Cohomology of Hawaiian earring?

Question

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).

Answer 1

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 $$H^1(\bigvee_{n \in \mathbb{N}} S^1;\mathbb{Z}) \cong \prod_{n \in \mathbb{N}} H^1(S^1;\mathbb{Z}) \cong \prod_{n \in \mathbb{N}} \mathbb{Z}.$$