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Any topological group $G$ has a classifying space, whose loopspace is a (homotopy) group which is homotopy equivalent to $G$ in a way that preserves the group structure. More generally, if $G$ is an $A_\infty$-group (a space with a binary operation which satisfies the group axioms up to coherent homotopy), it similarly can be delooped to a classifying space.

Now suppose you have a cogroup in the category of pointed spaces. If it is actually literally a cogroup, it's not hard to show it must be a point. However, up to (coherent) homotopy, the suspension of any space is a cogroup. Is the converse true? Can you desuspend any $A_\infty$-cogroup? Are there any examples of homotopy cogroups (possible not $A_\infty$) which are not suspensions? More generally, are there any criteria that you can use to prove that a space does not have the homotopy type of a suspension? The only one I know is that all cup products must vanish, but this also holds automatically for a homotopy cogroup (indeed, for any "co-H-space").

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    $\begingroup$ Well, there are obvious obstructions to desuspending in low dimensions (π_0 must be trivial and π_1 a free group). $\endgroup$
    – Reid Barton
    Commented Nov 4, 2009 at 17:44
  • $\begingroup$ True, but these seem to be exactly saying that \pi_0 and \pi_1 are themselves cogroups in the category of pointed spaces or groups, though I don't see how to prove that for the case of groups. $\endgroup$
    – Eric Wofsey
    Commented Nov 4, 2009 at 18:37
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    $\begingroup$ Fun fact: For any group G, the equalizer of the two maps G * G - > G (given by projecting to each factor) is a free group, generated by the elements g_1 g_2, where g is a nonidentity element of G and g_1 and g_2 are its two images in G * G. So a cogroup object is automatically a subgroup of a free group, hence free. $\endgroup$
    – Tyler Lawson
    Commented Nov 4, 2009 at 20:47
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    $\begingroup$ Related questions: Do we know anything about the relationship between A_infty cogroups and the suspension-loop cotriple C? If you have a coalgebra over C, does its cobar construction have suspension equivalent to the original space? $\endgroup$
    – Tyler Lawson
    Commented Nov 4, 2009 at 21:17
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    $\begingroup$ There is a "realization" which is the space of maps from the standard cosimplicial space Delta, whose n'th space is the n-simplex. It's the inverse limit of a tower of fibrations, and passing a suspension across said limit is going to be nontrivial if possible. The analogue (the loop space of a simplicial space can be done levelwise if the spaces are all connected) is a difficult point in May's iterated loop space work. $\endgroup$
    – Tyler Lawson
    Commented Nov 5, 2009 at 0:36

4 Answers 4

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Whilst the n-lab page on co-H-spaces could be described as a little meagre, it does nonetheless contain a reference to a paper in the Handbook of Algebraic Topology. Various parts of this book are in the "preview" in google books, in particular page 1153 which contains the magical sentence:

We can now construct cell complexes $S^n \cup_\alpha e^m$ which are co-H-spaces but not suspensions.

This follows a theorem which classifies when such spaces are co-H-spaces and suspensions.

The specific article is the aptly named "Co-H-spaces" by Martin Arkowitz, and is chapter 23 of Handbook of Algebraic Topology, edited by Ioan James. I would recommend this as a first place to look to find out more about these spaces.

Incidentally, I don't believe your remark about delooping. I'm almost certain that there are H-spaces that can't be delooped so either there are H-spaces that aren't $A_\infty$-groups or your claim is incorrect. Unfortunately, I'm not currently surrounded by my usual stock of algebraic topology textbooks so can't look this up (and you can do an internet search as well as I can). Can you supply a reference for the delooping claim?

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    $\begingroup$ An example of an H-space that is not A_\infty is S^7. It can't be delooped because its delooping would have cohomology a polynomial ring on a generator in degree 8, and this is impossible by mod p Steenrod operations for any odd p. A reference for delooping an A_\infty group is May's Geometry of Iterated Loopspaces $\endgroup$
    – Eric Wofsey
    Commented Nov 4, 2009 at 18:43
  • $\begingroup$ Thanks! So the H vs A_\infty is more subtle than I thought. I'll add that to my list of things to look through on a snowy day. $\endgroup$
    – Andrew Stacey
    Commented Nov 4, 2009 at 19:27
  • $\begingroup$ all spaces are $A_1$; H means $A_2$; $A_n$ is the playground Stasheff built, and where associahedra are from; he constructs a family of $H$ spaces that are $A_p$ but not $A_{p+1}$ for many (I think prime?) values of $p$ (or $p+1$, perhaps... this paper is sitting in my computer, why don't I look it up?!) Anyways, Stasheff, Transactions AMS Vol 108 No.2 pp 275-292 $\endgroup$
    – some guy on the street
    Commented Apr 1, 2013 at 1:05
  • $\begingroup$ @LoopSpace: Sorry if this is a silly question (this is not my area). The example given in Arkowitz's paper is a co-H-space which is not a suspension. However, it is not a cogroup (it does not admit an associative comultiplication). Wasn't the OP looking for an example of a cogroup (as Arkowitz defines it) which is not a suspension? I apologise if I have misunderstood something. $\endgroup$
    – Michael Albanese
    Commented Nov 11, 2015 at 1:28
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This is a question I've asked myself, but never found an answer. By analogy with the story for A-infinity spaces and topological groups, you would like to be able to start with a co-A-infinity space Y, build a "cobar complex" on it, then take the homotopy limit of that complex to obtain a space X, and then hope that ΣX=Y.

If this has any chance, it had better work when Y is already known to be a suspension. In the paper by Goerss, "Barratt's desuspension spectral sequence and the Lie ring analyzer", Quart. J. Math Oxford, 44 (1993), something like this is proved. He builds (actually, he describes how Barratt builds) a cosimplicial space whose n-th space is homotopy equivalent to an n-fold wedge of ΣX, and shows for simply connected X that the Tot of this complex recovers X.

Given this, I suspect it should be possible to prove that "co-A-infinity"="suspension", at least for 2-connected spaces.

Edit: some of these ideas get discussed in the commentary under the question; since they were hidden, I didn't notice them till now.

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  • $\begingroup$ Presumably in general there should be a map of co-A-infinity spaces ΣX -> Y which is an equivalence when Y is "cogrouplike". Any ideas as to what "cogrouplike" might be? From the remark above it seems that 2-connected implies cogrouplike, so it ought to be a condition on the 2-skeleton. $\endgroup$
    – Oscar Randal-Williams
    Commented Nov 5, 2009 at 10:27
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    $\begingroup$ Oh, I hadn't even though about the "cogrouplike" condition. I guess cogrouplike should mean (i,c): Y v Y --> Y v Y is a weak equivalence, where on the first factor this is inclusion into one of the summands, and on the other one is the comultiplication (this is just dual to the "grouplike" condition on an H-space). If Y is simply connected, we just have to check that the map induces iso in homology; but this is necessarily so, since co-A-infinity structure makes H_*Y into a comonoid in abelian groups, which automatically has to be a cogroup. $\endgroup$
    – Charles Rezk
    Commented Nov 5, 2009 at 16:44
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Mike Hopkins tells me it is indeed true that for any A∞ cogroup space Y, the homotopy limit X of the associated cobar complex is a desuspension of Y (Y = ΣX as A∞ cogroups)--we don't even need any connectivity assumptions on Y.

Edit: I think the way to prove this is to look at the spectral sequence for the homology of X. It's easy to see that algebraically it degenerates on the E2 page to something which could be the homology of a desuspension of Y. I don't know how to check that this algebraic convergence has anything to do with X, or how to handle π0 and π1 issues. However, in the case that Y is the suspension of a discrete pointed set, one can check by hand that X works out to be the original pointed set, which makes it plausible that these low-dimensional issues don't cause problems.

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    $\begingroup$ Cool. Any hint on how it's proved? Goerss only claimed to get it up to Z-completion issues. $\endgroup$
    – Charles Rezk
    Commented Nov 11, 2009 at 8:12
  • $\begingroup$ I think Goerss was doing something slightly different, namely trying to recover X from its suspension with its A∞ cogroup structure. That requires some conditions on X, and in general I think Mike said the result is something like the Z-nilpotent completion of X. $\endgroup$
    – Reid Barton
    Commented Nov 11, 2009 at 17:32
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    $\begingroup$ Oh. So part of the point is that desuspension isn't unique, even when the A-inifinity cogroup structure is prescribed. (After all, the point can be desuspended non-trivially.) $\endgroup$
    – Charles Rezk
    Commented Nov 11, 2009 at 17:58
  • $\begingroup$ Yes (just as the delooping of a topological group isn't unique; we could throw in non-basepoint components). $\endgroup$
    – Reid Barton
    Commented Nov 11, 2009 at 18:14
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Hopkins' result alluded to above gives a coordinate-free approach a lá Segal. The paper I wrote with Schwänzl and Vogt:

Comultiplication and suspension. Topology Appl. 77 (1997), no. 1, 1–18,

gives an actual co-operadic approach: we show that a 2-connected space which has a homotopy everything co-action by the Stasheff operad (i.e., a co-$A_\infty$-structure) is always a suspension.

This paper answers the first of the questions posed above.

There is an open question about this matter which I'd like to bring up. Hopkins claims his result holds for 1-connected spaces, but I and my co-authors could only get our result to work for 2-connected ones. Hopkins' proof, which was based on the Bousfield spectral sequence, never appeared (our proof involved the higher Blakers-Massey theorems).

Hopkin's result is stated, but not proved, in: Formulations of cocategory and the iterated suspension. Algebraic homotopy and local algebra (Luminy, 1982), 212-226, Astérisque, 113-114, Soc. Math. France, Paris, 1984. Thus we don't know if the 1-connected case has really been settled or not.

Additional Note: Schwänzl, Vogt and I also show that if the space is highly connected with respect to its CW dimension (i.e., $(n-1)$-metastable), then the existence of a co-$A_n$-structure on it is tantamount to the existence a co-$A_\infty$-structure. It follows that it too desuspends.

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