cup product in cohomology ring of a suspension

Question

Let $X$ be a CW-complex. Let $\Sigma$ be suspension. Let $R$ be a commutative ring. Is the cup product of $$ H^*(\Sigma X;R)$$ trivial? How to prove? Where can I find the result?

Answer 1

Yes. Actually there is a more general result. If $X=\bigcup_{i=1}^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $\Sigma X$ is a union of two cones, each of which is contractible, the result follows.

To establish the general result, simply note that $H^*(X)\cong H^*(X, A_i)$ as $A_i$ is contractible, and there is a cup product map \begin{eqnarray}H^*(X, A_1)\times \cdots \times H^*(X, A_n)\to H^*(X, \bigcup_{i=1}^n A_i)=H^*(X, X)=0\end{eqnarray}