$\pi_n((S^1 \vee S^n)\cup e^{n+1})$, Example 4.35, Hatcher, Algebraic Topology

Question

In Example 4.35, I can't see how the underlined statement holds. I see that $\pi_n(S^1\vee S^n)$ is isomorphic to $\Bbb Z[t,t^{-1}]$ as algebras, but why $\pi_n(X)$ is the ring $\Bbb Z[t,t^{-1}]/(2t-1)$?

Answer 1

Recall that if $f: A \rightarrow B$ has $\operatorname{im}(\pi_1(f))=0$ then $f$ lifts to the universal cover. Since $n>1$ the map we are attaching the disk by lifts to the universal cover of the wedge of spheres. Lifting it so the basepoint of the wedge lifts into the sphere corresponding to the element $1 \in \mathbb{Z}[t,t^{-1}]$, we see that this attaching map wraps negatively around the sphere at $1$ and wraps twice around the sphere at $t$. We can also lift to the sphere corresponding to $t^n$. In the covering space this is the sphere $n$ spheres to the right of the sphere corresponding to $1$. We get a similar attaching map along the sphere $t^n$ and then at $t^{n+1}$, wrapping negatively around the first and twice around the second.

I leave it to you to show that this is the universal cover of $X$. Calculating the nth homotopy group is as easy as taking the free abelian group generated by the spheres and quotienting out by the relations the disks impose. You'll see that the disk associated to the lift of the basepoint to the sphere at $t^n$ gives the relation $-t^n+2t^{n+1}=(2t-1)t^{n}$. The subgroup generated by these is precisely the ideal $(2t-1)$.