spaces with isomorphic homotopy groups, though not homotopy equivalent Is there a way to prove that $S^2$ isn't homotopy equivalent to $S^3 \times \mathbb{CP^\infty}$ without homology theory? The tricky part is that all homotopy groups are isomorphic and I don't know many other homotopy invariants.
 A: They can be distinguished by their Whitehead brackets. On any pointed space, there is a natural bilinear map
$$\pi_2(X) \times \pi_2(X) \to \pi_3(X)$$
and I claim that this map is nonzero on $S^2$ but zero on $S^3 \times \mathbb{CP}^{\infty}$. Roughly speaking, the point is that you've given the latter space nontrivial $\pi_2$ by just taking the direct product with $\mathbb{CP}^{\infty}$, and so this $\pi_2$ doesn't "talk to" the $\pi_3$ that you already had. The machinery of $k$-invariants is another way of making this precise, but this requires cohomology.
Proving this is harder than just computing homology or cohomology, though.

Edit: Actually, here's an easier argument that also makes precise the intuition that $\pi_2$ of the second space doesn't "talk to" any other part of it, but that doesn't use any additional machinery. 
Suppose $f : S^2 \to S^3 \times \mathbb{CP}^{\infty}$ is a homotopy equivalence. Then it must induce an isomorphism on $\pi_2$, and hence $f$ must in fact be a representative of a generator of $\pi_2(S^3 \times \mathbb{CP}^{\infty})$. But such a generator factors through a map
$$\mathbb{CP}^{\infty} \to S^3 \times \mathbb{CP}^{\infty}$$
given by picking a point in $S^3$, and so $f$ must also factor (up to homotopy) as a composite
$$S^2 \to \mathbb{CP}^{\infty} \to S^3 \times \mathbb{CP}^{\infty}.$$
But such a map cannot induce an isomorphism on $\pi_3$, and hence can't be a homotopy equivalence; contradiction. This seems like a good example for building some intuition about why the hypothesis of the Whitehead theorem, that a map induces an isomorphism on all homotopy groups, is strictly stronger than the hypothesis that all homotopy groups are isomorphic. 
