My question is related to this question, which I tried to post an answer. I think it is better to ask a question directly.
My question is from the book "Foundations of Quantum Theory: from classical concepts to operator algebras" by Klaas Landsman, on page 172, theorem 5.57.
I don't know how to prove surjectivity because the proof given in the book is clearly wrong.
To begin with, let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. Following Lie's third theorem, there exists a unique connected and simply connected Lie group $\widetilde{G}$ such that its Lie algebra is $\mathfrak{g}$, and $G\cong\widetilde{G}/\pi_{1}(G)$, where $\pi_{1}(G)$ is the first fundamental group of $G$. Thus, one has the following short exact sequence: $$1\xrightarrow{}\pi_{1}(G)\overset{\iota}{\hookrightarrow}\widetilde{G}\overset{\tilde{p}}{\twoheadrightarrow}G\xrightarrow{}1.$$
So the universal covering $\widetilde{G}$ is viewed as a central extension of $G$.
Theorem 5.57: Let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. If the second cohomology class of $\mathfrak{g}$ is trivial, i.e $\mathrm{H}^{2}(\mathfrak{g},\mathbb{R})=0$, then one has the isomorphism $$\widehat{\pi_{1}(G)}\cong\mathrm{H}^{2}(G,U(1)),$$ where $\pi_{1}(G)$ is the first fundamental group of $G$, and $\widehat{\pi_{1}(G)}=\mathrm{Hom}(\pi_{1}(G),U(1))$ is the Pontryagin dual of the first fundamental group.
Notice that the second cohomology group $\mathrm{H}^{2}(G,U(1))$ classifies central extensions of $G$ by $U(1)$, i.e $$1\xrightarrow{}U(1){\hookrightarrow}E\overset{\pi}{\twoheadrightarrow}G\xrightarrow{}1.$$
The above theorem states that the characters of the first fundamental group of $G$ is in one-to-one correspondence with the cohomology classes in $\mathrm{H}^{2}(G,U(1))$, i.e one has to find a map $$\chi\rightarrow[c_{\chi}],$$
where $\chi\in\widehat{\pi_{1}(G)}$, and $[c_{\chi}]\in\mathrm{H}^{2}(G,U(1))$, and show that it is an isomorphism.
The injectivity part is easy. One chooses a cross-section $\widetilde{s}:G\rightarrow\widetilde{G}$ of the canonical homomorphism $\widetilde{p}$, i.e $\widetilde{p}\circ\widetilde{s}=\mathrm{id}_{G}$, and $\widetilde{s}$ is assumed to be smooth in a neighborhood of the identity $1_{G}$, satisfying $\widetilde{s}(1_{G})=1_{\widetilde{G}}$.
With the above chosen cross-section, one can write down a cocycle $$\omega_{s}(g,h)=\widetilde{s}(g)\widetilde{s}(h)\widetilde{s}(gh)^{-1}\in\mathrm{Z}^{2}(G,\pi_{1}(G)),$$
where $g$, $h\in G$. Given a character $\chi\in\widehat{\pi_{1}(G)}$, with the above cocycle one can define $c_{\chi}:G\times G\rightarrow U(1)$ by $$c_{\chi}(g,h)=\chi\circ\omega_{s}(g,h)=\chi(\widetilde{s}(g)\widetilde{s}(h)\widetilde{s}(gh)^{-1}).$$
Then, it follows that $$c_{\chi}(g,h)c_{\chi}(gh,k)=c_{\chi}(g,hk)c_{\chi}(h,k),\quad\mathrm{and}\quad c(1_{G},g)=c(g,1_{G})=1,$$
where $g$, $h$, and $k\in G$. Thus, $c_{\chi}$ defines a cocycle in $\mathrm{Z}^{2}(G,U(1))$. Next, one can prove that the definition of $c_{\chi}$ is independent of the choice of the cross-section $\widetilde{s}$. One can pick up another cross-section $\widetilde{s}^{\prime}:G\rightarrow\widetilde{G}$ of the canonical homomorphism $\widetilde{p}$, with $\widetilde{s}^{\prime}(1_{G})=1_{\widetilde{G}}$. Then there exists a map $\alpha:G\rightarrow\widetilde{G}$ such that it is smooth in a neighborhood of $1_{G}$, and $\widetilde{s}=\alpha\widetilde{s}^{\prime}$. Then, $\mathrm{im}(\alpha)$ is in the center of $\widetilde{G}$. Now denote the cocycle induced by $\widetilde{s}^{\prime}$ as $c^{\prime}_{\chi}$, then \begin{align} c^{\prime}_{\chi}(g,h)c_{\chi}(g,h)^{-1}&=\chi(\widetilde{s}^{\prime}(g)\widetilde{s}^{\prime}(h)\widetilde{s}^{\prime}(gh)^{-1})\chi(\widetilde{s}(gh)\widetilde{s}(h)^{-1}\widetilde{s}(g)^{-1}) \\ &=\chi(\alpha(g)^{-1}\alpha(h)^{-1}\alpha(gh)) \\ &=\chi(\partial\alpha(g,h)). \end{align}
Thus, $c_{\chi}$ and $c_{\chi}^{\prime}$ induced by $\widetilde{s}$ and $\widetilde{s}^{\prime}$, respectively, differ by a coboundary, and therefore are cohomologous.
Finally, one has to show that the map $\chi\rightarrow[c_{\chi}]$ is surjective. To this end, one must show that given a central extension $$1\xrightarrow{}U(1){\hookrightarrow}E\overset{\pi}{\twoheadrightarrow}G\xrightarrow{}1,$$
for any given cohomology class $[c]\in\mathrm{H}^{2}(G,U(1))$, there is a character $\chi\in\widehat{\pi_{1}(G)}$ such that $c$ and $c_{\chi}$ differ by a coboundary, and so $[c]=[c_{\chi}]$.
This part of the proof given in the reference is clearly wrong. Can you tell me how to prove the surjectivity?
