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Hatcher exercise 3.1.5 contains the following: regarding a cochain $\phi \in C^1(X,G)$ as a function from paths in $X$ to $G$, show that if $\phi$ is a cocycle then (among other things) $\phi(f \cdot g) = \phi(f) + \phi(g)$ and $\phi(f) = \phi(g)$ is $f \simeq g$. Hatcher says that these two statements give a map $H^1(X,G) \to \text{Hom}(\pi_1(X), G)$, which the universal coefficient theorem says is an isomorphism if $X$ is path-connected. But I don't really understand why these two facts are needed to prove this statement; isn't the following enough?

Proof that $H^1(X;G) \cong \text{Hom}(\pi_1(X), G)$: By the universal coefficient theorem, $H^1(X;G)$ is naturally isomorphic to $\text{Hom}(H_1(X), G)$. If $X$ is path-connected, then $H_1(X)$ is naturally isomorphic to the abelianization of $\pi_1(X)$. By the universal property of abelianization, every map in $\text{Hom}(\pi_1(X), G)$ factors uniquely through $H_1(X)$. This shows that $H^1(X;G)$ is naturally isomorphic to $\text{Hom}(\pi_1(X), G)$.

Is there something wrong with my proof? Do we need the facts stated in the first paragraph? I've found this thread, but I'm still confused.

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No, there is nothing wrong with your proof. In Hatcher's exercise you just define explicitly the isomorphism $H^1(X, G) \to \operatorname{Hom}(\pi_1(X), G)$, instead of just knowing that it exists.

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