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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?

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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}

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The answer depends on which homology theory you are using. The statement fails for singular cohomology. However there is a fairly easy way to show that the cup product is trivial on reduced cohomology $\tilde{H}{\,}^\bullet(\Sigma X)=H^\bullet(\Sigma X,\text{pt})$.

Write $\Sigma X=\text{Cone}_+(X)\cup\text{Cone}_-(X)$ and let $\iota:\text{pt}\longrightarrow\text{Cone}(X)$ be the inclusion map. We get the following commutative diagram $\require{AMScd}$ \begin{CD} H^p(\Sigma X,\text{pt})\otimes H^q(\Sigma X,\text{pt}) @>{\smile}>> H^{p+q}(\Sigma X,\text{pt})\\ @A{H^p(\iota)\otimes H^q(\iota)}AA @AA{H^{p+q}(\iota)}A\\ H^p(\Sigma X,\text{Cone}_+(X))\otimes H^q(\Sigma X,\text{Cone}_-(X)) @>>{\smile}> H^{p+q}(\Sigma X,\Sigma X) \end{CD}

Now note that $H^\bullet(\iota):H^\bullet(\Sigma X,\text{Cone}(X))\longrightarrow H^\bullet(\Sigma X,\text{pt})$ is an isomorphism because $\text{Cone}(X)$ is contractible. Therefore we have found a factorisation of $a\smile b=H^{p+q}(\iota)\left(H^p(\iota)^{-1}(a)\smile H^q(\iota)^{-1}(b)\right)$ over $H^{p+q}(\Sigma X,\Sigma X)\cong0$, hence $a\smile b=0$ for any $a\otimes b\in H^p(\Sigma X,\text{pt})\otimes H^q(\Sigma X,\text{pt})$.

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