Recovering a character from a cohomology class Let $G$ be a finite group, and consider $\mathbf Z, \mathbf Q$ and $\mathbf Q/\mathbf Z$ as trivial $G$-modules. Then $H^1(G, \mathbf Q/\mathbf Z) = {\rm Hom}(G, \mathbf Q/\mathbf Z) = \widehat{G}$. Moreover, for a finite group, we have $H^i(G, \mathbf Q)=0$ for $i>0$. The long exact sequence of cohomology coming from $\mathbf Z \to \mathbf Q \to \mathbf Q/\mathbf Z$ therefore induces an isomorphism
$$\widehat{G} = H^2(G, \mathbf Z).$$
The group $H^2(G, \mathbf Z)$ is in bijection with central extensions of $G$ by $\mathbf Z$. Using this, we can describe the map $\epsilon: \widehat{G} \to H^2(G, \mathbf Z)$ realizing the above isomorphism: namely, given a character $\chi : G \to \mathbf Q/\mathbf Z$, we can pull-back the sequence $\mathbf Z \to \mathbf Q \to \mathbf Q/\mathbf Z$ along $\chi$ to get a sequence $\mathbf Z \to E \to G$, which is a central extension of $G$ by $\mathbf Z$.
Of course the map $\epsilon$ exists even when $G$ is not finite, but when $G$ is finite, $\epsilon$ is an isomorphism. I am wondering if there is a simple way of describing the inverse of $\epsilon$ in the case where $G$ is finite. Namely, given a central extension $\mathbf Z \to E \to G$, how can one construct the corresponding character of $G$?
 A: $\require{AMScd}\def\ZZ{\mathbb Z}\def\Tor{\operatorname{Tor}}\def\QQ{\mathbb Q}$
Let $0\to\ZZ\to E\to G\to0$ be a central extension, let $t$ be a generator of the kernel of $E\to G$, and let $\xi=t-1\in\ZZ E$. Then $\xi$ is a central non-zerodivisor of $\ZZ E$. In particular, we have an exact sequence $$0\to\ZZ E\xrightarrow{\xi} \ZZ E\to\ZZ G\to 0$$ which is a projective resolution of $\ZZ G$ as a left or right $\ZZ E$-module. Using it to compute, we immediately find that $$\Tor^{\ZZ E}_1(\ZZ G,\ZZ)\cong\ZZ.$$
Now let $\epsilon:\ZZ E\to\ZZ$ be the usual augmentation, $I$ its kernel and consider the short exact sequence $0\to I\to \ZZ E\to\ZZ\to 0$ of left $\ZZ E$-modules. The long exact sequence which results from applying the functor $\ZZ G\otimes_{\ZZ E}(\mathord-)$ to it gives us the exact sequence $$0\to\Tor^{\ZZ E}_1(\ZZ G,\ZZ)\to\ZZ G\otimes_{\ZZ E}I\to\ZZ G\otimes_{\ZZ G}\ZZ e\ZZ G\otimes_{\ZZ G}\ZZ\to 0$$ which is, up to standard isomorphisms and the isomorphism we found above the same as $$0\to\ZZ\to I/\xi I\to \ZZ G\to \ZZ\to0$$ The map $\ZZ G\to\ZZ$ appearing here is the augmentation of $\ZZ G$, whose kernel is the usual augmentation ideal $J\subseteq\ZZ G$, so this exact sequence gives us a short exact sequence $$0\to\ZZ\to I/\xi I\to J\to0$$ Now there is a projective presentation of $J$ as a left $\ZZ G$-module of the form $$\ZZ G\otimes \ZZ G\otimes\ZZ G\xrightarrow{f_1}\ZZ G\otimes \ZZ G\xrightarrow{f_0} J\to 0$$ with $f_0(g_1\otimes g_2)=g_1(g_2-1)$ and $f_1(g_1\otimes g_2\otimes g_3)=g_1g_2\otimes g_3-g_1\otimes g_2g_3+g_1\otimes g_2$, and $\ZZ G$ acting on $\ZZ G\otimes\ZZ G\otimes\ZZ G$ and on $\ZZ G\otimes\ZZ G$ «only on the first factor.» The usual properties of projective resolutions implies that there are vertical arrows of $\ZZ G$-modules which make the following diagram commute:
$$\begin{CD}
\ZZ G\otimes \ZZ G\otimes\ZZ G
  @>{f_1}>>
  \ZZ G\otimes \ZZ G
  @>{f_0}>> 
  J
  @>>> 0
  \\
@VVV @VVV @VVV
  \\
\ZZ
  @>>>
  I/\xi I
  @>>>
  J
  @>>>
  0
\end{CD}$$
Not let $\ZZ\to\QQ$ be the inclusion. We can complete the above diagram with a $\ZZ G$-module $X$ constructed so that the bottom left square is a pushout:
$$\begin{CD}
\ZZ G\otimes \ZZ G\otimes\ZZ G
  @>{f_1}>>
  \ZZ G\otimes \ZZ G
  @>{f_0}>> 
  J
  @>>> 0
  \\
@VVV @VVV @VVV
  \\
\ZZ
  @>>>
  I/\xi I
  @>>>
  J
  @>>>
  0
  \\
@VVV @VVV @VVV
  \\
\QQ
  @>>>
  X
  @>>> 
  J
  @>>>
  0
\end{CD}$$
The composition going from $\ZZ G\otimes \ZZ G\otimes\ZZ G$ to $\QQ$ in this diagram is a cocycle on the bar complex for $G$, so it is a coboundary, as $H^2(G,\QQ)=0$. This means that it factors as the composition of the map $f_1$ and some map of $\ZZ G$-modules $\phi:\ZZ G\otimes\ZZ G\to\QQ$. I claim that the function $\psi:G\to \QQ/\ZZ$ such that $$\psi(g)=\phi(1\otimes g)+\ZZ$$ is a group homomorphism, and it is the map you want.
