Let $X = \Sigma Y = Y \wedge S^1$, cup product $\tilde{H}^p(X) \otimes \tilde{H}^q(X) \to \tilde{H}^{p+q}(X)$ is the zero homomorphism?

Question

We take cohomology with coefficients in a commutative ring $R$ and we write $\otimes$ for $\otimes_R$.

Let $X = \Sigma Y = Y \wedge S^1$. How do I see that the cup product$$\tilde{H}^p(X) \otimes \tilde{H}^q(X) \to \tilde{H}^{p+q}(X)$$is the zero homomorphism?

Answer 1

Thinking about suspensions geometrically, it is pretty clear we can write $X$ as the union of two contractible open sets $U$ and $V$. If $x$, $y$ are cohomology classes in $\tilde{E}^*(X)$, then we can pull them back uniquely to relative classes $x'$, $y'$ in $\tilde{E}^*(X, U)$ and $\tilde{E}^*(X, V)$ by the long exact sequence of the pairs. Let$$i_U(X, x_0) \to (X, U),\text{ }i_V(X, x_0) \to (X, V),\text{ }i: (X, x_0) \to (X, U \cup V) = (X, X)$$be the inclusions, where $x_0 \in X$ is the basepoint. The relative cup product is natural, so we have$$x \cup y = i_U^*\left(x'\right) \cup i_V^*\left(y'\right) = i^*\left(x' \cup y'\right) = i^*(0) = 0,$$as the pair $(X, X)$ has trivial reduced cohomology.