So I'm trying to solve one of Hatcher's exercises and it seems that my proof would work if $\pi_n(X^n,X^{n-1})\simeq \pi_n(X^n/X^{n-1})$ for $n\geq 2$. I tried applying proposition $4.28$, but it doesn't work in this case. All I can gather is that the we have a surjection, but this is not enough to show that $\pi_n(X^n,X^{n-1})$ is free abelian using the fact that $\pi_n(X^n/X^{n-1})=\pi_n(\vee_\alpha S^n_\alpha)$ is free abelian, which is what I am trying to show. Does anyone see away around this issue? For reference I have included proposition $4.28$ ${}$ below.
