Let \(D\) be a closed connected subspace of \(S^2\) that separates \(S^2\) into \(n\) components.
If \(A\) is an arc in \(S^2\) whose intersection with \(D\) consists of one of its end points, show that \(D \cup A\) separates \(S^2\) into \(n\) components.
If \(A\) is an arc in \(S^2\) whose intersection with \(D\) consists of its end points, show that \(D \cup A\) separates \(S^2\) into \(n + 1\) components.
If \(C\) is a simple closed curve in \(S^2\) that intersects \(D\) in a single point, show \(D \cup C\) separates \(S^2\) into \(n + 1\) components.
Solution
Goal: Determine the number of connected components of \(S^2 \setminus (D \cup A)\) and \(S^2 \setminus (D \cup C)\) using the Mayer-Vietoris sequence in reduced homology.
1. Homological characterization of component count:For any closed subset \(K \subset S^2\), the number of path components of the open set \(S^2 \setminus K\) is \(1 + \operatorname{rank}(\widetilde{H}_0(S^2 \setminus K))\). In particular, \(S^2 \setminus D\) having \(n\) components means \(\widetilde{H}_0(S^2 \setminus D) \cong \mathbb{Z}^{n-1}\).
2. Mayer-Vietoris framework:Let \(U = S^2 \setminus D\) and let \(V = S^2 \setminus K'\) (where \(K'\) is \(A\) or \(C\)). Then \(U \cap V = S^2 \setminus (D \cup K')\) and \(U \cup V = S^2 \setminus (D \cap K')\).Since \(U, V\) are open subsets of \(S^2\), the Mayer-Vietoris sequence in reduced homology is exact:\[\widetilde{H}_1(U \cup V) \xrightarrow{\partial} \widetilde{H}_0(U \cap V) \xrightarrow{(i_U, i_V)} \widetilde{H}_0(U) \oplus \widetilde{H}_0(V) \xrightarrow{j_U - j_V} \widetilde{H}_0(U \cup V) \to 0.\]
3. Part (a): \(D \cap A = \{p\}\) (one endpoint of arc \(A\)).Proof:
3.1. Since \(A \cong [0, 1]\) is an arc, \(S^2 \setminus A\) is connected and simply connected, so \(\widetilde{H}_0(V) = 0\) and \(\widetilde{H}_1(V) = 0\).
3.2. \(D \cap A = \{p\}\), so \(U \cup V = S^2 \setminus \{p\} \cong \mathbb{R}^2\), which is contractible. Thus \(\widetilde{H}_0(U \cup V) = 0\) and \(\widetilde{H}_1(U \cup V) = 0\).
3.3. The Mayer-Vietoris sequence reduces to:\[0 \to \widetilde{H}_0(U \cap V) \xrightarrow{\cong} \widetilde{H}_0(U) \oplus 0 \to 0.\]
3.4. Hence \(\widetilde{H}_0(S^2 \setminus (D \cup A)) \cong \widetilde{H}_0(S^2 \setminus D) \cong \mathbb{Z}^{n-1}\).
3.5. Therefore, \(S^2 \setminus (D \cup A)\) has \(1 + (n - 1) = n\) components.
4. Part (b): \(D \cap A = \{p, q\}\) with \(p \neq q\) (both endpoints of arc \(A\)).Proof:
4.1. As in 3, \(V = S^2 \setminus A\) satisfies \(\widetilde{H}_0(V) = 0\) and \(\widetilde{H}_1(V) = 0\).
4.2. \(D \cap A = \{p, q\}\), so \(U \cup V = S^2 \setminus \{p, q\} \simeq S^1\).
4.3. Thus \(\widetilde{H}_0(U \cup V) = 0\) and \(\widetilde{H}_1(U \cup V) \cong \mathbb{Z}\).
4.4. The Mayer-Vietoris sequence gives the short exact sequence:\[0 \to \widetilde{H}_1(U \cup V) \xrightarrow{\partial} \widetilde{H}_0(U \cap V) \to \widetilde{H}_0(U) \oplus 0 \to 0\]\[0 \to \mathbb{Z} \to \widetilde{H}_0(S^2 \setminus (D \cup A)) \to \mathbb{Z}^{n-1} \to 0.\]
4.5. Since \(\mathbb{Z}^{n-1}\) is free abelian, the sequence splits, giving:\[\widetilde{H}_0(S^2 \setminus (D \cup A)) \cong \mathbb{Z}^{n-1} \oplus \mathbb{Z} \cong \mathbb{Z}^n.\]
4.6. Therefore, \(S^2 \setminus (D \cup A)\) has \(1 + n = n + 1\) components.
5. Part (c): \(C\) is a simple closed curve and \(D \cap C = \{p\}\) (a single point).Proof:
5.1. Since \(C \cong S^1\) is a simple closed curve, by the Jordan Curve Theorem \(V = S^2 \setminus C\) has two components, so \(\widetilde{H}_0(V) \cong \mathbb{Z}\) and \(\widetilde{H}_1(V) = 0\).
5.2. \(D \cap C = \{p\}\), so \(U \cup V = S^2 \setminus \{p\} \cong \mathbb{R}^2\), which has \(\widetilde{H}_0(U \cup V) = 0\) and \(\widetilde{H}_1(U \cup V) = 0\).
5.3. The Mayer-Vietoris sequence reduces to:\[0 \to \widetilde{H}_0(U \cap V) \xrightarrow{\cong} \widetilde{H}_0(U) \oplus \widetilde{H}_0(V) \to 0.\]
5.4. Hence \(\widetilde{H}_0(S^2 \setminus (D \cup C)) \cong \widetilde{H}_0(S^2 \setminus D) \oplus \widetilde{H}_0(S^2 \setminus C) \cong \mathbb{Z}^{n-1} \oplus \mathbb{Z} \cong \mathbb{Z}^n\).
5.5. Therefore, \(S^2 \setminus (D \cup C)\) has \(1 + n = n + 1\) components. Q.E.D.