= Solution
Writing the <suspension of a topological space> as two cones and applying the <Mayer-Vietoris theorem> gives the <reduced homology of a suspension>
$$
\widetilde H_i(\Sigma X;\mathbb Z)
\cong\widetilde H_{i-1}(X;\mathbb Z).
$$
Thus $H_0(\Sigma X)\cong\mathbb Z$ for nonempty $X$, $H_1(\Sigma X)\cong\widetilde H_0(X)$, and the same shift holds in every higher degree. For a space of finite CW type, the reduced <Euler characteristic> changes sign, so
$$
\chi(\Sigma X)=2-\chi(X),
\qquad
\chi(\Sigma^jX)=1+(-1)^j(\chi(X)-1).
$$
Since $\chi(\mathbb{CP}^2)=3$,
$$
\chi(\Sigma^j\mathbb{CP}^2)=1+2(-1)^j\in\{3,-1\}.
$$
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. Define
$$
\deg f=\deg f^+,
$$
using the <degree of a continuous mapping>. Since $f^+$ acts invertibly on $H_n(S^n;\mathbb Z)\cong\mathbb Z$, its degree is $\pm1$. Functoriality of induced homology maps gives
$$
\deg(f\circ g)=\deg f\deg g.
$$
For $A\in\operatorname{GL}(n,\mathbb R)$, 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
$$
\deg A=\operatorname{sign}(\det A).
$$
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 permutation
$$
\sigma(y_1,y_2,y_3,y_4)=(y_4,y_1,y_2,y_3)
$$
interchanges 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 $\deg(\sigma^2)=-1$. On the other hand, multiplicativity gives $\deg(\sigma^2)=(\deg\sigma)^2=1$, a contradiction. Therefore no such $Y$ exists.
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