(Michigan May ’09) Let \(X\) be the space obtained from \(S^1\times\mathbb R\) by removing the interior of \(k\) disjoint \(2\)-disks.
Compute the fundamental group \(\pi_1(X)\).
What would be your answer to part (a) if \(S^1\times\mathbb R\) is replaced by \(S^2\times\mathbb R\) and \(2\)-disks are replaced by \(3\)-balls?
Let \(Y\) be the union of two copies of the real projective plane \(\mathbb{RP}^2\) having exactly one point \(y\) in common. Compute \(\pi_1(Y,y)\).
Solution
Choose a compact annular region \(S^1\times[-N,N]\) containing all \(k\) deleted disks. The two noncompact ends deformation retract onto its boundary circles, so \(X\) deformation retracts onto a compact annulus with the interiors of \(k\) disks removed. This is an orientable genus-zero surface with \(k+2\) boundary components, hence it deformation retracts onto a wedge of \(k+1\) circles. Therefore \[ \pi_1(X)\cong F_{k+1}. \]
The answer is trivial. Start with \(S^2\times\mathbb R\), which is simply connected. Removing the interior of one embedded \(3\)-ball does not change the fundamental group: if \(M=(M\setminus\operatorname{int}B^3)\cup B^3\), then after thickening the intersection it deformation retracts to \(S^2\), so Seifert–van Kampen gives \[ \pi_1(M)\cong \pi_1(M\setminus\operatorname{int}B^3). \] Applying this successively to the \(k\) disjoint balls gives \[ \pi_1\bigl((S^2\times\mathbb R)\setminus\textstyle\bigcup_i\operatorname{int}B_i^3\bigr)=0. \]
The space \(Y\) is the wedge \(\mathbb{RP}^2\vee\mathbb{RP}^2\). Van Kampen therefore gives \[ \pi_1(Y,y)\cong \pi_1(\mathbb{RP}^2)*\pi_1(\mathbb{RP}^2) \cong (\mathbb Z/2)*(\mathbb Z/2) \cong \langle a,b\mid a^2=b^2=1\rangle. \]