Hatcher 3.3.21 (quoted below for completeness):
For a space $X$ , let $X^+$ be the one-point compactification. If the added point, denoted $\infty$, has a neighborhood in $X^+$ that is a cone with $∞$ the cone point, show that the evident map $H^n_c (X; G) \to H^n (X^+, \infty; G)$ is an isomorphism for all n. [Question: Does this result hold when $X = \mathbb Z×\mathbb R$?]
Here $H_c^n$ is cohomology with compact support. I'm specifically concerned about the question part, with coefficients $G = \mathbb Z$. It is straightforward to compute $H_c^1 (X) = \mathbb Z^{\oplus \mathbb Z}$ in this case, and the one-point compactification is the Hawaiian earring. So the question essentially amounts to asking about the cohomology of the Hawaiian earring.
I gathered that $H_1 (X^+) = \mathbb Z^{\mathbb Z} \oplus (\mathbb Z^{\mathbb Z} / \mathbb Z^{\oplus\mathbb Z})$. The first factor has a clear geometric description corresponding to paths that loop around each circle the specified number of times. Since $H_0 = \mathbb Z$ is free, the universal coefficient theorem says the obvious morphism $H^1(X^+) \to \hom(H_1(X^+), \mathbb Z)$ is an isomorphism. By Specker's theorem, $\hom(\mathbb Z^{\mathbb Z}, \mathbb Z) \cong \mathbb Z^{\oplus\mathbb Z}$, freely generated by the evaluation morphisms $\alpha \mapsto \alpha_n$. So the dual of the second factor must be zero. Therefore $H^1 (X^+) \cong \mathbb Z^{\oplus \mathbb Z}$, and considering the geometric interpretation, the isomorphism seems to be given by the obvious one $H_c^1 (X) \to H^1 (X^+)$.
Is this argument correct? This doesn't seem to be doable (unless Hatcher expects me to work out the homology of $X^+$ and prove Specker's theorem by myself), and some of the answer keys I read gave the opposite conclusion.