Let \(X\) be a connected CW complex and \(G\) a group such that every homomorphism \(\pi_1(X) \to G\) is trivial. Show that every map \(X \to K(G,1)\) is nullhomotopic.
Solution
Let \[ f:X\to K(G,1) \] be any map and choose basepoints.
1. The induced homomorphism\[ f_*:\pi_1(X)\to\pi_1(K(G,1))\cong G \]is trivial.
This is exactly the hypothesis that every homomorphism from \(\pi_1(X)\) to \(G\) is trivial.
2. The map \(f\) lifts to the universal covering space\[ p:\widetilde K\to K(G,1). \]
1.Hence there is a lift\[ \widetilde f:X\to\widetilde K \]with \(p\widetilde f=f\).
3. The universal cover \(\widetilde K\) is contractible.
Because \(K(G,1)\) is aspherical, \[ \pi_n(K(G,1))=0 \qquad(n\ge2). \] A covering induces isomorphisms on all homotopy groups in degrees \(n\ge2\), while \[ \pi_1(\widetilde K)=0. \] Thus all homotopy groups of the CW complex \(\widetilde K\) vanish. By Whitehead’s theorem, the map \(\widetilde K\to *\) is a homotopy equivalence, so \(\widetilde K\) is contractible.
4. The map \(f\) is nullhomotopic.
Since \(\widetilde K\) is contractible, the lift \(\widetilde f\) is homotopic to a constant map. Composing this homotopy with \(p\) gives a homotopy from \[ f=p\widetilde f \] to a constant map in \(K(G,1)\).
5. Hence every map\[ \boxed{X\to K(G,1)} \]is nullhomotopic.
The map \(f\) was arbitrary.