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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?

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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.

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