 Solution

Writing the suspension of a topological space as two cones and applying the Mayer-Vietoris theorem gives the reduced homology of a suspensionThus $H_0(\Sigma X)\cong \mathbb{Z}$ for nonempty $X$, $H_1(\Sigma X)\cong {\tilde{H}}_0(X)$, and the same shift holds in every higher degree. For a space of finite CW type, the reduced Euler characteristic changes sign, soSince $\chi ({\mathbb{CP}}^2)=3$,But the Euler characteristic of a product satisfies $\chi (A\times A)=\chi (A{)}^2$, a nonnegative perfect square. Neither $3$ nor $-1$ is such a square, so no ${\Sigma }^j{\mathbb{CP}}^2$ is homotopy equivalent to $A\times A$.

A homeomorphism $f:{\mathbb{R}}^n\to {\mathbb{R}}^n$ is a proper map, so it extends over the one-point compactification to a homeomorphism $f^{+}:S^n\to S^n$ fixing infinity. Defineusing the degree of a continuous mapping. Since $f^{+}$ acts invertibly on $H_n(S^n;\mathbb{Z})\cong \mathbb{Z}$, its degree is $\pm 1$. Functoriality of induced homology maps givesFor $A\in \mathrm{GL}(n,\mathbb{R})$, path connectedness of each determinant-sign component reduces $A$ to the identity when $\det A>0$ and to one coordinate reflection when $\det A<0$. Hence

Suppose $h:Y\times Y\to {\mathbb{R}}^{2n+1}$ were a homeomorphism, and put $d=2n+1$. Then $Y^4\cong {\mathbb{R}}^d\times {\mathbb{R}}^d\cong {\mathbb{R}}^{2d}$. The square of the cyclic permutationinterchanges the two $Y^2$ factors. Under $h\times h$, it is conjugate to the linear factor swap $(u,v)\mapsto (v,u)$ on ${\mathbb{R}}^d\times {\mathbb{R}}^d$, whose determinant has sign $(-1{)}^{d^2}=-1$. Thus $\mathrm{deg}({\sigma }^2)=-1$. On the other hand, multiplicativity gives $\mathrm{deg}({\sigma }^2)=(\mathrm{deg}\sigma {)}^2=1$, a contradiction. Therefore no such $Y$ exists.