Using cup products, show that every map \(S^{k+\ell} \to S^k \times S^\ell\) induces the trivial homomorphism \(H_{k+\ell}(S^{k+\ell}) \to H_{k+\ell}(S^k \times S^\ell)\), assuming \(k > 0\) and \(\ell > 0\).
Solution
1. Cohomology ring of the product \(S^k \times S^\ell\):
1.1. By the Künneth Theorem for cohomology with \(\mathbb{Z}\) coefficients, the cohomology groups of \(S^k \times S^\ell\) (with \(k, \ell > 0\)) are:- \(H^0(S^k \times S^\ell) \cong \mathbb{Z}\),- \(H^k(S^k \times S^\ell) \cong \mathbb{Z} \langle \alpha \rangle\),- \(H^\ell(S^k \times S^\ell) \cong \mathbb{Z} \langle \beta \rangle\),- \(H^{k+\ell}(S^k \times S^\ell) \cong \mathbb{Z} \langle \alpha \smile \beta \rangle\).
Künneth Theorem for product spaces.
1.2. In particular, the generator of top cohomology is the cup product \(\gamma = \alpha \smile \beta \in H^{k+\ell}(S^k \times S^\ell)\).
cross product and cup product isomorphism in top degree.
2. Induced maps on cohomology:
2.1. The cohomology of the sphere \(S^{k+\ell}\) vanishes in all intermediate dimensions:\[ H^i(S^{k+\ell}) = 0 \quad \text{for all } 0 < i < k + \ell. \]Because \(k > 0\) and \(\ell > 0\), we have \(0 < k < k + \ell\) and \(0 < \ell < k + \ell\), so \(H^k(S^{k+\ell}) = 0\) and \(H^\ell(S^{k+\ell}) = 0\).
cellular cohomology of spheres.
2.2. Therefore the induced homomorphisms in dimensions \(k\) and \(\ell\) must be zero:\[ f^*(\alpha) = 0 \in H^k(S^{k+\ell}), \qquad f^*(\beta) = 0 \in H^\ell(S^{k+\ell}). \]
unique homomorphism into the zero group.
2.3. Because \(f^*: H^*(S^k \times S^\ell) \to H^*(S^{k+\ell})\) is a ring homomorphism respecting cup products:\[ f^*(\alpha \smile \beta) = f^*(\alpha) \smile f^*(\beta) = 0 \smile 0 = 0 \in H^{k+\ell}(S^{k+\ell}). \]Thus \(f^*: H^{k+\ell}(S^k \times S^\ell) \to H^{k+\ell}(S^{k+\ell})\) is the trivial (zero) homomorphism.
ring homomorphism property of pullback on cohomology.
3. Dualization to homology:
3.1. By the Universal Coefficient Theorem for Cohomology, for any space \(X\) with free finitely generated homology, \(H^n(X) \cong \operatorname{Hom}(H_n(X), \mathbb{Z})\).Under this canonical duality, the induced cohomology map \(f^*\) is the transpose/dual of the induced homology map \(f_*\):\[ f^* = (f_*)^t: \operatorname{Hom}(H_{k+\ell}(S^k \times S^\ell), \mathbb{Z}) \to \operatorname{Hom}(H_{k+\ell}(S^{k+\ell}), \mathbb{Z}). \]
Universal Coefficient Theorem naturality.
3.2. Since \(f^* = 0\), its dual map on homology \(f_*: H_{k+\ell}(S^{k+\ell}) \to H_{k+\ell}(S^k \times S^\ell)\) is also the zero homomorphism.
dual of the zero map on free abelian groups is zero.
4. Conclusion:Every continuous map \(S^{k+\ell} \to S^k \times S^\ell\) induces the trivial homomorphism on top homology. Q.E.D.
1 through 3.