Show that \(\tilde{H}_i(S^n - X) \approx \tilde{H}_{n-i-1}(X)\) when \(X\) is homeomorphic to a finite connected graph. [First do the case that the graph is a tree.]
Solution
Let \(X\subset S^n\) be homeomorphic to a finite connected graph. Since all homology groups of a finite graph are free abelian, the universal coefficient theorem identifies \[ H^j(X;\mathbb Z)\cong \operatorname{Hom}(H_j(X;\mathbb Z),\mathbb Z) \] and hence, noncanonically but naturally in rank, \[ \widetilde H^j(X;\mathbb Z)\cong \widetilde H_j(X;\mathbb Z). \]
1. If \(X\) is a tree, then\[ \widetilde H_i(S^n-X)=0 \qquad\text{for all }i. \]
A finite tree is contractible, so its reduced cohomology vanishes. Alexander duality gives \[ \widetilde H_i(S^n-X)\cong \widetilde H^{\,n-i-1}(X)=0. \]
2. For an arbitrary finite connected graph,\[ \boxed{\widetilde H_i(S^n-X)\cong \widetilde H_{\,n-i-1}(X).} \]
Alexander duality gives \[ \widetilde H_i(S^n-X)\cong \widetilde H^{\,n-i-1}(X). \] The homology of a connected graph is free: \(H_0(X)=\mathbb Z\), \(H_1(X)\) is a free abelian group, and \(H_j(X)=0\) for \(j>1\). Therefore the universal coefficient theorem has no \(\operatorname{Ext}\) term, and \(\widetilde H^{j}(X)\) is a free abelian group of the same rank as \(\widetilde H_j(X)\). Choosing the basis of \(H_1(X)\) obtained by collapsing a maximal tree identifies these groups, yielding the displayed isomorphism.
In particular, if \(r=\operatorname{rank}H_1(X)\), then the only possible nonzero reduced homology group of the complement is \[ \widetilde H_{n-2}(S^n-X)\cong\mathbb Z^r. \]