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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeOct 14th 2010

    I split off an entry regular monomorphism in an (infinity,1)-category from regular monomorphism. But not much there yet.

    • CommentRowNumber2.
    • CommentAuthorMike Shulman
    • CommentTime7 days ago

    Is anything known about the notion of regular monomorphism in an (infinity,1)-category? E.g. what are these in Gpd\infty Gpd?

  1. I think in \infty-groupoids they are the same as monomorphisms, i.e., (1)(-1)-truncated maps. Here is a non-constructive argument. (I would expect that this also holds constructively, but I haven’t thought about it.)

    Any monomorphism of \infty-groupoids can be written as a coproduct inclusion AABA \to A \sqcup B. Consider the cosimplicial object mAB×[m]m \mapsto A \sqcup B \times [m] where [m][m] is seen as a discrete \infty-groupoid. Connected limits commute with coproducts so we have lim m(AB×[m])=(lim mA)(B×lim m[m])=AB×=A\lim_m (A \sqcup B \times [m]) = (lim_m A) \sqcup (B \times \lim_m [m]) = A \sqcup B \times \emptyset = A.

    • CommentRowNumber4.
    • CommentAuthorMike Shulman
    • CommentTime7 days ago

    That shows that every mono is a regular mono, but why would the converse be true?

    • CommentRowNumber5.
    • CommentAuthorDmitri Pavlov
    • CommentTime7 days ago

    What are some examples of regular monos that are not monos? (Not necessarily in ∞-Grpd.)

    • CommentRowNumber6.
    • CommentAuthorMike Shulman
    • CommentTime7 days ago

    I don’t know, not offhand.

    By the way, I think the construction of #3 coincides with the “Cech co-nerve”, which is probably the constructive way to express it, and shows that monos are in fact “effective” monos.

    • CommentRowNumber7.
    • CommentAuthorMike Shulman
    • CommentTime7 days ago

    Ok, here is an example. Let DD be any small category and F:DGpdF:D\to \infty Gpd, and build the cobar construction C(*,D,F)C(\ast,D,F), a cosimplicial space where

    C n(*,D,F)= d 0,,d nDF(d n) D(d 0,d 1)××D(d n1,d n).C_n(\ast,D,F) = \prod_{d_0,\dots,d_n \in D} F(d_n)^{D(d_0,d_1)\times \cdots \times D(d_{n-1},d_n)}.

    Then TotC n(*,D,F)=holim DFTot C_n(\ast,D,F) = holim^D F. Thus, holim DF dDF(d)holim^D F \to \prod_{d\in D} F(d) is a regular mono. For instance, if XX is a pointed space and D=(*X*)D = (\ast \to X \leftarrow \ast) is the cospan whose pullback is ΩX\Omega X, we find that the constant map ΩXX\Omega X \to X is a regular mono.

  2. There are warnings, e.g., here about different naming conventions in this area.

    • CommentRowNumber9.
    • CommentAuthorMike Shulman
    • CommentTime6 days ago

    Yeah, “regular monomorphism” is not really the best name. Maybe we can think of a better one.

  3. The example with cobar construction had me stumped, but also got me thinking and I am now quite convinced that every map of \infty-groupoids is a regular monomorphism. (In that case “regular monomorphism” would be a really bad name, but we wouldn’t have to invent a new one.)

    As Mike noted, the construction I proposed before is a sort of “Čech conerve” which should make sense for any map. I will compute it in topological spaces so that I don’t have to worry about fibrancy.

    Let f:ABf \colon A \to B be a cofibration between nice enough spaces (Hausdorff should be enough). Define X mX_m as the colimit of m+1m+1 copies of ff coming out of a single copy of AA. Since ff is a cofibration, this is also a homotopy colimit. X mX_ms form a cosimplicial object and since every space is fibrant, I can compute its homotopy limit as the limit weighted by the standard cosimplicial object Δ \Delta^\bullet. A point of holimX\holim X is a sequence (x 0,x 1,x 2,)(x_0, x_1, x_2, \ldots) where x m:Δ mX mx_m \colon \Delta^m \to X_m. This has to be compatible with the action of Δ\Delta. Compatibility with faces says that every vertex of x mx_m is an image of x 0x_0 in X mX_m, i.e., a copy of x 0Bx_0 \in B in one of m+1m+1 copies of BB in X mX_m. Compatibility with degeneracies says that when I collapse all these copies back to BB, x mx_m becomes constant at x 0x_0. This means that x mx_m takes values in the set of copies of x 0x_0 which is discrete since X mX_m is Hasudorff. Hence x mx_m itself is constant which in turn implies that x 0Ax_0 \in A. Thus holimX=A\holim X = A.

    Does that look correct?

    • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTime6 days ago

    Yes, I think that looks reasonable. Thanks!

    We do, I think, still need a name for the concept in other (,1)(\infty,1)-categories. In particular, can this result be generalized to any (,1)(\infty,1)-topos?

    • CommentRowNumber12.
    • CommentAuthorMarc Hoyois
    • CommentTime5 days ago
    • (edited 5 days ago)

    @Karol This cannot be correct in general, as applying it to X*X\to * says that XX is the limit of the cosimplicial diagram X X^\bullet where X n=ΣXΣXX^n=\Sigma X\vee \dots\vee \Sigma X is a wedge of nn copies of ΣX\Sigma X. But it can happen that ΣX\Sigma X is contractible and XX isn’t. It probably works under some simply connected assumptions…

    • CommentRowNumber13.
    • CommentAuthorMarc Hoyois
    • CommentTime5 days ago

    A wild guess is that epis and regular monos form a factorization system in ∞-groupoids. If so, regular mono = map whose fibers have hypoabelian π 1\pi_1. (Note that X*X\to * is epi iff ΣX\Sigma X is contractible.)

    • CommentRowNumber14.
    • CommentAuthorMike Shulman
    • CommentTime5 days ago

    Ah, #13 sounds like a more a priori plausible guess (although I don’t see yet exactly where Karol’s argument goes wrong). In fact, I think more generally than #12 we can say that if f:ABf:A\to B is epi, then its Cech conerve is constant at BB. And it seems likely that epis are left orthogonal to regular monos in any (,1)(\infty,1)-category, for essentially the same reasons as the corresponding fact in 1-categories. So all that remains to have a factorization system is to show that for any f:ABf:A\to B, the map from AA to the totalization of the Cech conerve of ff is epi.

    What exactly are the epimorphisms in Gpd\infty Gpd?

    • CommentRowNumber15.
    • CommentAuthorKarol Szumiło
    • CommentTime5 days ago
    • (edited 5 days ago)

    I agree that my claim must be wrong, but I am also baffled because I can’t find a mistake in my argument.

    There is a paper about epimorphisms which even discusses the factorization system, but doesn’t say that much about the other class.

  4. I think I know where my argument might have gone wrong.

    To compute the homotopy limit of a levelwise fibrant diagram as a weighted limit, we need a projectively cofibrant weight and I’m pretty sure that Δ \Delta^\bullet does not satisfy this condition. (This argument implies that the corresponding simplicial weight is not projectively cofibrant, so the topological one shouldn’t be either.) On the other hand, Δ \Delta^\bullet is Reedy cofibrant which, I believe, would suffice for computing homotopy limits of Reedy fibrant diagrams, but conerves (as I constructed them) certainly fail to be Reedy fibrant in general.

    • CommentRowNumber17.
    • CommentAuthorMike Shulman
    • CommentTime5 days ago

    Ah, right! That’s annoying though.

    Why have I not read that paper yet? It looks most excellent. So the suggestion in #13 would imply that given a map f:XYf:X\to Y of connected (for simplicity) spaces, the plus construction X P +X^+_P at the maximal perfect subgroup of ker(π 1(f))ker(\pi_1(f)) is equivalent to the (homotopy) totalization of the Cech conerve of ff. Is that plausible?

    • CommentRowNumber18.
    • CommentAuthorMarc Hoyois
    • CommentTime5 days ago

    It seems plausible, but I don’t know how to prove it. This paper shows that X*X\to * is a regular mono when XX is 22-connected, and Gijs’ comment in the homotopy theory chat suggests this is still true if XX is simply connected.

    For what it’s worth I have a note where I discuss Quillen’s plus construction in ∞-topoi (it doesn’t work very well though).

    • CommentRowNumber19.
    • CommentAuthorMike Shulman
    • CommentTime4 days ago

    Given that we already have a factorization system whose left class is the epis, and we know that regular monos are right orthogonal to epis, it suffices to show that any map in the right class (i.e. that is right orthogonal to epis – are these the “hypoabelian” maps?) is also a regular mono.

    Thanks for the links! By “doesn’t work very well” are you referring to your use of hypotheses like hypercompleteness and π 1\pi_1 preserving products?

    • CommentRowNumber20.
    • CommentAuthorMarc Hoyois
    • CommentTime4 days ago

    Yes, hypoabelian maps form the right class.

    Thanks for the links! By “doesn’t work very well” are you referring to your use of hypotheses like hypercompleteness and π 1\pi_1 preserving products?

    Indeed, π 1\pi_1 preserving products rules out most ∞-topoi. It seems that in a general ∞-topos a perfect subgroup of π 1\pi_1 isn’t quite the right input for the plus construction. More precisely, acyclic maps are still those that universally kill a perfect subgroup (at least in the hypercomplete case, this is Lemmas 2&3), but not every perfect subgroup can be universally killed (as far as I could see).

    • CommentRowNumber21.
    • CommentAuthorMike Shulman
    • CommentTime4 days ago

    It’s quite surprising to me, as a category theorist, that such a simple concept as “epimorphism” gets involved in complicated things like universally killing perfect subgroups. In 1Gpd1Gpd the epimorphisms are (unless I’m mistaken) just the essentially surjective and full maps, and I always assumed that this would continue to higher dimensions, but I guess not. Do we know what the epimorphisms are in nGpdn Gpd for 1<n1\lt n \le \infty?

    • CommentRowNumber22.
    • CommentAuthorMarc Hoyois
    • CommentTime4 days ago

    Well, the map ****\sqcup * \to * is not an epimorphism of 1-groupoids, since its pushout is the groupoid BB\mathbb{Z}. I think the epimorphisms of 1-groupoids are exactly the 0-connected maps, i.e., maps inducing iso on π 0\pi_0 and epi on π 1\pi_1. In general (n1)(n-1)-connected maps are epimorphisms in nn-groupoids, but the converse fails for n2n \geq 2, since there exist acyclic 1-truncated spaces, so that there are epimorphisms X*X\to * that are not 1-connected.

    I misread the paper I linked to, they actually prove that X*X\to * is a regular mono for XX simply connected. But in fact Bousfield already proves this when XX is nilpotent in “On the Homology Spectral Sequence of a Cosimplicial Space”. So I believe we at least know the implications

    nilpotent fibers ⇒ regular mono ⇒ hypoabelian fibers.

    • CommentRowNumber23.
    • CommentAuthorMike Shulman
    • CommentTime4 days ago

    I think the epimorphisms of 1-groupoids are exactly the 0-connected maps, i.e., maps inducing iso on π 0\pi_0 and epi on π 1\pi_1.

    Yes, I believe in categorical language that’s the same as what I said: essentially surjective and full. The map ****\sqcup * \to * is not full.

    In general (n1)(n-1)-connected maps are epimorphisms in nn-groupoids, but the converse fails for n2n \geq 2, since there exist acyclic 1-truncated spaces, so that there are epimorphisms X*X\to * that are not 1-connected.

    I see that there are maps X*X\to * that are not 1-connected and that are epimorphisms of \infty-groupoids, but I don’t think it follows (at least not a priori) that they are also epimorphisms of nn-groupoids for any n<n\lt\infty: the inclusion nGpdGpdn Gpd \hookrightarrow \infty Gpd need not reflect epimorphisms, since it doesn’t preserve pushouts.

    • CommentRowNumber24.
    • CommentAuthorMarc Hoyois
    • CommentTime4 days ago

    Ah yes, my bad.

    I see that there are maps X*X\to * that are not 1-connected and that are epimorphisms of \infty-groupoids, but I don’t think it follows (at least not a priori) that they are also epimorphisms of nn-groupoids for any n<n\lt\infty: the inclusion nGpdGpdn Gpd \hookrightarrow \infty Gpd need not reflect epimorphisms, since it doesn’t preserve pushouts.

    Fully faithful functors reflect epimorphisms, in fact any functor that reflects pushouts does.

    • CommentRowNumber25.
    • CommentAuthorMike Shulman
    • CommentTime4 days ago

    Right you are; my bad this time. What’s a simple example of an acyclic 1-truncated space? I might understand better if I worked out exactly what happens with a concrete example in 2-groupoids.

    • CommentRowNumber26.
    • CommentAuthorUlrik
    • CommentTime4 days ago
    • (edited 4 days ago)

    It seems the easiest example is given by the classifying space of Higman’s group; generated by x 0x_0, x 1x_1, x 2x_2, x 3x_3 subject to x i+1x ix i+1 1=x i 2x_{i+1} x_i x_{i+1}^{-1} = x_i^2 for each ii with i+1i+1 computed mod 4. See at encyclopedia of math.

    No finite group is acyclic. (As seen on this MO question.)

  5. Here are some references. It lists the simplest example, I’ve seen: BGB G where G=x 0,x 1,x 2,x 3x i+1x ix i+1 1=x i 2,i/4G = \langle x_0, x_1, x_2, x_3 \mid x_{i+1} x_i x^{-1}_{i+1} = x^2_i, i \in \mathbb{Z} / 4 \rangle, but maybe there are some simpler ones if you follow the references.

    • CommentRowNumber28.
    • CommentAuthorMike Shulman
    • CommentTime4 days ago

    Thanks. I am confused again however. If GG is an acyclic group, then BGB G is an acyclic space, and so BG*B G \to \ast is supposed to be an epimorphism of \infty-groupoids. But BGB G and *\ast are both 1-groupoids, and 1GpdGpd1 Gpd \hookrightarrow \infty Gpd is fully faithful, so it seems this should imply that BG*B G \to \ast is also an epimorphism even of 1-groupoids. Is that true? If so, then I must have been wrong already about the claim that the epimorphisms of 1-groupoids are the 0-connected maps, which would surprise me even more. Am I missing something?

    • CommentRowNumber29.
    • CommentAuthorMarc Hoyois
    • CommentTime3 days ago

    BG*BG\to * is 0-connected!

    • CommentRowNumber30.
    • CommentAuthorMike Shulman
    • CommentTime3 days ago

    Buh, of course it is. Sorry, I guess I’m just having trouble keeping track of the directions of all these implications. So when we embed

    1Gpd1Gpd2Gpd3GpdGpd 1 Gpd \hookrightarrow 1 Gpd \hookrightarrow 2 Gpd \hookrightarrow 3 Gpd \hookrightarrow \cdots \hookrightarrow \infty Gpd

    at each step epimorphisms are reflected but not preserved: the conditions to be an epimorphism get stronger as we go up. My intuition is that in nGpdn Gpd the condition to be an epimorphism involves 00 through (n1)(n-1)-cells, so that when we move to (n+1)Gpd(n+1)Gpd we add a condition on nn-cells for something to remain an epimorphism. The naive idea I had was that the condition on 00 through (n1)(n-1)-cells was essential surjectivity, but it seems that that’s only right for 0-cells and 1-cells; at 2-cells we start to get a weaker condition. But it’s extra confusing because the lack of essential surjectivity on 2-cells manifests as nontrivial 1-cells in the fiber, and the relevant condition is naturally expressed in terms of the fiber.

    So I think what I want now to see (or write down myself) is an explicit proof that when GG is an acyclic group and HH is a 2-groupoid, the map 2Gpd(*,H)2Gpd(BG,H)2 Gpd(\ast,H) \to 2 Gpd(B G, H) is a monomorphism of 2-groupoids. Ideally I would like some abstract group-theoretic property of acyclic groups that would make such a proof work, and which I could separately verify to be true in some examples like #26-27. However, I suspect that there isn’t any such property that differs significantly from the definition of “being acyclic”; is that the case? If so, then I guess what I have to do is pick an example like #26-27 and translate the proof of acyclicity into a proof of this statement about 2-groupoids.

    • CommentRowNumber31.
    • CommentAuthorTim Campion
    • CommentTime3 days ago

    Don’t let me sidetrack anyone, but I remember one question that came up when thinking about this stuff before is the following: are corepresentable functors SpacesSpacesSpaces \to Spaces closed under subobjects? That is, if XX is a space, and F() XF \to {(-)}^X is an \infty-monomorphism of \infty-functors SpacesSpacesSpaces \to Spaces, then must FF be corepresentable? Equivalently, is every subfunctor of a corepresentable functor () X(-)^X of the form “maps out of XX which kill PP” for some perfect normal subgroupoid PΠ 1(X)P \subseteq \Pi_1(X)?

    • CommentRowNumber32.
    • CommentAuthorMarc Hoyois
    • CommentTime2 days ago

    @Tim If XX has no perfect subgroups, wouldn’t that be saying that every map out of XX factors through every other?

    • CommentRowNumber33.
    • CommentAuthorTim Campion
    • CommentTime2 days ago

    @Marc Ah… you’re saying that I can generate subfunctors from maps just like generating sieves in ordinary category theory, so there are a lot of them. I guess that makes sense. Somehow I had gotten the impression that being an \infty-monomorphism was a very restrictive condition, but it’s not quite as restrictive as I was thinking. So the answer to my question is obviously no.