Show \(\tilde H_i(\Sigma X) \cong \tilde H_{i-1}(X)\)
- Show \(\Sigma S^n \cong S^{n+1}\)
Solution
Goal: Let \(X\) be a topological space. Prove that the suspension \(\Sigma X = (X \times [-1, 1]) / (X \times \{1\} \sim N, X \times \{-1\} \sim S)\) satisfies \(\widetilde{H}_i(\Sigma X) \cong \widetilde{H}_{i-1}(X)\) for all \(i \in \mathbb{Z}\), and prove that \(\Sigma S^n \cong S^{n+1}\) for all \(n \ge 0\).
1. Prove the suspension isomorphism \(\widetilde{H}_i(\Sigma X) \cong \widetilde{H}_{i-1}(X)\).
1.1. Decompose \(\Sigma X\) into two open cones:
\(U = (X \times (-1, 1]) / (X \times \{1\} \sim N) = C_+ X \setminus \{S\}\),
\(V = (X \times [-1, 1)) / (X \times \{-1\} \sim S) = C_- X \setminus \{N\}\). <2>2. \(U\) and \(V\) are open in \(\Sigma X\), and \(U \cup V = \Sigma X\). <2>3. \(U\) deformation retracts to the north cone vertex \(N\), and \(V\) deformation retracts to the south cone vertex \(S\). Thus \(\widetilde{H}_k(U) = 0\) and \(\widetilde{H}_k(V) = 0\) for all \(k \ge 0\). <2>4. The intersection \(U \cap V = X \times (-1, 1)\) deformation retracts onto \(X \times \{0\} \cong X\). Thus \(\widetilde{H}_k(U \cap V) \cong \widetilde{H}_k(X)\) for all \(k \ge 0\). <2>5. Write the Mayer-Vietoris sequence in reduced homology for \((\Sigma X; U, V)\): \[\cdots \to \widetilde{H}_i(U) \oplus \widetilde{H}_i(V) \to \widetilde{H}_i(\Sigma X) \xrightarrow{\partial} \widetilde{H}_{i-1}(U \cap V) \to \widetilde{H}_{i-1}(U) \oplus \widetilde{H}_{i-1}(V) \to \cdots\] <2>6. Substituting the vanishing groups from <2>3 gives exact sequences: \[0 \longrightarrow \widetilde{H}_i(\Sigma X) \xrightarrow{\partial} \widetilde{H}_{i-1}(X) \longrightarrow 0.\] <2>7. Therefore, the connecting homomorphism \(\partial \colon \widetilde{H}_i(\Sigma X) \xrightarrow{\cong} \widetilde{H}_{i-1}(X)\) is an isomorphism for all \(i \in \mathbb{Z}\).
1.8. Substituting \(\widetilde{H}_*(U) = \widetilde{H}_*(V) = 0\) into the Mayer–Vietoris sequence of the suspension open cover leaves the short exact sequence \(0 \to \widetilde{H}_i(\Sigma X) \xrightarrow{\partial} \widetilde{H}_{i-1}(X) \to 0\), so \(\partial\) is an isomorphism.
2. Prove \(\Sigma S^n \cong S^{n+1}\).
2.1. Realize \(S^n \subset \mathbb{R}^{n+1}\) as \(\{x \in \mathbb{R}^{n+1} \mid \|x\| = 1\}\), and \(S^{n+1} \subset \mathbb{R}^{n+2} = \mathbb{R}^{n+1} \times \mathbb{R}\) as \(\{(x, t) \in \mathbb{R}^{n+1} \times \mathbb{R} \mid \|x\|^2 + t^2 = 1\}\).
2.2. Define a map \(f \colon S^n \times [-1, 1] \to S^{n+1}\) by: \[f(u, t) = \left( \sqrt{1 - t^2} \, u, \, t \right) \in \mathbb{R}^{n+1} \times \mathbb{R}.\]
2.3. Verify the image and continuity of \(f\):
For any \(u \in S^n\) and \(t \in [-1, 1]\), \(\|\sqrt{1-t^2} u\|^2 + t^2 = (1-t^2)\|u\|^2 + t^2 = (1-t^2)(1) + t^2 = 1\).
Thus \(f(u, t) \in S^{n+1}\), and \(f\) is continuous as a composition of elementary continuous functions. <2>4. Check the fibers of \(f\):
At \(t = 1\): \(f(u, 1) = (0, 1) = N \in S^{n+1}\) for all \(u \in S^n\).
At \(t = -1\): \(f(u, -1) = (0, -1) = S \in S^{n+1}\) for all \(u \in S^n\).
For \(t \in (-1, 1)\): \(f(u, t) = (x, t) \implies \sqrt{1-t^2} u = x \implies u = \frac{x}{\sqrt{1-t^2}}\), which determines \(u\) uniquely because \(\sqrt{1-t^2} > 0\). <2>5. \(f\) is surjective: For any \((x, t) \in S^{n+1}\), \(|t| \le 1\). If \(|t| < 1\), set \(u = x / \sqrt{1-t^2} \in S^n\), then \(f(u, t) = (x, t)\). If \(t = \pm 1\), \((x, t) = (0, \pm 1) = f(u, \pm 1)\). <2>6. Thus \(f\) induces a continuous bijection from the quotient space \(\Sigma S^n = (S^n \times [-1, 1]) / (S^n \times \{1\} \sim N, S^n \times \{-1\} \sim S)\) onto \(S^{n+1}\). <2>7. Since \(\Sigma S^n\) is compact (quotient of compact \(S^n \times [-1, 1]\)) and \(S^{n+1}\) is Hausdorff, any continuous bijection \(\overline{f} \colon \Sigma S^n \to S^{n+1}\) is a homeomorphism.
2.8. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, so \(\overline{f}\) is a homeomorphism.
3. Q.E.D.
3.1. 1 establishes the suspension isomorphism and 2 establishes \(\Sigma S^n \cong S^{n+1}\).