# Is there a counterexample to this stronger version of James homotopy splitting?

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.

• Note: If the inclusion $X \to \Omega \Sigma X$ admits a retraction, then $X$ is an $H$-space. Dec 17 '11 at 23:13

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.