# the first cohomology of the shrinking wedge of circles is uncountable

The shrinking wedge of circles $$X$$, is the union of the circles $$C_n$$, $$n=1,2,3,\dots$$, where $$C_n$$ is the circle of radius $$1/n$$ and center $$(1/n,0)$$. I want to show that the first cohomology $$H^1(X;\Bbb Z)$$ is uncountable.

It is well-known that there is a surjective homomorphism $$\pi_1(X)\to \Bbb Z^\Bbb N$$, where $$\Bbb Z^\Bbb N$$ is the direct product of infinitely many copies of $$\Bbb Z$$, which is an uncountable group. Since $$H_1(X)$$ is the abelianization of $$\pi_1(X)$$, we also have a surjection $$H_1(X)\to \Bbb Z^\Bbb N$$, so $$H_1(X)$$ is also uncountable. Can we conclude from here that $$H^1(X;\Bbb Z)$$ is also uncountable? I tried to use the universal coefficient theorem, but it doesn't work so well, I think.

I believe I have read the first cohomology is countable. This is shown by computing the homology and explicitly taking its dual. See: https://web.math.rochester.edu/people/faculty/doug/otherpapers/eda-kawamura2.pdf for a computation of its first homology. Everything but the infinite product $$\Pi \mathbb{Z}$$ is killed off by taking the dual, and the dual of a countable infinite product is actually a countable free abelian group (see https://mathoverflow.net/questions/10239/is-it-true-that-as-bbb-z-modules-the-polynomial-ring-and-the-power-series-r/10249#10249).