Show that \(S^2 \not\cong S^3\).
Solution
Goal: Prove that the 2-sphere \(S^2\) is not homeomorphic to the 3-sphere \(S^3\) (\(S^2 \not\cong S^3\)).
1. Compute the homology groups of \(S^2\) and \(S^3\).
1.1. For the 2-sphere \(S^2\), the singular homology groups are:
\(H_0(S^2; \mathbb{Z}) \cong \mathbb{Z}\),
\(H_2(S^2; \mathbb{Z}) \cong \mathbb{Z}\),
\(H_k(S^2; \mathbb{Z}) = 0\) for all \(k \neq 0, 2\) (in particular, \(H_3(S^2; \mathbb{Z}) = 0\)). <2>2. For the 3-sphere \(S^3\), the singular homology groups are:
\(H_0(S^3; \mathbb{Z}) \cong \mathbb{Z}\),
\(H_3(S^3; \mathbb{Z}) \cong \mathbb{Z}\),
\(H_k(S^3; \mathbb{Z}) = 0\) for all \(k \neq 0, 3\) (in particular, \(H_2(S^3; \mathbb{Z}) = 0\)).
1.3. The sphere \(S^n\) has a CW structure with one 0-cell and one \(n\)-cell, so its cellular chain complex has \(H_0 \cong \mathbb{Z}\), \(H_n \cong \mathbb{Z}\), and all other homology groups zero.
2. Apply topological invariance of homology.
2.1. If two topological spaces \(X\) and \(Y\) are homeomorphic (\(X \cong Y\)), then their homology groups are isomorphic in every dimension: \(H_k(X; \mathbb{Z}) \cong H_k(Y; \mathbb{Z})\) for all \(k \ge 0\).
2.2. Comparing at dimension \(k = 2\): \[H_2(S^2; \mathbb{Z}) \cong \mathbb{Z} \not\cong 0 \cong H_2(S^3; \mathbb{Z}).\]
2.3. Comparing at dimension \(k = 3\): \[H_3(S^2; \mathbb{Z}) \cong 0 \not\cong \mathbb{Z} \cong H_3(S^3; \mathbb{Z}).\]
2.4. Therefore, \(S^2\) cannot be homeomorphic to \(S^3\).
2.5. A homeomorphism would induce isomorphisms on all homology groups, but \(H_2(S^2) \cong \mathbb{Z} \not\cong 0 \cong H_2(S^3)\); this contradiction shows \(S^2 \not\cong S^3\).
3. Alternative proof via point deletion and local homology / dimension invariance.
3.1. For any point \(p \in S^n\), \(H_k(S^n, S^n \setminus \{p\}) \cong \widetilde{H}_{k-1}(S^{n-1}) \cong \begin{cases} \mathbb{Z} & k = n, \\ 0 & k \neq n. \end{cases}\)
3.2. A homeomorphism \(S^2 \cong S^3\) would induce an isomorphism on local homology groups \(H_2(S^2, S^2 \setminus \{p\}) \cong H_2(S^3, S^3 \setminus \{f(p)\})\), meaning \(\mathbb{Z} \cong 0\), a contradiction.
3.3. Excision identifies \(H_k(S^n, S^n \setminus \{p\})\) with \(H_k(D^n, \partial D^n)\), and the long exact sequence of the pair gives \(\widetilde{H}_{k-1}(S^{n-1})\); this is \(\mathbb{Z}\) only when \(k = n\), so a homeomorphism \(S^2 \cong S^3\) would force \(\mathbb{Z} \cong 0\).
4. Q.E.D.
4.1. Both 2 and 3 show \(S^2 \not\cong S^3\).