Show that the map \(\langle X, Y \rangle \to \operatorname{Hom}(\pi_n(X), \pi_n(Y))\), \([f] \mapsto f_*\), is a bijection if \(X\) is an \((n-1)\)-connected CW complex and \(Y\) is a path-connected space with \(\pi_i(Y) = 0\) for \(i > n\). Deduce that CW complexes \(K(G, n)\)’s are uniquely determined, up to homotopy type, by \(G\) and \(n\).
Solution
We use cellular obstruction theory. Since \(X\) is \((n-1)\)-connected, it is homotopy equivalent to a CW complex with one \(0\)-cell and no cells in dimensions \(1,\dots,n-1\). We may therefore assume this form.
A based map \(f:X\to Y\) is homotopic to the constant map on \(X^{n-1}\). On each \(n\)-cell its restriction determines an element of \(\pi_n(Y)\). The condition that these choices extend over the \((n+1)\)-cells is exactly that the resulting homomorphism on \(n\)-cycles factor through \[ H_n(X)\cong\pi_n(X) \] (the Hurewicz isomorphism). Thus a map determines a homomorphism \[ \pi_n(X)\to\pi_n(Y). \] Conversely, any such homomorphism prescribes compatible maps on the \(n\)-cells. All higher extension obstructions lie in groups \(\pi_i(Y)\) with \(i>n\), which are zero by hypothesis, so the map extends over all higher cells. The same vanishing says that two extensions inducing the same map on \(\pi_n\) are homotopic rel the \(n\)-skeleton.
Hence \[ \boxed{\langle X,Y\rangle\xrightarrow{\cong}\operatorname{Hom}(\pi_nX,\pi_nY).} \] For \(n=1\) the identical cellular argument is the usual classification of based maps into a \(K(G,1)\) by homomorphisms of fundamental groups.
Now let \(X\) and \(Y\) be CW complexes of type \(K(G,n)\). The identity homomorphism of \(G\) determines maps \[ f:X\to Y, \qquad g:Y\to X. \] The composites \(gf\) and \(fg\) induce the identity on \(\pi_n\). By the bijection just proved they are homotopic to the identity maps. Therefore \(f\) and \(g\) are homotopy inverses, so \[ \boxed{K(G,n)\text{ is unique up to homotopy type}.} \]