I'm following this paper, trying to understand Corollary 5.
Essentially, there is a Lie Bracket $[,]:C^*\otimes C^*\to C^*$ on a cochain complex $C^*$ such that the the differential $d$ of the complex is a derivation of the bracket, i.e.
$$d[x,y]=[dx,y]+(-1)^{|x|}[x,dy].$$
The author says that from this it is easy to see that this map takes cocycles to cocycles and is independent of the choice of representative of the cohomology classes. I can see that if $dx=0=dy$ (both $x$ and $y$ are cocycles) then $d[x,y]=0$ by the above identity, so the bracket takes cocycles to cocycles. My problem comes when trying to show that the value on different representatives of the same cohomology class is the same on cohomology.
I can compute the bracket of another representative using linearity
$[x+da,y+db]=[x,y]+[x,db]+[da,y]+[da,db]$
If I assume that $[x,da]=[x,0]=0$, then I would be done, but I think $[x,da]=[x,0]$ is part of what I want to prove. Otherwise I would need to show that $[x+da,y+db]-[x,y]$ is a coboundary. But this difference is
$[x,db]+[da,y]+[da,db]$
and I don't see how this is $d(something)$.
I found out how to describe the $[x,db]+[da,y]+[da,db]$ as a coboundary.
It is $d((-1)^{|x|}[x,b]+[a,y]+[a,db])$. Here's the computation using the derivation property, note that $x$ and $y$ are cocycles, so they vanish when $d$ acts on them
$$d((-1)^{|x|}[x,b]+[a,y]+[a,db])=(-1)^{|x|}[dx,b]+[x,db]+[da,y]+(-1)^{|a|}[a,dy]+[da,db]+(-1)^{|a|}[a,ddb]=$$
$$[x,db]+ [da,y]+[da,db]$$
as I wanted.
P.S.: mathematicians, if something is easy, do it or at least indicate it, this is easy once you see it, but I don't think it's so obvious.