functor. | |
| Apr 27, 2023 at 6:16 | comment | added | display llvll | @DmitriPavlov Nice answer. I was wondering, does this functor preserve all limits (and enjoys the other good properties) as Kan’s Ex_infty ? | |
| Oct 19, 2020 at 1:04 | comment | added | Manuel Rivera | @DmitriPavlov The morphisms $\mathfrak{C}(X)(x,y)$ involve certain identifications but they also have a description which is conceptually useful: for every necklace in $T\to X$ from $x$ to $y$ you consider a simplicial cube $(\Delta^{1})^{\times n}$ where $n$ is the dimension of the necklace $T$ (sum of the dim of beads minus the number of beads) and then you glue all of these cubes according to how necklaces fit inside $X$. Some of the identifications involve noting that there are different necklaces which give rise to the same simplicial cubes. | |
| Oct 19, 2020 at 0:58 | comment | added | Manuel Rivera | @DmitriPavlov ok, I see, so you are just providing a model which seems a bit simpler. | |
| Oct 18, 2020 at 23:08 | comment | added | Dmitri Pavlov | @ManuelRivera: C^nec has mapping simplicial sets that are nerves of categories, C does not. For C^nec, simplices are chains of necklaces, whereas for C simplices are certain equivalence classes of such chains. Thus, C^nec is a bit simpler in that it avoids equivalence classes. | |
| Oct 18, 2020 at 23:05 | comment | added | Dmitri Pavlov | @TimCampion: It appears to me that C Δ^n→C^nec X does not preserve composition already for n=3 or n=4, so is not a simplicial functor. Also, Dugger and Spivak give an explicit description of C in terms of equivalence classes of necklaces with additional data (Corollary 4.4 of their paper), and your construction, if correct, would provide a functorial way to extract a necklace from such an equivalence class, which seems unlikely. | |
| Oct 18, 2020 at 20:42 | comment | added | Manuel Rivera | I am not sure I understood the point of using $\mathfrak{C}^{nec}$ instead of $\mathfrak{C}$ in the answer to the question. Can someone explain? Thank you! | |
| Oct 17, 2020 at 16:40 | comment | added | Tim Campion | Assuming this is a levelwise equivalence, the zigzag through $N Ex^\infty \mathfrak C^{hoc}$ can be avoided, in favor of a direct map $1 \to N Ex^\infty \mathfrak C^{nec}$, which obviates the main advantage of using $\mathfrak C$ instead of $\mathfrak C^{nec}$. | |
| Oct 17, 2020 at 16:39 | comment | added | Tim Campion | Although Dugger and Spivak don't say it, I believe there is a natural transformation $\mathfrak C \Rightarrow \mathfrak C^{nec}$, i.e. $1 \Rightarrow N \mathfrak C^{nec}$. It sends a simplex $\sigma \in X_n$ to the simplicial functor $\mathfrak C \Delta^n \to C^{nec} X$ which does the obvious thing on objects, sends the "free" 1-morphism $f_{ij}$ from $i$ to $j$ to the composite $\Delta^{j-i} \to \Delta^n \xrightarrow \sigma X$, sends the homotopy $f_{jk} f_{ij} \to f_{ik}$ to the obvious map $\Delta^{k-j} \vee \Delta^{j-i} \to \Delta^{k-i}$, and is otherwise determined by 2-coskeletality. | |
| Oct 13, 2020 at 19:19 | history | answered | Dmitri Pavlov | CC BY-SA 4.0 |