$SO(3)$ 2-cocycle trivialized to a 2-coboundary in $SU(2)$?

Question

I was trying to understand this interesting question 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 $SU(2)$.

Answer 1

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 $\delta\beta=inf(\alpha)$.