I was trying to understand [this interesting question](https://mathoverflow.net/questions/250287/g-cocycle-split-to-a-coboundary-in-j-via-a-group-extension) by example.


Let me follow their previous discussion and ask: Let a generic nontrivial 2-cocycle $\omega\_2^G(g\_1,g\_2) \in H^2(G,\mathbb{R}/\mathbb{Z})$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 2-cocycle $\omega\_2^G$ is a complex $U(1)=\mathbb{R}/\mathbb{Z}$ function with the norm $|\omega\_2^G|=1$ but with a $U(1)$ complex phase satisfying the cocycle condition.


I like to ask how can we trivialize the 2-cocycle $\omega\_2(g\_1,g\_2)$ of $G$ into 2-coboundary if we lift $G$ into a larger group $J$, and given that we know the group homomorphism $r$:
$$J \overset{r}{\rightarrow} G.$$
In particular I like to focus on:



> 
> $$SU(2) \overset{r}{\rightarrow} SO(3).$$
> 
> 
> 


So that  

$$\omega\_2^J(j\_1,j\_2)=\omega\_2^G(r(j\_1),r(j\_2))=\omega\_2^G(g\_1,g\_2) \text{ is trivial in } H^2(J,U(1)).$$
Namely $\omega\_2^G(r(j\_1),r(j\_2))$ becomes a 2-coboundary in $H^2(J,\mathbb{R}/\mathbb{Z})$ for the cohomology group of $J$, but $\omega\_2^G(g\_1,g\_2)$ originally was not a 2-coboundary but was a 2-cocycle for the cohomology group of $G$. We can explicitly write 
$$
\omega\_2^G(g\_1,g\_2)=\omega\_2^G(r(j\_1),r(j\_2))=
\frac{\beta\_1^J(j\_2)\beta\_1^J(j\_1)}{\beta\_1^J(j\_1 j\_2)}. 
$$
Here $\beta\_1^J(j\_1)$ is a 1-cochain for $j\_1, j\_2 \in J$, and that 
$g\_1=r(j\_1)$, $g\_2=r(j\_2)\in G$. 



> 
> Is that true that the 2-cocycle in
>  $$H^2[SO(3),\mathbb{R}/\mathbb{Z}]=\mathbb{Z}/2\mathbb{Z}$$
>  for $G=SO(3)$ can be trivialize in $J=SU(2)$? How can it be shown?
>  Here we consider the cohomology group of the Lie group $SO(3)$ and $SU(2)$.
> 
> 
> 



$SO(3)$ 2-cocycle trivialized to a 2-coboundary in $SU(2)$?
Let $G$ be a group, and let $$1\to A\to J\to G\to 1$$ be an extension of groups with an abelian kernel. Choose a set-theoretical lifting $s:G\to J$ of the quotient map $p:J\to G$. Now define a function $\beta:G^2\to A$ by the formula $$\alpha(g,h) = s(g)s(h)s(gh)^{-1}.$$
This formula defines a two cocycle, and this is the two cocycle which corresponds to the above extension. Let us now show that the inflation of $\alpha$ to $J$ is trivial: define $\beta(j) = j\cdot((sp)(j))^{-1}\in A$. Then a direct calculation shows that $\delta\beta=inf(\alpha)$. 


