Omar Antolín-Camarena
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Registered User
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May 1 |
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For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring? @Noah: Whoops! You're right, I got confused. |
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May 1 |
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For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring? given a Boolean algebra structure, the poset is recovered by setting $a\le b$ iff $a\wedge b=a$. |
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Apr 30 |
revised |
What are some examples of weak ω-categories? added 251 characters in body |
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Apr 28 |
awarded | ● Nice Question |
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Apr 28 |
revised |
What are some examples of weak ω-categories? added previous question of which this is a duplicate; deleted 258 characters in body |
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Apr 27 |
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What are some examples of weak ω-categories? Did you mean $\infty$-groupoid instead of contractible, @JeremyHahn? I mean an ordinary groupoid has all adjoints, right? I certainly don't want it to be contractible as an $(\infty,\infty)$-category. (Or do I?) If this second homotopy theory of $(\infty,\infty)$-categories makes either (1) a all homotopy types contractible, or (2) homotopy inverses not count as adjoints, it doesn't sound like such a great idea. |
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Apr 27 |
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What are some examples of weak ω-categories? Yes, I think that's right @SamGunningham. I've added a remark about this to the question. |
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Apr 27 |
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What are some examples of weak ω-categories? incorporated Sam Gunningham's correction |
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Apr 26 |
asked | What are some examples of weak ω-categories? |
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Apr 18 |
accepted | Connected groupoids and action groupoids |
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Apr 18 |
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Connected groupoids and action groupoids I hope that is clear enough, @MikhailBorovoi. If not let me know. As you can see, the proof is pretty much just what Sam Cunningham wrote (only slightly generalized). If you want an explicit isomorphism between A and G/H, you can just compose the isomorphisms A -> H x X -> G/H given in the proof of Claim 1. |
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Apr 18 |
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Connected groupoids and action groupoids added 354 characters in body |
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Apr 18 |
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Connected groupoids and action groupoids added 899 characters in body |
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Apr 17 |
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Connected groupoids and action groupoids @Mikhail: Pick a set of representatives for the cosets of H, and pick a bijection between those representatives and the objects of A. I can add more detail later when I'm at a computer with a real keyboard. |
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Apr 17 |
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Connected groupoids and action groupoids These are some of the groups you can use, but in general you don't need $G$ to be a product, just to have a subgroup isomorphic to $G_0$ with index $|X|$. |
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Apr 17 |
answered | Connected groupoids and action groupoids |
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Apr 17 |
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Connected groupoids and action groupoids This is easier than everyone is making it sound. |
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Apr 17 |
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Connected groupoids and action groupoids I answered on math.stack exchange. |
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Apr 3 |
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How should one understand orbifold fundamental groups? I've reworded the last paragraph to make it clearer, since Dan Petersen's question made me see I wasn't being very clear. |
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Apr 3 |
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How should one understand orbifold fundamental groups? added explanation of generalization mentioned at the end |
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Apr 3 |
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How should one understand orbifold fundamental groups? No Dan, the argument I gave works whenever you have a continuous partition of unity subordinate to the cover (I.e.,for paracompact spaces M even if M is not a manifold) and gives you a homotopy equivalence. What I'm saying in the last paragraph is that without a subordinate partition of unity you don't get a homotopy equivalence anymore, but you still get a weak equivalence. |
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Apr 3 |
answered | How should one understand orbifold fundamental groups? |
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Apr 2 |
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Are subfunctors of left exact functors also left exact? Everyone can stop down-voting my answer: I've corrected the spelling of 'necessarily'. :P (Seriously now, I agree that Eric's example is much nicer and closer to the context Samuel Mf is interested in.) |
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Apr 2 |
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Are subfunctors of left exact functors also left exact? corrected spelling |
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Apr 1 |
answered | Are subfunctors of left exact functors also left exact? |
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Mar 31 |
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Are Lurie’s operads special SMCs? Yes, that's right. |
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Mar 29 |
accepted | Are Lurie’s operads special SMCs? |
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Mar 29 |
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Are Lurie’s operads special SMCs? fixed grammar |
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Mar 29 |
answered | Are Lurie’s operads special SMCs? |
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Mar 27 |
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Are Lurie’s operads special SMCs? I think the construction you are looking for, that given an operad produces a symmetric monoidal category, is the monoidal envelope described in Higher Algebra, section 2.2.4. |
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Mar 27 |
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Are Lurie’s operads special SMCs? For what you want, given an operad $p : C \to Fin_{\ast}$, you would need to construct a $q : D \to Fin_{\ast}$ where the objects of $q^{-1}\langle 1 \rangle$ come from all of $C$, not just from $p^{-1} \langle 1 \rangle$. |
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Mar 27 |
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Are Lurie’s operads special SMCs? When you think of an non-symmetric operad as special kind of monoidal in the first paragraph, the objects of the monoidal category are not the objects of the corresponding operad but rather strings of these objects. But for both Lurie's way of encoding operads and SMC the objects are the objects of $p^{-1} \langle 1 \rangle$. The two definitions Lurie gives are much more suited to seeing symmetric monoidal category as special kinds of operads than the other way around. |
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Mar 14 |
accepted | Simplicial sets from bisimplicial sets, and their realisations. |
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Mar 7 |
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About the Cole-Ström model category structure with a locally presentable category I think you probably wrote "Tobi Barthels" as a mashup of Tobias Barthel and Toby Bartels. Tobias Barthel is the one you meant. |
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Mar 4 |
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The main theorems of category theory and their applications I've heard people say before that Cayley's theorem for groups is a special case of the Yoneda Lemma, but I think it's more correct to say that the permutation representation of G as acting on itself by translations is a special case of the Yoneda embedding. The Yoneda lemma, then says that for any G-set X, G-equivariant maps G → X are in bijection with the elements of X. |
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Feb 26 |
answered | problems from the scottish book |
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Feb 26 |
awarded | ● Popular Question |
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Feb 25 |
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(Co-)Limits and fibrations of DG-Categories? The Grothendieck construction is the "oplax-colimit", to get the 2-colimit, you must invert the Cartesian morphisms. |
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Feb 23 |
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realization of maps between classifying spaces of categories @MarkGrant: I think it's fairly standard. @RonnieBrown: I agree with the advantages of $\pi_1(X,A)$, but for the case $X=A$, I find $\pi_1(X,X)$ too long, and, of course most people would assume $\pi-1(X)$ is a group, so I use $\pi_{\le 1}(X)=\pi_1(X,X)$. |
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Feb 23 |
answered | realization of maps between classifying spaces of categories |
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Feb 22 |
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realization of maps between classifying spaces of categories The remark is correct: the homotopy category of 1-types is equivalent to the homotopy category of groupoids, so the answer is affirmative if both $\mathcal{C}_1$ and $\mathcal{C}_2$ are groupoids. |
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Feb 20 |
answered | Simplicial sets from bisimplicial sets, and their realisations. |
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Feb 20 |
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Why is the the classifying space of the natural numbers homotopy equivalent to the circle? (3) You talk about a minimal fibration, but I don't know which map you mean: there is only the constant map BZ -> BN (simplicial maps like that are just monoid homomorphisms Z->N), and the inclusion BN->BZ is not a fibration. (Also, a fibrant replacement for a simplicial set X is a Kan complex Y with an acyclic co*fibration X -> Y.) Did you mean instead that you wanted the thing you get by adding simplices to BM to be a minimal *simplicial set? |
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Feb 20 |
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Why is the the classifying space of the natural numbers homotopy equivalent to the circle? I don't understand several things about this answer: (1) You seem to be saying you can get BKM from BM by just adding fillers of outer horns, but of course, once you've added a few inverses of elements of M you need to deal with their compositions too. (2) Even if you didn't just mean fillers for outer horns in the original BM, you still seem to be saying that you get BKM from BM by just adding simplices; but having left and right cancelation does not guarantee that M->KM is injective! |
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Feb 19 |
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Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG? Yes, that is correct. Alternatively, instead of using the universal cover argument, I think you can use Quillen's theorem B to show the slice category is simply connected, and thus contractible iff it has trivial homology, in which case Theorem B implies the canonical map is an equivalence BM->BG. |
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Feb 19 |
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Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG? That's a nice paper, thanks for the reference, @BenjaminSteinberg! |
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Feb 18 |
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Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG? Plus, it id's not true that just adding fillers for the outer horns of NM gets you NKM, since you also have compositions of elements of KM\M. |
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Feb 18 |
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Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG? I'm also confused by talk of "minimal fibrations" in Spice the bird's answer since there is no map NZ->NN and the inclusion NN->NZ is not a fibration, of course. I think Spice might have meant NKM is a minimal simplicial set or something like that instead... |
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Feb 18 |
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Is there a (discrete) monoid M injecting into its group completion G for which BM is not homotopy equivalent to BG? I don't understand Spice the bird's argument. He or she seems to be saying that for a "cancelable" M, NKM is obtained from NM by attaching files for outer horns, in particular, by only adding simplices, so it would seem the argument would imply that M injects into KM, but Malcev's example shows that's not always true. (Maybe "cancelable" means "injects into KM" rather than "has left and right cancellation" in Spice's answer?) |
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Feb 18 |
awarded | ● Nice Question |
