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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?

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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 $H\to G$.

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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.

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  • $\begingroup$ Do you mean "commutator subgroup" instead of "centralizer"? $\endgroup$ – Joe Johnson 126 Jul 9 '14 at 1:16
  • $\begingroup$ Yes, that is it. I have forgotten more about math than most people will ever know... $\endgroup$ – Rene Schipperus Jul 9 '14 at 1:37

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