Construct a space \(X\) such that \(H_*(X; \ZZ) = (\ZZ, \ZZ_6, \ZZ_{12}, \ZZ \oplus \ZZ_4, 0 \cdots)\) Compute \(H^*(X; \ZZ)\)
Solution
1. For an abelian group \(A\) and \(r\ge1\), let \(M(A,r)\) denote a Moore space with\[ \widetilde H_i(M(A,r);\mathbb Z)\cong \begin{cases} A,&i=r,\\ 0,&i\ne r. \end{cases} \]For \(A=\mathbb Z/m\), one may take\[ M(\mathbb Z/m,r)=S^r\cup_m e^{r+1}, \]where the top cell is attached by a degree-\(m\) map.
Its cellular chain complex in the two positive dimensions is \[ 0\to\mathbb Z\xrightarrow{m}\mathbb Z\to0, \] so the only reduced homology is \(\mathbb Z/m\) in degree \(r\).
2. Take\[ X=M(\mathbb Z/6,1)\vee M(\mathbb Z/12,2) \vee S^3\vee M(\mathbb Z/4,3). \]Then\[ H_i(X;\mathbb Z)\cong \begin{cases} \mathbb Z,&i=0,\\ \mathbb Z/6,&i=1,\\ \mathbb Z/12,&i=2,\\ \mathbb Z\oplus\mathbb Z/4,&i=3,\\ 0,&i\ge4. \end{cases} \]
Reduced homology takes wedges of connected CW complexes to direct sums. Each summand contributes exactly the indicated group in its designated degree.
3. The cohomological universal coefficient theorem gives\[ 0\to\operatorname{Ext}(H_{i-1}(X),\mathbb Z) \to H^i(X;\mathbb Z) \to\operatorname{Hom}(H_i(X),\mathbb Z)\to0. \]
This is the universal coefficient short exact sequence for integral cohomology.
4. Therefore\[ \boxed{H^i(X;\mathbb Z)\cong \begin{cases} \mathbb Z,&i=0,\\ 0,&i=1,\\ \mathbb Z/6,&i=2,\\ \mathbb Z\oplus\mathbb Z/12,&i=3,\\ \mathbb Z/4,&i=4,\\ 0,&i\ge5. \end{cases}} \]
Use \[ \operatorname{Hom}(\mathbb Z/m,\mathbb Z)=0, \qquad \operatorname{Ext}(\mathbb Z/m,\mathbb Z)\cong\mathbb Z/m, \] and \[ \operatorname{Hom}(\mathbb Z\oplus\mathbb Z/4,\mathbb Z)\cong\mathbb Z, \qquad \operatorname{Ext}(\mathbb Z\oplus\mathbb Z/4,\mathbb Z)\cong\mathbb Z/4. \] In degree \(3\) the short exact sequence \[ 0\to\mathbb Z/12\to H^3(X)\to\mathbb Z\to0 \] splits because \(\mathbb Z\) is free, yielding \(H^3(X)\cong\mathbb Z\oplus\mathbb Z/12\).