# Is there an isomorphism $\mathrm{Hom}(H_1(X),G)\simeq\mathrm{Hom}(\pi_1(X),G)$ when $X$ is path connected?

In Hatcher 3.1.5 on pg. 205, one proves that if $\varphi\in C^1(X;G)$ is a cocycle, where $X$ a space and $G$ an abelian group, then for paths $f$ and $g$ one has various properties $\varphi(f\cdot g)=\varphi(f)+\varphi(g)$, $\varphi$ sends constant paths to $0$, if $f\simeq g$, then $\varphi(f)=\varphi(g)$, and $\varphi$ is a coboundary iff $\varphi(f)$ only depends on the endpoints of $f$.

Using this, there is a map $H^1(X:G)\to\mathrm{Hom}(\pi_1(X),G)$. I see that since there is a map from the set of cocyles $\ker\delta_1\to\mathrm{Hom}(\pi_1(X),G)$ given by $\varphi\mapsto \bar{\varphi}$ where $\bar{\varphi}([f])=\varphi(f)$, which is well-defined, and which factors through the set of coboundaries to give a map on the cohomology group.

But Hatcher remarks that the universal coefficient theorem says this is an isomorphism if $X$ is path connected, which confuses me since $\pi_1(X)$ does not appear in the UCT. The UCT in the text says that there is an exact sequence $$0\to\mathrm{Ext}(H_{n-1}(X),G)\to H^n(X;G)\to\mathrm{Hom}(H_n(X),G)\to 0.$$ So if $X$ is path connected, I know $H_0(X)\simeq\mathbb{Z}$, which is free, so the Ext term vanishes, and we get an isomorphism $$H^1(X;G)\simeq\mathrm{Hom}(H_1(X);G).$$ Since $X$ is path connected, I know $H_1(X)$ is the abelianization of $\pi_1(X)$, does this imply $\mathrm{Hom}(H_1(X),G)\simeq\mathrm{Hom}(\pi_1(X),G)$ to get the conclusion?

Your question has nothing really to do with topology. For any abelian group $G$ and any group $H$ you would have $Hom(H,G)\simeq Hom(H/[H,H],G)$ since any homomorphism $H\to G$ kills $[H,H]$ and any homomorphism $H/[H,H]\to G$ can be lifted (in the obvious way) to $H\to G$.
Since $G$ is abelian any map $\pi_1(X)\rightarrow G$ will send the commutator to zero and define a map $H_1(X)\rightarrow G$. Conversely any map from $H_1(X)$ defines a map from the fundamental group by composition with the quotient map.