A $2$-cocycle of $\mathbb{R}$ (valued in the unitary group $U(1)$) is a map $c:\mathbb{R}\times \mathbb{R}\to U(1)$ such that
- $ \ \ \ \ c(x+y,z)c(x,y)=c(x,y+z)c(y,z)$
- $ \ \ \ \ c(x,0)=c(0,x)=1$
I'm trying to prove that every $2$-cocycle is a $2$-coboundary, i.e. there is a function $\alpha:\mathbb{R}\to U(1)$ such that
$$c(x,y)=\frac{\alpha(x)\alpha(y)}{\alpha(x+y)}$$
Here's my attempt
My attempt
There are basically two candidates for $\alpha$, in fact
$$c(x,y)=\frac{\alpha(x)\alpha(y)}{\alpha(x+y)}\Rightarrow c(x,0)=\alpha(0)\Rightarrow \alpha(0)=1$$
So:
$$c(x,-x)=\alpha(x)^2\Rightarrow \alpha(x)=\pm \sqrt{c(x,-x)}$$
Basically, I've to prove that $c(x,y)^2=\frac{c(x,-x)c(y,-y)}{c(x+y,-x-y)}$.
Using the cocycles first property (with $z=-x-y$), we get:
$$c(x,y)^2=\frac{c(x,-x)c(y,-x-y)c(x,-x)c(y,-x-y)}{c(x+y,-x-y)^2}$$
Using again the cocycles first property (with $z=-x-y$), we have $c(x,-x)c(y,-x-y)=c(x+y,-x-y)c(x,y)$, so:
$$c(x,y)^2=\frac{c(x+y,-x-y)c(x,y)c(x,-x)c(y,-x-y)}{c(x+y,-x-y)^2}=\frac{c(x,-x)c(x,y)c(y,-x-y)}{c(x+y,-x-y)}$$
I'm almost there, but I can't conclude the proof. Could you help me?