For any space \(X\), show that \(H^1(X)\) is free abelian
Solution
1. As stated for an arbitrary space \(X\), the claim is false.
By the universal coefficient theorem, \[ H^1(X;\mathbb Z)\cong\operatorname{Hom}(H_1(X;\mathbb Z),\mathbb Z) \] when \(H_0(X)\) is free (as it always is), but the dual of an arbitrary abelian group need not be a free abelian group. For example, there exist torsion-free abelian groups \(A\) whose dual \(\operatorname{Hom}(A,\mathbb Z)\) is not free; such groups can be realized as \(H_1\) of connected CW complexes.
2. The standard valid version is: if \(H_1(X;\mathbb Z)\) is finitely generated (for example, if \(X\) is a finite CW complex), then \(H^1(X;\mathbb Z)\) is free abelian.
Write the finitely generated abelian group as \[ H_1(X;\mathbb Z)\cong\mathbb Z^r\oplus T \] with \(T\) finite. Then \[ H^1(X;\mathbb Z)\cong\operatorname{Hom}(H_1(X),\mathbb Z) \cong\mathbb Z^r, \] since every homomorphism from the torsion group \(T\) to the torsion-free group \(\mathbb Z\) is zero.