Timeline for Does the classification diagram localize a category with weak equivalences?
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13 events
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| Mar 1, 2015 at 17:48 | comment | added | Aaron Mazel-Gee | @Denis-CharlesCisinski: Actually, I'm still a little confused about your proof here too. I certainly see that a map $C \to X$ taking $W$ to equivalences induces a map $N(C,W) \to \iota (\Delta^*,X)$, but I don't see why every such map should be obtained in that way. | |
| Feb 24, 2015 at 5:12 | comment | added | Mike Shulman | @AaronMazel-Gee - drat. I can't think of a quick fix; maybe we can show that the left adjoint of $G$ takes $N(C,W)$ to something equivalent to the marked simplicial nerve of $(C,W)$? That's starting to look more like Denis-Charles' original argument too. | |
| Feb 21, 2015 at 21:20 | comment | added | Aaron Mazel-Gee | @MikeShulman: Could you explain a little further? It looks like your functor $G$ is a sort of nerve functor, so I'd expect it to be a right adjoint. But if it is a right Quillen functor, it only necessarily has the correct behavior on fibrant objects of $\mathrm{sSet}^+$, whereas the whole point is that a relative (quasi)category won't generally define a fibrant object. So how do we know that $G(C,W)$ selects the correct weak equivalence class in the complete Segal space model structure? | |
| Apr 3, 2012 at 21:50 | vote | accept | Mike Shulman | ||
| Apr 3, 2012 at 21:50 | comment | added | Mike Shulman | You're right, it does! | |
| Apr 3, 2012 at 21:23 | comment | added | D.-C. Cisinski | Your reformulation is quite nice. Moreover, all this shows that you can replace $N(C)$ by any quasi-category. | |
| Apr 3, 2012 at 19:39 | comment | added | Mike Shulman | ... The factorization is via the functor $G$ from marked simplicial sets to bisimplicial sets defined by $G(X)_{n,m} = MSSet((\Delta^n)^\flat \times (\Delta^m)^\sharp, X)$; thus $G$ also induces an equivalence of homotopy categories. (In fact, it appears that $G$ is even part of a Quillen equivalence.) But $G$ takes the "marked simplicial nerve" of $(C,W)$ to the bisimplicial set $N(C,W)$, and the marked simplicial nerve clearly represents the localization of $C$ at $W$. | |
| Apr 3, 2012 at 19:39 | comment | added | Mike Shulman | Okay, I see! How about the following for a reformulation in terms of marked simplicial sets? Your functor $i(\Delta^*,-)$ from quasicategories to complete Segal spaces induces an equivalence of homotopy categories (it is homotopy equivalent to the functor $k^!$ from arxiv.org/abs/math.AT/0607820). But $i(\Delta^*,-)$ also factors through the functor $X\mapsto (X,i(X)_1)$ from quasicategories to marked simplicial sets, which also induces an equivalence of homotopy categories. ... | |
| Apr 3, 2012 at 19:04 | comment | added | D.-C. Cisinski | Dear Mike, I didn't understood your comment: it seems that I identified $X$ with the wrong evaluation at zero of $i(\Delta^\ast,X)$. This is corrected. | |
| Apr 3, 2012 at 19:02 | history | edited | D.-C. Cisinski | CC BY-SA 3.0 |
added 5 characters in body
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| Apr 3, 2012 at 11:48 | comment | added | D.-C. Cisinski | I see quasi-categories as constant simplicial quasi-categories. Also, for a quasi-category $X$, I see $i(\Delta^\ast,X)$ as a simplicial object in the category of quasi-categories: for any fixed integer $q$, $i(\Delta^\ast,X)_q$ is a quasi-category (and for $q=0$, this is precisely $X$). | |
| Apr 3, 2012 at 0:03 | comment | added | Mike Shulman | This looks like a promising approach, but I don't follow it yet. When you write $N(C)$, you mean to regard it as a bisimplicial set discrete in which direction? And it seems to me that $i(\Delta^0,X) = i(X)$, not $X$. | |
| Apr 2, 2012 at 23:53 | history | answered | D.-C. Cisinski | CC BY-SA 3.0 |