(Michigan May ’09) Let \[S^1=\{(x,y,0,0)\in\mathbb R^4\mid x^2+y^2=1\}\] be the unit circle and consider \(M=\mathbb R^4\setminus S^1\). Compute the fundamental group \(\pi_1(M)\) and the homology groups \(H_*(M)\) of \(M\).
Solution
First compute the fundamental group. Any loop in \[ M=\mathbb R^4\setminus S^1 \] bounds a disk in \(\mathbb R^4\), since \(\mathbb R^4\) is simply connected. By smooth or PL general position, the disk can be perturbed relative to its boundary to be disjoint from the embedded circle: the expected intersection dimension is \[ 2+1-4=-1. \] Thus every loop contracts in \(M\), and \[ \pi_1(M)=0. \]
For homology, compactify \(\mathbb R^4\) to \(S^4\). Then \[ M\cong S^4\setminus A, \qquad A=S^1\sqcup\{\infty\}. \] Alexander duality gives \[ \widetilde H_i(M;\mathbb Z) \cong \widetilde H^{\,3-i}(A;\mathbb Z). \] The only nonzero reduced cohomology groups of \(A\) are \[ \widetilde H^0(A)\cong\mathbb Z, \qquad H^1(A)\cong\mathbb Z. \] Hence \[ \widetilde H_3(M)\cong\mathbb Z, \qquad \widetilde H_2(M)\cong\mathbb Z, \] and all other reduced homology groups vanish. Therefore \[ H_i(M)\cong \begin{cases} \mathbb Z,&i=0,2,3,\\ 0,&\text{otherwise}. \end{cases} \]