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Question: Is there a simple example of a space $X$ possessing torsion in its first integral cohomology group $H^1(X,\mathbb{Z})$?

For reasonable spaces $X$, e.g. CW-complexes, one has $H^1(X,\mathbb{Z}) \cong [X,\mathbb{T}]$, the group of homotopy classes of maps from $X$ to the circle group $\mathbb{T}$. I would be happiest if I could also see an explicit mapping $u : X \to \mathbb{T}$ whose homotopy class is torsion.

Mike F
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1 Answers1

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By the universal coefficient theorem $H^1(X)\cong\operatorname{Hom}(H_1(X),\Bbb Z)$ as $H_0$ is always free. Morphism groups to torsion-free groups are always torsion-free.

PVAL-inactive
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  • I see, thanks. Any thoughts as to how one might see $H^1(X)$ has no torsion directly from the description $H^1(X) \cong [X,\mathbb{T}]$? – Mike F Aug 07 '14 at 22:28
  • Well there's an obvious way to map $[X,\mathbb{T}]$ to $\operatorname{Hom}(\pi_1(X),\Bbb Z)$. Showing this map is an isomorphism likely boils down to a proof isomorphic to the proof of the universal coefficient theorem. – PVAL-inactive Aug 07 '14 at 22:54