Show there is a Moore space \(M(G, 1)\) with \(\pi_1\bigl(M(G, 1)\bigr) \approx G\) iff \(H_2(K(G, 1); \mathbb{Z}) = 0\). In particular, there is no \(M(\mathbb{Z}^n, 1)\) with fundamental group \(\mathbb{Z}^n\), free abelian of rank \(n\), if \(n \geq 2\).
Solution
1. A Moore space \(M(G, 1)\) is a connected CW complex with \(\pi_1 \cong G\) and \(\widetilde{H}_i = 0\) for \(i \neq 1\).
definition of a Moore space of type \((G, 1)\).
2. If \(M(G,1)\) exists, then \(H_2(K(G,1);\ZZ) = 0\).
2.1. \(M(G,1)\) is a \(K(G,1)\).
\(\pi_1 \cong G\) and \(\pi_i = 0\) for \(i \ge 2\) (since \(\widetilde{H}_i = 0\) for \(i \ge 2\) and \(\pi_1\) acts trivially, Hurewicz gives \(\pi_i = 0\)).
2.2. \(H_2(K(G,1);\ZZ) = H_2(M(G,1);\ZZ) = 0\).
1 and the Moore-space condition \(\widetilde{H}_2 = 0\).
3. Conversely, if \(H_2(K(G,1);\ZZ) = 0\), then \(M(G,1)\) exists.
3.1. Attach \(3\)-cells to \(K(G,1)\) to kill \(\pi_2\), then \(4\)-cells to kill \(\pi_3\), and so on.
standard construction of a Moore space from a \(K(G,1)\) by killing higher homotopy groups.
3.2. This does not change \(H_2\) (attaching cells of dimension \(\ge 3\) only affects homology in degrees \(\ge 2\) via the boundary maps, and \(H_2\) remains \(0\) since it was already \(0\)).
attaching a \(3\)-cell can only kill \(H_2\) or create \(H_3\); since \(H_2 = 0\) already, it stays \(0\).
3.3. The result is a \(K(G,1)\)-based complex with \(\pi_1 = G\) and \(\widetilde{H}_i = 0\) for \(i \neq 1\), i.e. a Moore space \(M(G,1)\).
1 and 2.
4. For \(G = \ZZ^n\) with \(n \ge 2\), \(H_2(K(\ZZ^n,1);\ZZ) = H_2(T^n;\ZZ) = \ZZ^{\binom{n}{2}} \neq 0\).
\(K(\ZZ^n,1) = T^n\) (the \(n\)-torus), and \(H_2(T^n) \cong \bigwedge^2 \ZZ^n \cong \ZZ^{\binom{n}{2}}\).
5. Hence there is no \(M(\ZZ^n, 1)\) for \(n \ge 2\).
2 (contrapositive) and 4.
6. Q.E.D.
2, 3, and 5.