# Example showing that $\pi_n(S^1 \vee S^n)$ for $n \ge 2$ is free abelian on a countably infinite number of generators.

This is Example $4.27$ of Hatcher's Algebraic Topology. I am having difficulty understanding a step that determines the basis of $\pi_n$ of the universal cover of $S^1 \vee S^n$.

Let us show that $\pi_n(S^1 \vee S^n)$ for $n \ge 2$ is free abelian on countably infinite number of generators. For $i \ge 2$, $\pi_i(S^1 \vee S^n)$ is isomorphic to $\pi_i$ of its universal cover. The universal cover consists of a copy of $\mathbb{R}$ with a sphere $S_k^n$ attached to each integer point $k \in \mathbb{R}$, thus $\pi_i$ of the universal cover is isomorphic to $\pi_i( \vee_k S_k^n)$ where we have shown that $\pi_i( \vee_k S_k^n)$ is free abelian with basis the inclusion $S^n \hookrightarrow \vee_k S_k^n$.

Thus a basis for $\pi_n$ of the universal cover of $S^1 \vee S^n$ is represented by maps that lift the maps obtained from the inclusion $S^n \hookrightarrow \vee_k S_k^n$ by the action of the various elements of $\pi_1(S^1 \vee S^n) \cong \mathbb{Z}$.

Why does this give a basis of the universal cover? What does the action of the fundamental group have to do with the basis elements of the universal cover?

Here's perhaps an easier approach. Explicitly find the universal cover and see that it deformation retracts to a chain of copies of $S^n$ glued end to end. Now use Mayer-Vietoris to show $H_i(\tilde X)=0$ for $i < n$ and $H_n(\tilde X) = \oplus_{n=1}^{\infty}\Bbb Z$. Hurewicz and the fact that covering maps induce isomorphisms on higher homotopy groups should give everything you need from here.