 Solution

The assertion is false. Take $m=3$ and $n=2$. For the constant attaching map, the cell attachment gives $S^2\vee S^4$. Its degree-two generator has square zero: restricting the square to either sphere gives zero, and restrictions identify its degree-four cohomology with that of the $S^4$ summand.

For the Hopf fibration $\eta :S^3\to S^2={\mathbb{CP}}^1$, the attachment instead gives ${\mathbb{CP}}^2$. One can verify the attaching map explicitly with the characteristic mapIts interior maps homeomorphically to the complement of ${\mathbb{CP}}^1$; on the boundary it sends $(z_1,z_2)$ to $[z_1:z_2:0]$, exactly the Hopf fibration. By part (a), the degree-two generator of $H^{\ast }({\mathbb{CP}}^2;\mathbb{Z})$ has nonzero square generating degree four. HenceThe additive groups agree, but multiplication distinguishes the attachments.

The dimension condition explains why this example is the relevant one. In general the only positive-degree additive generators lie in degrees $n$ and $m+1$. A potentially nonzero product can only be the square of the degree-$n$ generator, and only when $m+1=2n$. Its coefficient is the Hopf invariant of the attaching map. Here the constant map has invariant zero, whereas the complex Hopf fibration has invariant one.