 Solution

The standard CW complex structure on Complex projective space ${\mathbb{CP}}^3$ has one cell in each dimension $0,2,4,6$ and no other cells. Its cellular chain complex therefore has zero differentials: each boundary has an odd-dimensional target group, which is zero. The product of the usual two-cell structures on $S^2$ and $S^4$ likewise has exactly one cell in dimensions $0,2,4,6$. Thus cellular homology gives, for both spaces,The cohomology rings detect more than these additive groups. Let $c\in H^2({\mathbb{CP}}^3;\mathbb{Z})$ be the hyperplane class. The cohomology ring of complex projective space isFor the normalization, a hyperplane is Poincare dual to $c$; intersecting three generic hyperplanes gives one point with positive complex orientation, so $⟨c^3,[{\mathbb{CP}}^3]⟩=1$. Together with the computed additive groups this makes $c,c^2,c^3$ generators in their respective degrees.

Let $u$ and $v$ be the pullbacks of the orientation classes of $S^2$ and $S^4$. The Künneth theorem and naturality of the cup product givewhere $uv$ generates degree six. In particular, every degree-two class on the product has square zero, whereas $c^2\ne 0$. Any graded ring isomorphism would send $c$ to $\pm u$, contradicting this product relation. A homotopy equivalence induces an isomorphism of cohomology rings, so the spaces have isomorphic integral homology but are not homotopy equivalent.