For a self-map \(f:S^n\to S^n\), show that \(\deg(\Sigma f)=\deg(f)\).
Conclude that for every \(n\ge1\) and every integer \(d\in\ZZ\), there exists a self-map \(S^n\to S^n\) of degree \(d\).
Solution
Goal: Let \(f \colon S^n \to S^n\) be continuous with \(n\ge1\). Prove \(\deg(\Sigma f)=\deg(f)\) and deduce that every integer occurs as the degree of a self-map of every sphere \(S^n\), \(n\ge1\).
1. Definition of degree and suspension isomorphism in homology.
1.1. For any continuous map \(g \colon S^k \to S^k\) (\(k \ge 1\)), the degree \(\deg(g) \in \mathbb{Z}\) is defined by the induced map on top homology: \(g_*(\alpha) = (\deg g) \cdot \alpha\) for any generator \(\alpha \in \widetilde{H}_k(S^k) \cong \mathbb{Z}\).
1.2. Decompose \(\Sigma S^n\) as the union of two open cones \(C_+ S^n\) and \(C_- S^n\) intersecting along \(S^n \times (-1, 1) \simeq S^n\).
1.3. The suspension isomorphism \(\sigma \colon \widetilde{H}_n(S^n) \xrightarrow{\cong} \widetilde{H}_{n+1}(\Sigma S^n)\) is the connecting isomorphism in the Mayer-Vietoris sequence of the suspension:\[\widetilde{H}_{n+1}(C_+ S^n) \oplus \widetilde{H}_{n+1}(C_- S^n) \to \widetilde{H}_{n+1}(\Sigma S^n) \xrightarrow{\partial} \widetilde{H}_n(S^n) \to \widetilde{H}_n(C_+ S^n) \oplus \widetilde{H}_n(C_- S^n).\]Since cones are contractible, \(\partial \colon \widetilde{H}_{n+1}(\Sigma S^n) \to \widetilde{H}_n(S^n)\) is an isomorphism, and \(\sigma = \partial^{-1}\).
1.4. In the Mayer–Vietoris sequence of the suspension, the terms \(\widetilde{H}_*(C_+ S^n)\) and \(\widetilde{H}_*(C_- S^n)\) vanish because cones are contractible, so the connecting homomorphism \(\partial\) is an isomorphism.
2. Prove \(\deg(\Sigma f) = \deg(f)\).
2.1. The suspension map \(\Sigma f \colon \Sigma S^n \to \Sigma S^n\) preserves the cones \(C_+ S^n, C_- S^n\) and restricts on the equator \(S^n\) to \(f\).
2.2. By naturality of the Mayer-Vietoris connecting homomorphism \(\partial\) (proved in P-AMD-2IFZW3JK), the following square commutes:\[ \begin{CD} \widetilde{H}_{n+1}(\Sigma S^n) @>{\partial}>{\cong}> \widetilde{H}_n(S^n) \\ @VV{(\Sigma f)_*}V @VV{f_*}V \\ \widetilde{H}_{n+1}(\Sigma S^n) @>{\partial}>{\cong}> \widetilde{H}_n(S^n) \end{CD} \]
2.3. Let \(\beta \in \widetilde{H}_{n+1}(\Sigma S^n)\) be a generator such that \(\partial(\beta) = \alpha \in \widetilde{H}_n(S^n)\) is a generator.
2.4. Computing along the diagram:\[\partial((\Sigma f)_*(\beta)) = f_*(\partial(\beta)) = f_*(\alpha) = (\deg f) \alpha.\]
2.5. Applying \(\sigma = \partial^{-1}\) yields:\[(\Sigma f)_*(\beta) = (\deg f) \beta.\]
2.6. By definition of degree, \((\Sigma f)_*(\beta) = (\deg \Sigma f) \beta\), hence \(\deg(\Sigma f) = \deg(f)\).
2.7. The naturality square of 2 commutes, so \(\partial((\Sigma f)_*(\beta)) = f_*(\partial(\beta))\); combining this with <2>4–<2>5 gives \((\Sigma f)_*(\beta) = (\deg f)\beta\), and comparing with the definition of degree yields \(\deg(\Sigma f) = \deg(f)\).
3. For every \(n\ge1\) and \(d\in\mathbb Z\), construct a self-map of \(S^n\) of degree \(d\).
3.1. On \(S^1\subset\mathbb C\), the map\[ g_d(z)=z^d \]has degree \(d\) for \(d\ge0\); for \(d<0\) the same formula means \(z^d=\overline z^{\,|d|}\) on \(S^1\) and again has degree \(d\). For \(d=0\), it is the constant map \(1\).
Writing \(z=e^{i\theta}\) gives \(g_d(e^{i\theta})=e^{id\theta}\). Thus the induced map on \(H_1(S^1)\cong\mathbb Z\) is multiplication by \(d\).
3.2. For \(n>1\), define\[ f_{n,d}=\Sigma^{\,n-1}g_d:S^n\cong\Sigma^{\,n-1}S^1\longrightarrow\Sigma^{\,n-1}S^1\cong S^n. \]
Repeatedly use the canonical homeomorphisms \(\Sigma S^k\cong S^{k+1}\).
3.3. Then \(\deg(f_{n,d})=d\).
Apply 2 successively to the \(n-1\) suspensions: \[ \deg(\Sigma^{\,n-1}g_d)=\deg(g_d)=d. \]
4. Q.E.D.
4.1. 2 proves suspension preserves degree, and 3 constructs a self-map of every prescribed degree.