# Example of a cogroup in $\mathsf{hTop}_{\bullet}$ which is not a suspension

Let $\mathsf{hTop}_{\bullet}$ denote the homotopy category of pointed topological spaces. More precisely, the objects are pointed topological spaces and for two objects $X$ and $Y$, the morphisms from $X$ to $Y$ are the basepoint-preserving homotopy classes of maps from $X$ to $Y$ which are denoted by $[X, Y]_{\bullet}$.

A cogroup object in $\mathsf{hTop}_{\bullet}$ is an object $X$ such that for every object $Y$, $[X, Y]_{\bullet}$ is a group which is natural in $Y$. The positive dimensional spheres are examples of cogroup objects in $\mathsf{hTop}_{\bullet}$. More generally, the (reduced) suspension of any pointed topological space is a cogroup object in $\mathsf{hTop}_{\bullet}$.

According to nLab (see point 2), there are cogroup objects in $\mathsf{hTop}_{\bullet}$ (the page I link to writes $\mathsf{hTop}$, but I believe this is a typo) which are not suspensions, but no reference is given.

What is an example of a cogroup object in $\mathsf{hTop}_{\bullet}$ which is not a suspension?

If possible, I'd like to see a construction of such an object, but I'd also be satisfied with a reference.

• Isn't the wedge of two cogroups also a cogroup? – user98602 Nov 11 '15 at 22:48
• You want the additional condition that the group structure is natural in $Y$. – Qiaochu Yuan Nov 11 '15 at 23:16
• @MikeMiller: Is that true? It seems believable. Can the smash product of two suspensions always be written as a suspension? If not, this would provide an example. – Michael Albanese Nov 12 '15 at 1:27
• It's true but it's not helpful, I was being silly... $\Sigma(X \vee Y)$ is homotopy equivalent to $\Sigma X \vee \Sigma Y$ (collapse the suspension of the wedge point). – user98602 Nov 12 '15 at 1:30
• @QiaochuYuan: Thanks. I have tried to edit the question accordingly. Let me know if the statement is still incorrect. – Michael Albanese Nov 12 '15 at 1:54

## 1 Answer

I don't know much about this, but Berstein and Harper construct some such examples in this paper (which I haven't actually read). In particular, they construct an infinite family of 3-cell complexes which are cogroups but not suspensions, and also a 2-cell complex with one 5-cell and one 35-cell which is a cogroup but not a suspension. There is also a survey of this and related topics in Chapter 23 of the Handbook of Algebraic Topology (some of which you can read on Google books).

You may also be interested in the other answers I received when I asked a similar question many years ago on MathOverflow, though none of the answers say anything directly about how to construct an example.

• Thanks. I saw your question and the article by Arkowitz. The section that Loop Space seems to be referring to is Example 3.4 of Arkowitz's article. This is a co-H-space that is not a suspension, but if I'm not mistaken, this co-H-space is not a cogroup (the remark following the example states that the space does not admit an associative comultiplication). – Michael Albanese Nov 12 '15 at 1:10
• I can't access Example 3.4 (or any of page 1153) of Arkowitz's article on Google books at the moment, so I'll just take your word for it. However, Proposition 5.5 on page 1158 is definitely talking about cogroups which are not suspensions (namely, the ones constructed by Berstein and Harper). – Eric Wofsey Nov 12 '15 at 1:13
• I see, I didn't look that far into the article. I'll have to take a look at that paper. Thanks again. – Michael Albanese Nov 12 '15 at 1:25