Are $S(S^n\times S^m)$ and $S(S^n \vee S^m \vee S^{n+m})$ homotopy equivalent?

Question

For $n,m\geq 1$ the spaces $X=S^n\times S^m$ and $Y=S^n \vee S^m \vee S^{n+m}$ have isomorphic homology and cohomology groups. The cohomology groups are given by $H^i(X)=H^i(Y)=\mathbb{Z}$ for $i=0,n,m,n+m$ and zero else. The spaces $X$ and $Y$ are not homotopy equivalent however, since their cohomology rings are not isomorphic. For $X$, the cup product of a generator of the $n$-th cohomology with a generator of the $m$-th cohomology yields a generator of the $(m+n)$-th cohomology. The space $Y$ has trivial cup products in positive degree.

Using the suspension isomorphism we find that the unreduced suspensions $S(X)$ and $S(Y)$ have isomorphic cohomology and homology. The cup product on a suspension is trivial, so the cohomology rings of $S(X)$ and $S(Y)$ are isomorphic too. In other words, (co)homology does not seem to be able to distinguish these spaces. Are $S(X)$ and $S(Y)$ homotopy equivalent?

Answer 1

Answer from the comments. Thanks to @Charlie for helping me out.

Recall the following facts (for topological spaces $X,Y,Z$):

• $\Sigma X \simeq S^1 \land X,$
• $(X \vee Y)\wedge Z \simeq (X \wedge Z) \vee (Y \wedge Z)$ as proved here,
• $\Sigma (X\times Y) \simeq \Sigma X \lor \Sigma Y \lor \Sigma (X\land Y)$ as proved here,
• If $X$ is a CW-complex, then $\Sigma X\simeq SX$ as proved here,
• $S^n \land S^m \simeq S^{n+m}.$

Therefore

$$S(S^n\times S^m)\simeq \Sigma(S^n\times S^m)\simeq \Sigma S^n \vee \Sigma S^m \vee \Sigma (S^n \land S^m)\simeq (S^1 \land S^n) \vee (S^1 \land S^m) \vee (S^1 \land S^{n+m}) \simeq S^1\land (S^n \vee S^m \vee S^{n+m})\simeq \Sigma (S^n \vee S^m \vee S^{n+m})\simeq S(S^n \vee S^m \vee S^{n+m}).$$