Show that for a short exact sequence of abelian groups \(0 \to A \to B \to C \to 0\), a Moore space \(M(C, n)\) can be realized as a quotient \(M(B, n)/M(A, n)\). Applying the long exact sequence of cohomology for the pair \((M(B,n), M(A,n))\) with any coefficient group \(G\), deduce an exact sequence
\[0 \to \operatorname{Hom}(C,G) \to \operatorname{Hom}(B,G) \to \operatorname{Hom}(A,G) \to \operatorname{Ext}(C,G) \to \operatorname{Ext}(B,G) \to \operatorname{Ext}(A,G) \to 0.\]
Solution
Choose Moore spaces functorially enough for the given short exact sequence \[ 0\to A\xrightarrow{i}B\xrightarrow{q}C\to0. \] For \(n\ge1\), realize \(i\) by a cellular map \[ M(A,n)\longrightarrow M(B,n) \] inducing \(i\) on the only nonzero reduced homology group. Replace it by a mapping-cylinder inclusion, so we may regard \(M(A,n)\) as a subcomplex of \(M(B,n)\). Let \[ Q=M(B,n)/M(A,n). \] The reduced homology long exact sequence of the pair gives \[ \widetilde H_j(Q)=0\quad(j\ne n), \qquad \widetilde H_n(Q)\cong B/A\cong C. \] Thus \(Q\) is a Moore space \(M(C,n)\) (up to the standard Moore-space homotopy uniqueness in this range), so \[ M(C,n)\simeq M(B,n)/M(A,n). \]
Apply cohomology with coefficients \(G\) to the pair \((M(B,n),M(A,n))\). By the universal coefficient theorem for Moore spaces, \[ \widetilde H^n(M(D,n);G)\cong\operatorname{Hom}(D,G), \] \[ \widetilde H^{n+1}(M(D,n);G)\cong\operatorname{Ext}(D,G), \] and all other reduced cohomology groups vanish. The long exact sequence of the pair, together with \[ H^*(M(B,n),M(A,n);G)\cong\widetilde H^*(M(C,n);G), \] therefore becomes exactly \[ 0\to\operatorname{Hom}(C,G)\to\operatorname{Hom}(B,G)\to \operatorname{Hom}(A,G)\to\operatorname{Ext}(C,G)\to \operatorname{Ext}(B,G)\to\operatorname{Ext}(A,G)\to0. \]