 Solution

For a continuous map $f:S^2\times S^4\to {\mathbb{CP}}^3$, use the classes from part (a). Since $H^2(S^2\times S^4;\mathbb{Z})=\mathbb{Z}u$, there is an integer $k$ with $f^{\ast }c=ku$. Naturality of the cup product and $u^2=0$ yieldWith the product orientation on the source and the complex orientation on the target, the mapping degree $d$ is defined by $f_{\ast }[S^2\times S^4]=d[{\mathbb{CP}}^3]$. Evaluation on the fundamental class givesThus the induced group homomorphism on $H_6$ is zero. This is a cup-power obstruction to nonzero degree, rather than a consequence of the additive homology groups alone.