If \(X\) is a \(K(G, 1)\) CW complex, show that \(\pi_n(X^n)\) is free abelian for \(n \geq 2\).
Solution
Let \(p:\widetilde X\to X\) be the universal cover. Since \(X\) is a \(K(G,1)\), \(\widetilde X\) is contractible. The inverse image \(\widetilde X^{\,n}=p^{-1}(X^n)\) is the \(n\)-skeleton of \(\widetilde X\).
The pair \((\widetilde X,\widetilde X^{\,n})\) has no relative cells in dimensions at most \(n\). Hence \[ \pi_i(\widetilde X,\widetilde X^{\,n})=0\qquad(i\le n). \] Since \(\widetilde X\) is contractible, the long exact sequence gives \[ \pi_i(\widetilde X^{\,n})=0\qquad(i<n). \] Thus \(\widetilde X^{\,n}\) is \((n-1)\)-connected. Hurewicz gives \[ \pi_n(\widetilde X^{\,n})\cong H_n(\widetilde X^{\,n};\mathbb Z). \] Because \(\widetilde X^{\,n}\) has no \((n+1)\)-cells, \[ H_n(\widetilde X^{\,n})=\ker\bigl(d_n:C_n\to C_{n-1}\bigr), \] a subgroup of the free abelian group \(C_n\), hence itself free abelian.
Covering maps induce isomorphisms on \(\pi_n\) for \(n\ge2\), so \[ \boxed{\pi_n(X^n)\cong\pi_n(\widetilde X^{\,n})\text{ is free abelian}.} \]