A space \(X\) is called unicoherent if whenever \(X=U\cup V\) with \(U,V\) nonempty, open, and path-connected, then \(U\cap V\) is path-connected. Use the Mayer–Vietoris sequence to show that, for \(n\ge 2\), \(S^n\) is unicoherent.
Solution
Fix \(n\ge2\) and suppose \(S^n=U\cup V\) with \(U\), \(V\) nonempty, open, and path-connected.
1. \(U\cap V\neq\varnothing\).
Otherwise \(U\) and \(V\) would be disjoint nonempty open sets covering the connected space \(S^n\).
2. \(\widetilde H_0(U\cap V)=0\).
By step 1 the reduced Mayer–Vietoris sequence of the open cover \(\{U,V\}\) applies and contains the exact segment \[\widetilde H_1(S^n)\to\widetilde H_0(U\cap V)\to\widetilde H_0(U)\oplus\widetilde H_0(V).\] \(\widetilde H_1(S^n)=0\) because \(n\ge2\), and \(\widetilde H_0(U)=\widetilde H_0(V)=0\) because \(U\) and \(V\) are nonempty and path-connected.
3. Q.E.D.
For a nonempty space \(Y\), \(H_0(Y)\) is free abelian on the path components of \(Y\), so \(\widetilde H_0(Y)=0\) exactly when \(Y\) is path-connected. By steps 1 and 2, \(U\cap V\) is path-connected, so \(S^n\) is unicoherent.