Show that for a Moore space \(M(G, n)\) the Bockstein long exact sequence in cohomology associated to the short exact sequence of coefficient groups \(0 \to A \to B \to C \to 0\) reduces to an exact sequence
\[0 \to \operatorname{Hom}(G,A) \to \operatorname{Hom}(G,B) \to \operatorname{Hom}(G,C) \to \operatorname{Ext}(G,A) \to \operatorname{Ext}(G,B) \to \operatorname{Ext}(G,C) \to 0.\]
Solution
1. For a Moore space \(M(G, n)\), \(\widetilde H_n(M(G,n)) = G\) and \(\widetilde H_i(M(G,n)) = 0\) for \(i \neq n\).
definition of a Moore space.
2. By the universal coefficient theorem, \(H^n(M(G,n); A) \cong \operatorname{Hom}(G, A)\) and \(H^{n+1}(M(G,n); A) \cong \operatorname{Ext}(G, A)\), with all other cohomology groups (in positive degrees) zero.
1 and the universal coefficient theorem.
3. The Bockstein long exact sequence associated to \(0 \to A \to B \to C \to 0\) is\[\cdots \to H^i(M; A) \to H^i(M; B) \to H^i(M; C) \xrightarrow{\beta} H^{i+1}(M; A) \to \cdots.\]
the long exact sequence in cohomology from a short exact sequence of coefficients.
4. Since \(H^i(M(G,n); \cdot) = 0\) except for \(i = n\) and \(i = n+1\), the long exact sequence collapses to the exact sequence\[0 \to H^n(M;A) \to H^n(M;B) \to H^n(M;C) \to H^{n+1}(M;A) \to H^{n+1}(M;B) \to H^{n+1}(M;C) \to 0.\]
2 and 3 (all other terms vanish).
5. Substituting 2, this is exactly\[0 \to \operatorname{Hom}(G,A) \to \operatorname{Hom}(G,B) \to \operatorname{Hom}(G,C) \to \operatorname{Ext}(G,A) \to \operatorname{Ext}(G,B) \to \operatorname{Ext}(G,C) \to 0.\]
4.
6. Q.E.D.
5.