Let $X$ be a $K(G,1)$ which is a CW complex. We want to show that $\pi_n(X^n)$ is free (abelian when $n\geq 2$ where $X^n$ is the $n$ skeleton of $X$.
I thought I could choose any model for a $K(G,1)$ to prove my claim, but it turns out homotopic spaces don't necessairly have homotopic skeletons. And my ideas were based on that, hence I am quite lost.
I am thinking I should work instead on the universal cover, and then use the higher and higher connectdenss to allow an application of Hurecwiz which could lead to the desired result? I have yet to figure out if this idea can lead me to what I want. The (first) problem I am experiencing with this method is that it isn't clear that the $n$ skeleton of the universal cover is the universal cover of the $n$ skeleton.
Any method is accepted, I just wanted to share my ideas. Thank you.
It turns out that passing to the universal cover $+$ $n$ skeleton $+$ using Hurecwiz works.
Let $X$ be a $K(G,1)$, we want to show $\pi_n(X^n)$ is free abelian where $X^n$ is the $n$ skeleton of $X$. $X^n$ admits a universal cover, call it $Y^n$ which we can take to be the $n$ skeleton of $Y$ the universal cover of $X$ (Thank you Thorgott).
That being said we can see that $\pi_i(Y^n)\cong \pi_i(Y)\cong 0$ for $i\leq n-1$ where we used that $Y$ is contractible as the universal cover of an Eilenberg Maclane space. So $Y_n$ is $n-1$ connected, so we can use Hurecwiz to get $\pi_n(Y^n)\cong H_n(Y^n)$. Now because $Y^n$ has no $n+1$ cell we can see that $H^n(Y^n)$ is free with basis the $n$ cells of $Y^n$.
Chaining the isomorphisms together, we get the desired result $$\pi_n(X^n)\cong \pi_n(Y^n)\cong H_n(Y^n)\cong \mathbb{Z}^K$$