when cup product is a zero homomorphism How to see that the cup products vanish on suspensions?
 A: The cup product is a gussied-up version of the map induced by the reduced diagonal map
$\bar \Delta$, 
which is the composite of the ordinary diagonal $\Delta: X\to X\times X$
and the quotient map $q: X\times X \to X\wedge X$; note that $q$ is the cofiber of the 
inclusion $i: X\vee X\to X\times X$.
If $X$ is a suspension, then the map $\Delta$ has a lift (up to homotopy) 
$\lambda: X\to X\vee X$  through the inclusion 
$i : X\vee X \to X\times X$ of the wedge, and hence $\bar \Delta \simeq q\circ \Delta \simeq
q\circ i \circ \lambda \simeq *$.
One way to see the lifting is to lift the adjoint; and this is easy because
$\Omega(X\vee X) \to \Omega (X\times X)\cong (\Omega X) \times (\Omega X)$
has a homotopy section given by $s: (\omega, \tau) \mapsto (i_1\circ \omega) * (i_2\circ \tau)$, where $i_1, i_2: X\to X\vee X$ are the inclusions of the summands and $*$ denotes concatenation of paths.
A: This is a special case of the fact that the cup-length is a lower bound for the Lusternik–Schnirelmann category. Using those two terms as keywords should get you the standard arguments.
A: 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.
