(Arizona Aug ’06) Compute the singular homology groups \(H_*(X,\mathbb Z)\) of the space \(X=\mathbb R^3\setminus A\), where \(A\) is a subset of \(\mathbb R^3\) homeomorphic to the disjoint union of two unlinked circles.
Solution
1. Regard \(X=\RR^3\setminus A\) as a complement in \(S^3\).
Let \(\infty\) denote the point at infinity in the one-point compactification \(S^3=\RR^3\cup\{\infty\}\) and put \[ K=A\sqcup\{\infty\}. \] Then \[ X=S^3\setminus K. \] The set \(K\) is compact and locally contractible, so Alexander duality gives \[ \widetilde H_k(X;\ZZ) \cong \widetilde H^{\,2-k}(K;\ZZ) \] for every \(k\).
2. The reduced cohomology of \(K=S^1\sqcup S^1\sqcup\{\infty\}\) is\[ \widetilde H^j(K;\ZZ) \cong \begin{cases} \ZZ^2,&j=0,\\ \ZZ^2,&j=1,\\ 0,&j\ge2. \end{cases} \]
The space \(K\) has three connected components, so \(\widetilde H^0(K;\ZZ)\cong\ZZ^{3-1}=\ZZ^2\). Each circle contributes one copy of \(\ZZ\) in degree \(1\), while the isolated point contributes none, hence \(H^1(K;\ZZ)\cong\ZZ^2\). There is no cohomology in higher degrees.
3. Alexander duality therefore gives\[ \widetilde H_0(X;\ZZ)=0, \qquad H_1(X;\ZZ)\cong\ZZ^2, \qquad H_2(X;\ZZ)\cong\ZZ^2, \]
Apply 1 to the groups in 2: \(\widetilde H_0(X)\cong\widetilde H^2(K)=0\), \(H_1(X)=\widetilde H_1(X)\cong\widetilde H^1(K)=\ZZ^2\), and \(H_2(X)=\widetilde H_2(X)\cong\widetilde H^0(K)=\ZZ^2\). For \(k\ge3\), the corresponding cohomological degree is negative, so the homology group is zero.
4. Thus the singular homology groups of \(X=\RR^3\setminus(S^1\sqcup S^1)\) are\[ H_k(X;\ZZ) \cong \begin{cases} \ZZ,&k=0,\\ \ZZ^2,&k=1,\\ \ZZ^2,&k=2,\\ 0,&k\ge3. \end{cases} \]
Since \(\widetilde H_0(X)=0\), one has \(H_0(X)\cong\ZZ\); the remaining groups are exactly those computed in 3.