I'm trying to solve the following exercise from Algebraic Topology by Hatcher, self-study.
Let $ X $ be obtained from a lens space of dimension $ 2n+1 $ by deleting a point. Compute $ \pi_{2n}(X) $ as a module over $ \mathbb{Z}[\pi_{1}(X)] $.
Here are my thoughts: I want to use the covering space $ p: S^{2n+1} \rightarrow L_m $. Let $ x $ be the point removed from $L_m$ to obtain $ X $. I have $p: S^{2n+1} - p^{-1}(x) \rightarrow X $ is a fiber bundle. Using the long exact sequence of the fiber bundle, I get the isomorphism $ \pi_{2n}(S^{2n+1} - p^{-1}(x)) = \pi_{2n}(X) $. But $ S^{2n+1} - p^{-1}(x) $ is homotopy-equivalent to the wedge sum of $m - 1$ copies of $S^{2n}$. I know from the book that $\pi_{2n}$ of this space is free abelian.
Is this correct? How can I make this a module over $ \mathbb{Z}[\pi_{1}(X)] $? Thank you
One way to obtain this action of $\pi_{1}(X)$ on $\pi_{2n}(X)$ is to observe that $\pi_{1}$ acts on the universal covering $\widetilde{X}$, hence also on $\pi_{2n}(\widetilde{X}) \simeq \pi_{2n}(X)$.
You are correct that $S^{2n+1}$ with some points removed will be homotopy equivalent to a wedge of lower dimensional spheres. You can deduce the action on $\pi_{2n}$ by observing that $\pi_{1}(X) \simeq \mathbb{Z}_{m}$ acts by cyclically permuting the removed points.
Perhaps this computation will be more accessible when you observe that $H_{2n}(S^{2n+1} \setminus \{p_{1}, \ldots, p_{m} \}) \simeq \pi_{2n}(S^{2n+1} \setminus \{p_{1}, \ldots, p_{m} \})$ by Hurewicz, so you can work with homology which is sometimes easier.