Are maps inducing the same cohomology homomorphisms homotopic?

Solution 1:

This is not true; for instance, every map $S^3 \rightarrow S^2$ induces the trivial map on cohomology. However, you can detect nontriviality by taking the (homotopy) cofiber of the map, i.e. attach a 4-disk to $S^2$ along the image of $S^3$. For the trivial map this gives you $S^2 \vee S^4$, whereas for instance the Hopf fibration will give you $\mathbb{C}P^2$. To be totally precise, you can tell that this is distinct from $S^2 \vee S^4$ by checking that the self-cup product of the 2-dimensional generator $\alpha$ is nontrivial -- in fact, it's a 4-dimensional generator $\beta$. In general you get that $\alpha \smile \alpha = n\beta$ for some $n \in \mathbb{Z}$. This $n$ is called the Hopf invariant of the map; the Hopf invariant can be defined for any map $S^{2k-1} \rightarrow S^k$, and in fact defines a homomorphism $\pi_{2k-1}(S^k) \rightarrow \mathbb{Z}$. It's rather easy to show that this always hits the even integers when $k$ is even and is trivial when $k$ is odd. With a little more work, you can show that if this is surjective, it must be that $k=2^t$. But in fact, it's surjective precisely when $k \in \{1 , 2, 4, 8\}$, and moreover this is actually equivalent to the statement that the only real division algebras are the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, the quaternions $\mathbb{H}$, and the octonions $\mathbb{O}$!

If you think this is as cool as I do, you should check out Mosher & Tangora's excellent (and incredibly inexpensive) book Cohomology Operations and Applications to Homotopy Theory.