How to see that the cup products vanish on suspensions?



 when cup product is a zero homomorphism
13.66 in Switzer's Algebraic Topology: Homotopy and Homology. The idea is to use the fact that $\Sigma X$ decomposes into two copies of $CX$, say $A$ and $B$, glued along the common boundary of $X$. For any two cohomology classes $x$ and $y$ in $\tilde{E}^\* \Sigma X$, you can uniquely pull $x$ back to a class $x'$ on the relative pair $(\Sigma X, A)$ and $y$ back to a class $y'$ on $(\Sigma X, B)$. Cupping is natural w.r.t the two relative inclusions $i\_A: (\Sigma X, \{x\_0\}) \to (\Sigma X, A)$ and $i\_B: (\Sigma X, \{x\_0\}) \to (\Sigma X, B)$, and so you get the calculation $x \smile y = i\_A^\*(x') \smile i\_B^\*(y') = i^\*(x' \smile y')$, where $i: (\Sigma X, \{x\_0\}) \to (\Sigma X, \Sigma X)$ is another relative inclusion and $x' \smile y'$ a class on the pair $(\Sigma X, \Sigma X)$ --- but that guy has trivial reduced cohomology.


