Let \(\theset{A_i}^n \in \mathbf{Ab}\) be finitely generated, show \(\exists X \mid H_i(X) \cong A_i\) for \(i\leq n\) and 0 otherwise.
Solution
1. Write each finitely generated abelian group as\[ A_i\cong \mathbb Z^{r_i}\oplus\bigoplus_{j=1}^{s_i}\mathbb Z/d_{ij}, \qquad d_{ij}\ge2. \]
This is the structure theorem for finitely generated abelian groups.
2. For \(i\ge1\), realize the free summand \(\mathbb Z^{r_i}\) by a wedge of \(r_i\) copies of \(S^i\).
The reduced homology of \(S^i\) is \(\mathbb Z\) in degree \(i\) and zero elsewhere.
3. Realize each torsion summand \(\mathbb Z/d_{ij}\) in degree \(i\) by the Moore space\[ M(\mathbb Z/d_{ij},i)=S^i\cup_{d_{ij}}e^{i+1}. \]
Its cellular chain complex in positive degrees has \[ 0\to\mathbb Z\xrightarrow{d_{ij}}\mathbb Z\to0 \] concentrated in degrees \(i+1\) and \(i\), hence reduced homology \(\mathbb Z/d_{ij}\) in degree \(i\) and zero elsewhere.
4. Let \(X\) be the wedge of all these spheres and Moore spaces over \(1\le i\le n\).
There are finitely many summands because every \(A_i\) is finitely generated and only finitely many \(i\) occur.
5. Then for every \(1\le i\le n\),\[ \widetilde H_i(X;\mathbb Z)\cong A_i, \]and \(\widetilde H_k(X;\mathbb Z)=0\) for \(k>n\).
Reduced homology takes wedges of based CW complexes to direct sums. The only cells above dimension \(n\) are the \((n+1)\)-cells used in Moore spaces for torsion in \(A_n\), and their boundaries are injective multiplication maps, so they create no homology in degree \(n+1\).
6. If the requested \(A_0\) is \(\mathbb Z^c\) for some finite \(c\ge1\), take the disjoint union of \(c\) connected constructions, distributing the positive-degree summands among the components; then \(H_0\cong\mathbb Z^c\).
For any space, \(H_0\) is free abelian on its path components. Hence only free abelian groups can occur as \(H_0\). In the usual connected formulation one necessarily has \(A_0\cong\mathbb Z\).