Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a pointed space X, let $J(X)$ denote its James construction. There is a natural inclusion $X\rightarrow J(X)$ which can equivalently be described as the unit map $\eta_X:X\rightarrow \Omega \Sigma X$. The homotopy fiber of the counit map $\epsilon_X:\Sigma \Omega X\rightarrow X$ is known to be $\Omega X * \Omega X$. Is there an analogous result for the homotopy fiber of $\eta_X$ in terms of $X$?

share|improve this question
I have the feeling that an analogous result would involve the homotopy cofiber of $\eta$, not the homotopy fiber. By the way, is $\ast$ the wedge product? –  Konrad Voelkel Apr 17 '13 at 17:09
* is the join, ie the suspension of the smash product. –  Fernando Muro Apr 17 '13 at 18:19
I believe for $X=S^{n}$ then one can get some answers p-locally by looping the EHP sequence. In general I am not sure what can be said. –  Callan McGill Apr 17 '13 at 20:38
There is not even a known identification of the homotopy cofiber $C$ of $e_X$ in general. There are some partial results: its suspension $\Sigma C$ is weak equivalent to $\Sigma (X\wedge \Omega \Sigma X)$. In the metastable range (approximately 3 times the connectivity of $X$), the cofiber coincides with the "co-join," i.e., the holim of the diagram $\Sigma X \to \Sigma X \vee \Sigma X \leftarrow \Sigma$ given by the two inclusions. –  John Klein Apr 17 '13 at 22:57
Thanks for your comments. It seems there is not a known answer for this question, as John implicitly claimed above. –  MatanP Apr 19 '13 at 13:53

1 Answer 1

This might not be a satisfactory answer, but let me point out that S.-c. Wong constructed a "little cube" model for the homotopy fiber of $\Omega^{k-1}E^n : \Omega^{k-1}\Sigma^{k}X \to \Omega^{k-1+n}\Sigma^{k+n} X$ in this paper in 1994.

Note that May's little cube model for $\Omega\Sigma X$ is essentially the James construction.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.