Let \(P\) be a \(3\)-manifold whose fundamental group has order \(120\) and whose universal cover is \(S^3\). Compute \(\pi_3\) of the one-point union \(P \vee S^3\).
Solution
1. Set\[ X=P\vee S^3. \]Then \(\pi_1(X)=\pi_1(P)\) has order \(120\).
The sphere summand is simply connected.
2. The universal cover \(\widetilde X\) is obtained from the universal cover \(S^3\to P\) by attaching one copy of \(S^3\) at each of the \(120\) lifts of the wedge point.
The fiber of the universal cover over the wedge point has cardinality equal to the order of the deck group, namely \(120\). The simply connected wedge summand lifts independently at every point of this fiber.
3. Therefore\[ \widetilde X\simeq\bigvee^{121}S^3. \]
There is the original covering \(S^3\) plus \(120\) attached \(S^3\) summands. Moving their attachment points together along a tree in the original \(S^3\) and collapsing that tree gives the wedge up to homotopy.
4. The universal cover is \(2\)-connected and\[ H_3(\widetilde X;\mathbb Z)\cong\mathbb Z^{121}. \]
A wedge of \(121\) copies of \(S^3\) has these properties.
5. By Hurewicz and covering-space invariance of higher homotopy groups,\[ \boxed{\pi_3(P\vee S^3)\cong\mathbb Z^{121}.} \]
Hurewicz gives \(\pi_3(\widetilde X)\cong H_3(\widetilde X)\), and the universal covering map induces an isomorphism on \(\pi_3\).