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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeOct 14th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexI split off an entry [[regular monomorphism in an (infinity,1)-category]] from [[regular monomorphism]]. But not much there yet.

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

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeMay 14th 2018
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexIs anything known about the notion of [[regular monomorphism in an (infinity,1)-category]]? E.g. what are these in $\infty Gpd$?

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

3. • CommentRowNumber3.
   • CommentAuthorKarol Szumiło
   • CommentTimeMay 14th 2018
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   Author: Karol Szumiło
   Format: MarkdownItexI think in $\infty$-groupoids they are the same as monomorphisms, i.e., $(-1)$-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 $A \to A \sqcup B$. Consider the cosimplicial object $m \mapsto A \sqcup B \times [m]$ where $[m]$ is seen as a discrete $\infty$-groupoid. Connected limits commute with coproducts so we have $\lim_m (A \sqcup B \times [m]) = (lim_m A) \sqcup (B \times \lim_m [m]) = A \sqcup B \times \emptyset = A$.

   I think in $\infty$-groupoids they are the same as monomorphisms, i.e., $(-1)$-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 $A \to A \sqcup B$. Consider the cosimplicial object $m \mapsto A \sqcup B \times [m]$ where $[m]$ is seen as a discrete $\infty$-groupoid. Connected limits commute with coproducts so we have $\lim_m (A \sqcup B \times [m]) = (lim_m A) \sqcup (B \times \lim_m [m]) = A \sqcup B \times \emptyset = A$.

4. • CommentRowNumber4.
   • CommentAuthorMike Shulman
   • CommentTimeMay 14th 2018
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexThat shows that every mono is a regular mono, but why would the converse be true?

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

5. • CommentRowNumber5.
   • CommentAuthorDmitri Pavlov
   • CommentTimeMay 14th 2018
   • PermaLink
   Author: Dmitri Pavlov
   Format: MarkdownItexWhat are some examples of regular monos that are not monos? (Not necessarily in ∞-Grpd.)

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

6. • CommentRowNumber6.
   • CommentAuthorMike Shulman
   • CommentTimeMay 14th 2018
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   Author: Mike Shulman
   Format: MarkdownItexI 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.

   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.

7. • CommentRowNumber7.
   • CommentAuthorMike Shulman
   • CommentTimeMay 14th 2018
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexOk, here is an example. Let $D$ be any small category and $F:D\to \infty Gpd$, and build the cobar construction $C(\ast,D,F)$, a cosimplicial space where $$C_n(\ast,D,F) = \prod_{d_0,\dots,d_n \in D} F(d_n)^{D(d_0,d_1)\times \cdots \times D(d_{n-1},d_n)}.$$ Then $Tot C_n(\ast,D,F) = holim^D F$. Thus, $holim^D F \to \prod_{d\in D} F(d)$ is a regular mono. For instance, if $X$ is a pointed space and $D = (\ast \to X \leftarrow \ast)$ is the cospan whose pullback is $\Omega X$, we find that the constant map $\Omega X \to X$ is a regular mono.

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

   $$C_n(\ast,D,F) = \prod_{d_0,\dots,d_n \in D} F(d_n)^{D(d_0,d_1)\times \cdots \times D(d_{n-1},d_n)}.$$

   Then $Tot C_n(\ast,D,F) = holim^D F$. Thus, $holim^D F \to \prod_{d\in D} F(d)$ is a regular mono. For instance, if $X$ is a pointed space and $D = (\ast \to X \leftarrow \ast)$ is the cospan whose pullback is $\Omega X$, we find that the constant map $\Omega X \to X$ is a regular mono.

8. • CommentRowNumber8.
   • CommentAuthorDavid_Corfield
   • CommentTimeMay 14th 2018
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   Author: David_Corfield
   Format: MarkdownItexThere are warnings, e.g., [here](https://ncatlab.org/nlab/show/regular+monomorphism#Infty1Version) about different naming conventions in this area.

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

9. • CommentRowNumber9.
   • CommentAuthorMike Shulman
   • CommentTimeMay 14th 2018
   • PermaLink
   Author: Mike Shulman
   Format: MarkdownItexYeah, "regular monomorphism" is not really the best name. Maybe we can think of a better one.

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

10. • CommentRowNumber10.
    • CommentAuthorKarol Szumiło
    • CommentTimeMay 15th 2018
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    Author: Karol Szumiło
    Format: MarkdownItexThe 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 $f \colon A \to B$ be a cofibration between nice enough spaces (Hausdorff should be enough). Define $X_m$ as the colimit of $m+1$ copies of $f$ coming out of a single copy of $A$. Since $f$ is a cofibration, this is also a homotopy colimit. $X_m$s 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 $\Delta^\bullet$. A point of $\holim X$ is a sequence $(x_0, x_1, x_2, \ldots)$ where $x_m \colon \Delta^m \to X_m$. This has to be compatible with the action of $\Delta$. Compatibility with faces says that every vertex of $x_m$ is an image of $x_0$ in $X_m$, i.e., a copy of $x_0 \in B$ in one of $m+1$ copies of $B$ in $X_m$. Compatibility with degeneracies says that when I collapse all these copies back to $B$, $x_m$ becomes constant at $x_0$. This means that $x_m$ takes values in the set of copies of $x_0$ which is discrete since $X_m$ is Hasudorff. Hence $x_m$ itself is constant which in turn implies that $x_0 \in A$. Thus $\holim X = A$. Does that look correct?

    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 $f \colon A \to B$ be a cofibration between nice enough spaces (Hausdorff should be enough). Define $X_m$ as the colimit of $m+1$ copies of $f$ coming out of a single copy of $A$. Since $f$ is a cofibration, this is also a homotopy colimit. $X_m$s 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 $\Delta^\bullet$. A point of $\holim X$ is a sequence $(x_0, x_1, x_2, \ldots)$ where $x_m \colon \Delta^m \to X_m$. This has to be compatible with the action of $\Delta$. Compatibility with faces says that every vertex of $x_m$ is an image of $x_0$ in $X_m$, i.e., a copy of $x_0 \in B$ in one of $m+1$ copies of $B$ in $X_m$. Compatibility with degeneracies says that when I collapse all these copies back to $B$, $x_m$ becomes constant at $x_0$. This means that $x_m$ takes values in the set of copies of $x_0$ which is discrete since $X_m$ is Hasudorff. Hence $x_m$ itself is constant which in turn implies that $x_0 \in A$. Thus $\holim X = A$.

    Does that look correct?

11. • CommentRowNumber11.
    • CommentAuthorMike Shulman
    • CommentTimeMay 15th 2018
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    Author: Mike Shulman
    Format: MarkdownItexYes, I think that looks reasonable. Thanks! We do, I think, still need a name for the concept in *other* $(\infty,1)$-categories. In particular, can this result be generalized to any $(\infty,1)$-topos?

    Yes, I think that looks reasonable. Thanks!

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

12. • CommentRowNumber12.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 16th 2018
    • (edited May 16th 2018)
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItex@Karol This cannot be correct in general, as applying it to $X\to *$ says that $X$ is the limit of the cosimplicial diagram $X^\bullet$ where $X^n=\Sigma X\vee \dots\vee \Sigma X$ is a wedge of $n$ copies of $\Sigma X$. But it can happen that $\Sigma X$ is contractible and $X$ isn't. It probably works under some simply connected assumptions...

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

13. • CommentRowNumber13.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 16th 2018
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    Author: Marc Hoyois
    Format: MarkdownItexA wild guess is that epis and regular monos form a factorization system in ∞-groupoids. If so, regular mono = map whose fibers have hypoabelian $\pi_1$. (Note that $X\to *$ is epi iff $\Sigma X$ is contractible.)

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

14. • CommentRowNumber14.
    • CommentAuthorMike Shulman
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexAh, #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 $f:A\to B$ is epi, then its Cech conerve is constant at $B$. And it seems likely that epis are left orthogonal to regular monos in any $(\infty,1)$-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 $f:A\to B$, the map from $A$ to the totalization of the Cech conerve of $f$ is epi. What exactly are the epimorphisms in $\infty Gpd$?

    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 $f:A\to B$ is epi, then its Cech conerve is constant at $B$. And it seems likely that epis are left orthogonal to regular monos in any $(\infty,1)$-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 $f:A\to B$, the map from $A$ to the totalization of the Cech conerve of $f$ is epi.

    What exactly are the epimorphisms in $\infty Gpd$?

15. • CommentRowNumber15.
    • CommentAuthorKarol Szumiło
    • CommentTimeMay 16th 2018
    • (edited May 16th 2018)
    • PermaLink
    Author: Karol Szumiło
    Format: MarkdownItexI 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](https://arxiv.org/abs/1711.08898) about epimorphisms which even discusses the factorization system, but doesn't say that much about the other class.

    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.

16. • CommentRowNumber16.
    • CommentAuthorKarol Szumiło
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Karol Szumiło
    Format: MarkdownItexI 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 $\Delta^\bullet$ does not satisfy this condition. (This [argument](https://mathoverflow.net/a/127187) implies that the corresponding simplicial weight is not projectively cofibrant, so the topological one shouldn't be either.) On the other hand, $\Delta^\bullet$ 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.

    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 $\Delta^\bullet$ 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, $\Delta^\bullet$ 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.

17. • CommentRowNumber17.
    • CommentAuthorMike Shulman
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexAh, 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 $f:X\to Y$ of connected (for simplicity) spaces, the plus construction $X^+_P$ at the maximal perfect subgroup of $ker(\pi_1(f))$ is equivalent to the (homotopy) totalization of the Cech conerve of $f$. Is that plausible?

    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 $f:X\to Y$ of connected (for simplicity) spaces, the plus construction $X^+_P$ at the maximal perfect subgroup of $ker(\pi_1(f))$ is equivalent to the (homotopy) totalization of the Cech conerve of $f$. Is that plausible?

18. • CommentRowNumber18.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 16th 2018
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    Author: Marc Hoyois
    Format: MarkdownItexIt seems plausible, but I don't know how to prove it. [This paper](https://www.sciencedirect.com/science/article/pii/S0166864196001034) shows that $X\to *$ is a regular mono when $X$ is $2$-connected, and Gijs' comment in the [homotopy theory chat](https://chat.stackexchange.com/transcript/message/44484906#44484906) suggests this is still true if $X$ is simply connected. For what it's worth I have a [note](http://www-bcf.usc.edu/~hoyois/papers/acyclic.pdf) where I discuss Quillen's plus construction in ∞-topoi (it doesn't work very well though).

    It seems plausible, but I don’t know how to prove it. This paper shows that $X\to *$ is a regular mono when $X$ is $2$-connected, and Gijs’ comment in the homotopy theory chat suggests this is still true if $X$ 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).

19. • CommentRowNumber19.
    • CommentAuthorMike Shulman
    • CommentTimeMay 16th 2018
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    Author: Mike Shulman
    Format: MarkdownItexGiven 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 $\pi_1$ preserving products?

    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 $\pi_1$ preserving products?

20. • CommentRowNumber20.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexYes, 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 $\pi_1$ preserving products? Indeed, $\pi_1$ preserving products rules out most ∞-topoi. It seems that in a general ∞-topos a perfect subgroup of $\pi_1$ 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).

    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 $\pi_1$ preserving products?

    Indeed, $\pi_1$ preserving products rules out most ∞-topoi. It seems that in a general ∞-topos a perfect subgroup of $\pi_1$ 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).

21. • CommentRowNumber21.
    • CommentAuthorMike Shulman
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexIt'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 $1Gpd$ 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 $n Gpd$ for $1\lt n \le \infty$?

    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 $1Gpd$ 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 $n Gpd$ for $1\lt n \le \infty$?

22. • CommentRowNumber22.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexWell, the map $*\sqcup * \to *$ is not an epimorphism of 1-groupoids, since its pushout is the groupoid $B\mathbb{Z}$. I think the epimorphisms of 1-groupoids are exactly the 0-connected maps, i.e., maps inducing iso on $\pi_0$ and epi on $\pi_1$. In general $(n-1)$-connected maps are epimorphisms in $n$-groupoids, but the converse fails for $n \geq 2$, since there exist acyclic 1-truncated spaces, so that there are epimorphisms $X\to *$ that are not 1-connected. I misread the paper I linked to, they actually prove that $X\to *$ is a regular mono for $X$ simply connected. But in fact Bousfield already proves this when $X$ 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.

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

    I misread the paper I linked to, they actually prove that $X\to *$ is a regular mono for $X$ simply connected. But in fact Bousfield already proves this when $X$ 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.

23. • CommentRowNumber23.
    • CommentAuthorMike Shulman
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItex> I think the epimorphisms of 1-groupoids are exactly the 0-connected maps, i.e., maps inducing iso on $\pi_0$ and epi on $\pi_1$. 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 $(n-1)$-connected maps are epimorphisms in $n$-groupoids, but the converse fails for $n \geq 2$, since there exist acyclic 1-truncated spaces, so that there are epimorphisms $X\to *$ that are not 1-connected. I see that there are maps $X\to *$ 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 $n$-groupoids for any $n\lt\infty$: the inclusion $n Gpd \hookrightarrow \infty Gpd$ need not reflect epimorphisms, since it doesn't preserve pushouts.

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

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

    I see that there are maps $X\to *$ 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 $n$-groupoids for any $n\lt\infty$: the inclusion $n Gpd \hookrightarrow \infty Gpd$ need not reflect epimorphisms, since it doesn’t preserve pushouts.

24. • CommentRowNumber24.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexAh yes, my bad. > I see that there are maps $X\to *$ 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 $n$-groupoids for any $n\lt\infty$: the inclusion $n Gpd \hookrightarrow \infty Gpd$ need not reflect epimorphisms, since it doesn't preserve pushouts. Fully faithful functors reflect epimorphisms, in fact any functor that reflects pushouts does.

    Ah yes, my bad.

    I see that there are maps $X\to *$ 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 $n$-groupoids for any $n\lt\infty$: the inclusion $n Gpd \hookrightarrow \infty Gpd$ need not reflect epimorphisms, since it doesn’t preserve pushouts.

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

25. • CommentRowNumber25.
    • CommentAuthorMike Shulman
    • CommentTimeMay 16th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexRight 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.

    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.

26. • CommentRowNumber26.
    • CommentAuthorUlrik
    • CommentTimeMay 17th 2018
    • (edited May 17th 2018)
    • PermaLink
    Author: Ulrik
    Format: MarkdownItexIt seems the easiest example is given by the classifying space of Higman's group; generated by $x_0$, $x_1$, $x_2$, $x_3$ subject to $x_{i+1} x_i x_{i+1}^{-1} = x_i^2$ for each $i$ with $i+1$ computed mod 4. See at [encyclopedia of math](https://www.encyclopediaofmath.org/index.php/Acyclic_group). No finite group is acyclic. (As seen on this [MO question](https://mathoverflow.net/questions/291786/acyclic-finite-groups?noredirect=1&lq=1).)

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

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

27. • CommentRowNumber27.
    • CommentAuthorKarol Szumiło
    • CommentTimeMay 17th 2018
    • PermaLink
    Author: Karol Szumiło
    Format: MarkdownItex[Here](https://www.encyclopediaofmath.org/index.php/Acyclic_group) are some references. It lists the simplest example, I've seen: $B G$ where $G = \langle x_0, x_1, x_2, x_3 \mid x_{i+1} x_i x^{-1}_{i+1} = x^2_i, i \in \mathbb{Z} / 4 \rangle$, but maybe there are some simpler ones if you follow the references.

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

28. • CommentRowNumber28.
    • CommentAuthorMike Shulman
    • CommentTimeMay 17th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexThanks. I am confused again however. If $G$ is an acyclic group, then $B G$ is an acyclic space, and so $B G \to \ast$ is supposed to be an epimorphism of $\infty$-groupoids. But $B G$ and $\ast$ are both 1-groupoids, and $1 Gpd \hookrightarrow \infty Gpd$ is fully faithful, so it seems this should imply that $B G \to \ast$ 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?

    Thanks. I am confused again however. If $G$ is an acyclic group, then $B G$ is an acyclic space, and so $B G \to \ast$ is supposed to be an epimorphism of $\infty$-groupoids. But $B G$ and $\ast$ are both 1-groupoids, and $1 Gpd \hookrightarrow \infty Gpd$ is fully faithful, so it seems this should imply that $B G \to \ast$ 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?

29. • CommentRowNumber29.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 18th 2018
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    Author: Marc Hoyois
    Format: MarkdownItex$BG\to *$ is 0-connected!

    $BG\to *$ is 0-connected!

30. • CommentRowNumber30.
    • CommentAuthorMike Shulman
    • CommentTimeMay 18th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexBuh, 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 $$ 1 Gpd \hookrightarrow 1 Gpd \hookrightarrow 2 Gpd \hookrightarrow 3 Gpd \hookrightarrow \cdots \hookrightarrow \infty Gpd$$ 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 $n Gpd$ the condition to be an epimorphism involves $0$ through $(n-1)$-cells, so that when we move to $(n+1)Gpd$ we add a condition on $n$-cells for something to remain an epimorphism. The naive idea I had was that the condition on $0$ through $(n-1)$-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 $G$ is an acyclic group and $H$ is a 2-groupoid, the map $2 Gpd(\ast,H) \to 2 Gpd(B G, H)$ 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.

    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

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

    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 $n Gpd$ the condition to be an epimorphism involves $0$ through $(n-1)$-cells, so that when we move to $(n+1)Gpd$ we add a condition on $n$-cells for something to remain an epimorphism. The naive idea I had was that the condition on $0$ through $(n-1)$-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 $G$ is an acyclic group and $H$ is a 2-groupoid, the map $2 Gpd(\ast,H) \to 2 Gpd(B G, H)$ 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.

31. • CommentRowNumber31.
    • CommentAuthorTim Campion
    • CommentTimeMay 18th 2018
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    Author: Tim Campion
    Format: MarkdownItexDon'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 $Spaces \to Spaces$ closed under subobjects? That is, if $X$ is a space, and $F \to {(-)}^X$ is an $\infty$-monomorphism of $\infty$-functors $Spaces \to Spaces$, then must $F$ be corepresentable? Equivalently, is every subfunctor of a corepresentable functor $(-)^X$ of the form "maps out of $X$ which kill $P$" for some perfect normal subgroupoid $P \subseteq \Pi_1(X)$?

    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 $Spaces \to Spaces$ closed under subobjects? That is, if $X$ is a space, and $F \to {(-)}^X$ is an $\infty$-monomorphism of $\infty$-functors $Spaces \to Spaces$, then must $F$ be corepresentable? Equivalently, is every subfunctor of a corepresentable functor $(-)^X$ of the form “maps out of $X$ which kill $P$” for some perfect normal subgroupoid $P \subseteq \Pi_1(X)$?

32. • CommentRowNumber32.
    • CommentAuthorMarc Hoyois
    • CommentTimeMay 18th 2018
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItex@Tim If $X$ has no perfect subgroups, wouldn't that be saying that every map out of $X$ factors through every other?

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

33. • CommentRowNumber33.
    • CommentAuthorTim Campion
    • CommentTimeMay 18th 2018
    • PermaLink
    Author: Tim Campion
    Format: MarkdownItex@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_.

    @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.

34. • CommentRowNumber34.
    • CommentAuthorMarc Hoyois
    • CommentTimeOct 15th 2018
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexIn his paper "On the Homology Spectral Sequence of a Cosimplicial Space" Bousfield actually shows that for $X$ connected the limit of the associated cosimplicial space is the $\mathbb{Z}$-completion $\mathbb{Z}_\infty X$. For $X=\mathbb{R}P^2$ it is known that $X\to \mathbb{Z}_\infty X$ is not an equivalence (not even a homology equivalence), so $\mathbb{R}P^2\to*$ is not a regular monomorphism. Since $\pi_1(\mathbb{R}P^2)$ is abelian, this shows that epis and regular monos do not form a factorization system.

    In his paper “On the Homology Spectral Sequence of a Cosimplicial Space” Bousfield actually shows that for $X$ connected the limit of the associated cosimplicial space is the $\mathbb{Z}$-completion $\mathbb{Z}_\infty X$. For $X=\mathbb{R}P^2$ it is known that $X\to \mathbb{Z}_\infty X$ is not an equivalence (not even a homology equivalence), so $\mathbb{R}P^2\to*$ is not a regular monomorphism. Since $\pi_1(\mathbb{R}P^2)$ is abelian, this shows that epis and regular monos do not form a factorization system.

35. • CommentRowNumber35.
    • CommentAuthorMike Shulman
    • CommentTimeOct 15th 2018
    • PermaLink
    Author: Mike Shulman
    Format: MarkdownItexInteresting! I think my only question is did we solve the problem of whether every regular mono (map that is the limit of *some* cosimplicial object) must be the limit of its cosimplicial conerve?

    Interesting! I think my only question is did we solve the problem of whether every regular mono (map that is the limit of some cosimplicial object) must be the limit of its cosimplicial conerve?

36. • CommentRowNumber36.
    • CommentAuthorMarc Hoyois
    • CommentTimeOct 16th 2018
    • PermaLink
    Author: Marc Hoyois
    Format: MarkdownItexI'm not sure. If we're looking at $X\to *$, being regular mono implies that $X$ is a retract of the limit of the Cech conerve which is $\mathbb{Z}_\infty X$. In particular $X$ is $H\mathbb{Z}$-local (i.e. local with respect to homology equivalences), but for all I know that $\mathbb{Z}$-completion could still be different.

    I’m not sure. If we’re looking at $X\to *$, being regular mono implies that $X$ is a retract of the limit of the Cech conerve which is $\mathbb{Z}_\infty X$. In particular $X$ is $H\mathbb{Z}$-local (i.e. local with respect to homology equivalences), but for all I know that $\mathbb{Z}$-completion could still be different.