Show that the functors \(h^n(X) = \operatorname{Hom}(H_n(X), \mathbb{Z})\) do not define a cohomology theory on the category of CW complexes.
Solution
The functors \[ h^n(X)=\operatorname{Hom}(H_n(X),\mathbb Z) \] fail the exactness axiom.
1. Let \(X=M(\mathbb Z_m,1)\) be obtained from \(S^1=A\) by attaching a \(2\)-cell by a map of degree \(m>1\).Then\[ H_1(A)=\mathbb Z,\qquad H_1(X)=\mathbb Z_m,\qquad H_2(X,A)=\mathbb Z. \]The boundary map\[ \partial:H_2(X,A)\to H_1(A) \]is multiplication by \(m\).
The relative group is generated by the attached \(2\)-cell, and its relative boundary is the attaching map, which has degree \(m\) on the circle.
2. Applying \(\operatorname{Hom}(-,\mathbb Z)\) gives\[ h^1(X)=0,\qquad h^1(A)=\mathbb Z,\qquad h^2(X,A)=\mathbb Z,\qquad h^2(X)=0, \]and the map induced by \(\partial\) is multiplication by \(m\):\[ \mathbb Z\xrightarrow{\,m\,}\mathbb Z. \]
One has \(\operatorname{Hom}(\mathbb Z_m,\mathbb Z)=0\) and \(\operatorname{Hom}(\mathbb Z,\mathbb Z)=\mathbb Z\). Contravariance sends multiplication by \(m\) to multiplication by \(m\).
3. If the \(h^n\) formed a cohomology theory, the long exact sequence of the pair would contain\[ 0=h^1(X)\longrightarrow h^1(A)=\mathbb Z \xrightarrow{\,m\,}h^2(X,A)=\mathbb Z \longrightarrow h^2(X)=0, \]which would have to be exact.
Exactness at \(h^2(X,A)\) would require multiplication by \(m\) to be surjective, which it is not for \(m>1\).
Hence these functors do not define a cohomology theory.