I'm having trouble doing exercise 3.2.11 at page 231 of Hatcher. It states:
Using cup products, show that every map $S^{k+l}\to S^k\times S^l$ induces the trivial homomorphism $H_{k+l}(S^{k+l})\to H_{k+l}(S^k\times S^l)$.
I don't know how to use the cup product in order to do this kind of computations. Any help would be appreciated
Recall that $H^*(S^n)$ is an exterior algebra $\Lambda(x,n)$ with $x$ in degree $n$. If you pick a continuous map $f : S^n \to S^p \times S^q$ with $p+q=n$, there is induced a map in cohomology $$f^* : \Lambda(x',p)\otimes \Lambda(x'',q) \longrightarrow \Lambda(x,n)$$ by the Künneth formula. If $\pi_1$ and $\pi_2$ are the projections to $S^p$ and $S^q$, respectively, this corresponds to the map
$$a\otimes b\longmapsto (\pi_1f)^*(a) \smile (\pi_2f)^*(b).$$
Now what are the degree preserving maps $\Lambda(x',p)\to \Lambda(x,n)$ for $p<n$? Well, simply by degree considerations, there is only the trivial map (since $\Lambda(x,n)$ has nonzero parts only in degree $0$ and $n$) meaning that $\pi_i f$ both induce the trivial map on cohomology, and then so does $f^*$.