Let \(X\) be a Moore space \(M(\mathbb{Z}_m, n)\) obtained from \(S^n\) by attaching a cell \(e^{n+1}\) by a map of degree \(m\).
Show that the quotient map \(X \to X/S^n = S^{n+1}\) induces the trivial map on \(\widetilde{H}_i(-; \mathbb{Z})\) for all \(i\), but not on \(H^{n+1}(-; \mathbb{Z})\). Deduce that the splitting in the universal coefficient theorem for cohomology cannot be natural.
Show that the inclusion \(S^n \hookrightarrow X\) induces the trivial map on \(\widetilde{H}^i(-; \mathbb{Z})\) for all \(i\), but not on \(H_n(-; \mathbb{Z})\).
Solution
The cellular chain complex of \[ X=M(\mathbb Z_m,n)=S^n\cup_m e^{n+1} \] in the two relevant degrees is \[ 0\to\mathbb Z\xrightarrow{\,m\,}\mathbb Z\to0. \] Hence \[ \widetilde H_n(X;\mathbb Z)=\mathbb Z_m, \qquad \widetilde H_i(X;\mathbb Z)=0\quad(i\ne n), \] while the cellular cochain complex gives \[ \widetilde H^{n+1}(X;\mathbb Z)=\mathbb Z_m, \qquad \widetilde H^i(X;\mathbb Z)=0\quad(i\ne n+1). \]
1. Let\[ q:X\to X/S^n\cong S^{n+1} \]be the quotient map. Then \(q_*\) is zero on all reduced integral homology groups.
The sphere has reduced homology only in degree \(n+1\), but \(\widetilde H_{n+1}(X)=0\). In degree \(n\), the target sphere has zero homology. All other reduced groups on both sides vanish.
2. Nevertheless\[ q^*:H^{n+1}(S^{n+1};\mathbb Z)=\mathbb Z \longrightarrow H^{n+1}(X;\mathbb Z)=\mathbb Z_m \]is reduction modulo \(m\), hence is nonzero.
On cellular cochains, the quotient identifies the unique \((n+1)\)-cell of \(X\) with the top cell of \(S^{n+1}\). Thus the cochain map in degree \(n+1\) is the identity \(\mathbb Z\to\mathbb Z\) before passing to cohomology. The target cohomology is the cokernel of multiplication by \(m\), so the induced map is the quotient \(\mathbb Z\to\mathbb Z_m\).
3. Therefore the splitting in the cohomological universal coefficient theorem cannot be natural.
A natural splitting would express \(H^{n+1}\) functorially as the direct sum of the Hom and Ext terms. Since \(q_*\) is zero on all homology groups, naturality would force both induced Hom and Ext maps, and hence \(q^*\), to be zero. This contradicts 2.
4. For the inclusion\[ i:S^n\hookrightarrow X, \]the induced map on reduced integral cohomology is zero in every degree.
The only nonzero reduced cohomology of \(X\) is in degree \(n+1\), while the sphere has nonzero reduced cohomology only in degree \(n\). Thus there is no degree in which both source and target of \(i^*\) are nonzero.
5. But\[ i_*:H_n(S^n;\mathbb Z)=\mathbb Z\to H_n(X;\mathbb Z)=\mathbb Z_m \]is reduction modulo \(m\), hence nonzero.
The \(n\)-cell of the sphere is the \(n\)-cell of \(X\). Passing from the cellular chain group to \(H_n(X)\) quotients by the image \(m\mathbb Z\) of the \((n+1)\)-cell boundary. Thus \(i_*\) is the quotient map \(\mathbb Z\to\mathbb Z_m\).