I've been preparing for a qualifying exam in topology. I'm struggling with a recurring question to do with computing higher homotopy groups of the wedge of spaces.
Example: Let $P$ be the Poincare homology sphere, a $3$-manifold whose fundamental group has order $120$ and whose universal cover is $S^3$. Compute $\pi_3(P\vee S^3)$.
Attempt: One strategy is to compute a simply connected covering space $\widetilde{X}\rightarrow P\vee S^3$ and then to use the fact that $\pi_3(\widetilde{X})\cong \pi_3(P\vee S^3)$. In this thread
https://mathoverflow.net/questions/12606/universal-covering-space-of-wedge-products
Greg Kuperberg describes a general construction of the universal cover $\widetilde{X\vee Y}$ of the space $X\vee Y$ using $|\pi_1(X)|$, $|\pi_1(Y)|$, $\widetilde{X}$ and $\widetilde{Y}$. This yields a solution to my problem but I'm confused by the fact that his construction relies of some relatively recent results which seem well beyond the scope of first year graduate algebraic topology.
Can anyone see a more elementary way of solving this problem?
Well, in this case, since $S^3$ is simply connected, the construction is a lot simpler. Just take the universal cover $\widetilde{P}$ of $P$ and attach a copy of $S^3$ at each preimage of the basepoint to form a space $Y$. There is then an obvious covering map $Y\to P\vee S^3$ (map $\widetilde{P}$ to $P$ by the universal cover map and map the attached copies of $S^3$ to $S^3$ by homeomorphisms). Since $Y$ is homotopy equivalent to a wedge of $121$ $S^3$'s ($\widetilde{P}$, and then one at each of the $120$ preimages of the basepoint of $P$), we see that $Y$ is the universal cover of $P\vee S^3$ and $\pi_3(Y)\cong\mathbb{Z}^{121}$.