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I have two questions:

  1. Is there a way to tell if a functor $F:C \to D$ between two small categories is an equivalence in terms of the map $$N(F):N(C) \to N(D)$$ between simplicial sets? More generally, can we test separately when a functor $F$ is: essentially surjective, full, faithful etc., by an easy to verify property of $N(F)$? I am not interested in any answer involving applying the left-adjoint to $N;$ I am really looking for a description using simplicial sets.

  2. If $\varphi:X \to Y$ are quasi-categories, is there way of saying when $\varphi$ is a weak equivalence in the Joyal model structure (categorical equivalence in the language of Lurie) akin to "full and faithful and essentially surjective?". I am most interested in definitions which are easily checkable, so not definitions like "the induced map on mapping complexes..." since these are hard to compute in practice. The definition I have seen just says that $\varphi$ becomes a weak equivalence of simplicial categories after applying the left-adjoint to the homotopy coherent nerve, which is somehow not a very satisfying definition. Is there a more explicit description, not involving simplicial categories?

I hope this question is not too vague. Any comments to improve the wording etc. are welcome. Thanks!

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Are we allowed to use a factorisation system on sSet? Or how about translating the canonical model structure on Cat (where equivalences are the weak equivalences) to sSet? Fully faithful should be easy - consider maps right orthogonal to the nerve of the inclusion {1,2} --> {1->2} –  David Roberts Feb 3 '12 at 0:06
    
Thanks for the comment about fully faithful. Do you know if this holds for quasi-categories too? As far as translating the canonical model structure, do you mean transferring along the nerve-adjunction? If so, I'd rather avoid applying the left adjoint to $N$... –  David Carchedi Feb 3 '12 at 0:09
    
We wouldn't need the full model structure, just the result that an equivalence of catgories factors as an acyclic cofibration followed by an acyclic fibration. An acyclic fibration should be easy to characterise, it is an isofibration which is fully faithful. An isofibration is, at the level of simplicial sets, a map with the right lifting property with respect to the nerve of {1} --> {1\simeq 2}. An acyclic cofibration is an equivalence which is injective on objects. So we just need to characterise injective-on-0-cells maps which are ess. surj. (cont..) –  David Roberts Feb 3 '12 at 6:34
    
(essentially surjective isofibrations are surjective on objects) –  David Roberts Feb 3 '12 at 6:34
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1 Answer

Just a few ideas/observations...

For 1:

a. A functor is an isomorphism if and only if the induced map is an isomorphism of simplicial sets.

b. So a functor is an equivalence if and only if the induced map on the nerves of skeleta is an isomorphism. (This isn't very helpful though...)

c. Any map of simplicial sets $NC \rightarrow ND$ gives us a functor $C \rightarrow D$ (just look at the map on 0-simplices, 1-simplices, and 2-simplices to see what to do). So a map $NC \times \Delta^1 \rightarrow ND$ corresponds to functor $F: C \times [1] \rightarrow D$ (since $N[1] = \Delta^1$), but this is precisely a natural transformation of functors $C \rightarrow D$. Thus, if we have $NC \rightarrow ND$ and $ND \rightarrow NC$ such that the composites are "homotopic" to the identity (where I mean, use $\Delta^1$), then the original functor is an equivalence.

d. None of what we have said so far is any easier than just proving your original map is an equivalence. One thing you could do, that allows for homotopy theory, is associate to $C$ a stronger invariant. For example, take the bisimplicial set call it $\mathfrak{N}C$ given by: $\mathfrak{N}C_k = N(\text{iso }C^{[k]})$ (where "iso C" means "maximal groupoid" or "just take isomorphisms as your morphisms." This is a special case of a construction of Rezk's). It turns out that a functor is an equivalence if and only if the induced map of these special nerves is a weak equivalence.

e. I don't think that checking it's a weak equivalence works for the usual nerve, since (for example) any category with a final object has a nerve that's weakly equivalent to a point. However it is probably the case that if the induced map on nerves is a categorical equivalence then the original categories are equivalent. This brings us to point 2...

For 2

a. I'm not sure if there is any way of showing that a functor between $\infty$-categories is an equivalence (in general) that does not amount to showing the induced map on mapping spaces is an equivalence. However, you can do all of this without mentioning simplicial categories. Given an $\infty$-category, $\mathcal{C}$, we can define something (or several things) equivalent to the mapping space between two objects in $\mathfrak{C}[\mathcal{C}]$; let Hom_C^R(x,y) be the simplicial set defined by requiring that a map $\Delta^n \rightarrow \text{Hom}_{\mathcal{C}}^R(x,y)$ be a map $z: \Delta^{n+1} \rightarrow \mathcal{C}$ such that $d_0z$ is the constant diagram on $y$ and $z$ evaluated on the initial vertex is $x$. (Actually this might be $\text{Hom}^L$, I can't remember off the top of my head). Anyway, this turns out to be a Kan complex (i.e. a space), and Lurie/Joyal show that it's the same as what you get after doing $\mathfrak{C}$. See Lurie for more.

b. Given the description above, we can say a functor is an equivalence if it induces a weak equivalence of Kan complexes for all mapping spaces as we defined above (i.e. fully faithful), and essentially surjective (i.e. every object is equivalent to something in the image). This isn't easy to do but...

c. In general, an equivalence of $\infty$-categories is kind of a big deal. In particular, it should be at least as hard as writing down various Quillen equivalences of model categories, right? Maybe slightly easier, but still quite hard.

Anyway, I realize this probably isn't what you want, but it's the best I can do at the moment! I'll let you know if I think of anything more helpful (but probably someone brighter will end up posting something more helpful soon...)

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I should point at that in 1d, only $\mathfrak{N}C_0$ and $\mathfrak{N}C_1$ really matter, so that might cut down on computation. –  Dylan Wilson Feb 3 '12 at 5:25
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(A proof of my suspicion in (e) can be found, along with many relevant fun facts, in this paper of by Emily Riehl: math.uchicago.edu/~eriehl/topic.pdf) –  Dylan Wilson Feb 3 '12 at 6:26
    
Thanks Dylan. This does go in the right direction. And besides, my question was vague :p. –  David Carchedi Feb 3 '12 at 17:49
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