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).