James proved the homotopy decomposition $\Sigma\Omega\Sigma X\simeq \bigvee_{n=1}^\infty \Sigma X^{\wedge n}$. This is a natural homotopy equivalence for a pointed connected CW complex $X$. Here $X^{\wedge n}$ is the $n$-fold self-smash product of $X$.

Is there a counterexample to the stronger assertion that $\Omega\Sigma X\simeq \bigvee_{n=1}^\infty X^{\wedge n}$? This assertion implies James' decomposition as suspension, being left adjoint to looping, commutes with wedge sum.

  • $\begingroup$ Note: If the inclusion $X \to \Omega \Sigma X$ admits a retraction, then $X$ is an $H$-space. $\endgroup$
    – John Klein
    Dec 17, 2011 at 23:13

2 Answers 2


The space $X = S^2$ gives a counterexample. In this case, we're comparing $\Omega S^3$ with $\vee S^{2n}$. These can be distinguished by their cohomology rings. The cohomology ring of the latter has all products in positive degrees equal to zero.

The Pontrjagin ring structure on the homology $H_* \Omega S^3$ makes it isomorphic to a polynomial ring $\mathbb{Z}[x]$ on a generator $x$ in degree 2, coming from $H_* S^2$. There are no other generators in lower degree, so the diagonal map must send $x$ to $x \otimes 1 + 1 \otimes x$ in $H_*(\Omega S^3) \otimes H_*(\Omega S^3)$. Since this map respects the ring structure, we have $\Delta x^n = (x \otimes 1 + 1 \otimes x)^n$ for all $n \geq 0$.

As a consequence, taking duals tells us that the cohomology ring of $\Omega S^3$ is a divided power algebra $\mathbb{Z}\left[\frac{x^k}{k!}\right]$, and definitely does not have all products in positive degrees equal to zero.


Yes: $\Omega S^k$ has non-trivial cup products.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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