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Do the infinite wedge of circles and the Hawaiian earring have the same cohomology? I am happy that they have different homologies (the first is countably generated, the second uncountably).

Jeremy Judge
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    I doubt it, because I don't think $$\text{Hom}\left(\prod_{n=1}^\infty \mathbb Z, \mathbb Z\right)$$ is even $2^{\aleph_0}$-generated. –  Jun 04 '14 at 19:17
  • It may seem counterintuitive, but Specker's theorem says that $\mathrm{Hom}{\mathbb{Z}}(\prod{n \in \mathbb{N}} \mathbb{Z}, \mathbb{Z})$ is a free abelian group on countably many generators: $\mathrm{Hom}{\mathbb{Z}}(\prod{n \in \mathbb{N}} \mathbb{Z}, \mathbb{Z}) \cong \bigoplus_{n \in \mathbb{N}} \mathbb{Z}$. A nice write-up can be found here. – Martin Frankland Feb 17 '17 at 00:07

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No. Let $E$ denote the Hawaiian earring. Then the first cohomology groups are respectively $$H^1(E;\mathbb{Z}) \cong \bigoplus_{n \in \mathbb{N}} \mathbb{Z},$$ as explained here, and $$H^1(\bigvee_{n \in \mathbb{N}} S^1;\mathbb{Z}) \cong \prod_{n \in \mathbb{N}} H^1(S^1;\mathbb{Z}) \cong \prod_{n \in \mathbb{N}} \mathbb{Z}.$$

Martin Frankland
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