Lack of homotopy equivalence between given spaces

Question

I am trying to show that spaces $\mathbb{S}^3\times \mathbb{CP}^\infty$ and $\mathbb{S}^2$ are not homotopy equivalent. (My goal is to use then as examples of non-homotopy equivalent spaces of the same homotopy groups). I still do not clearly see simple argument why it is the case. Maybe it is obvious, but after some time I need at least a hint.

Answer 1

Here is an argument using only homotopy theory, with no homological considerations. In fact it uses nothing more than homotopy groups. It ends up being far more complicated than a homological argument, however, so I would suggest broadening your critera for accepting such an argument as an answer.

Assume there is a homotopy equivalence $\Phi:S^3\times\mathbb{C}P^\infty\xrightarrow\simeq S^2$. Then for each $n$ we have a map $\mathbb{C}P^n\hookrightarrow \mathbb{C}P^\infty\hookrightarrow S^3\times\mathbb{C}P^\infty\xrightarrow\simeq S^2$ which induces an isomorphism on $\pi_2$ and the trivial map on all other homotopy groups. In particular we have a map

$$\phi:S^2\cong\mathbb{C}P^1\rightarrow S^2$$

which induces an isomorphism on $\pi_2$ and the trivial map on all other homotopy groups. But this is absurd, since $\pi_2S^2\cong\mathbb{Z}$ is generated by the identity map, so we must have $\phi=\pm id_{S^2}$, and this leads to a contradiction. For instance, if $\phi=+id_{S^2}$ then by functorality it induces the identity map on all homotopy groups. Therefore the only possibility is that $\phi=-id_{S^2}\in\pi_2S^2$. But $-id_{S^2}=A$ is the antipodal map, from which it follows that $A\circ\phi$ is a map which induces an isomorphism on $\pi_2$ and the trivial map on all other homotopy groups. Since then $A\circ \Phi=A\circ A=id_{S^2}$ we are back in the last situation.

Hence $\phi$ cannot exist, and this prohibits the existence of the homotopy equivalence $\Phi$.