An extension of a ring $C'$ by an ideal $N$ can be expressed by an exact sequence $0\to N\to C\to C'\to 0$. Such an extension is called a Hochschild extension if $C$ is isomorphic to $C'\oplus N$ as additive group, while the multiplication is given by $(a,x)(b,y)=(ab,ay+bx+f(a,b))$ $(a,b\in C';x,y\in N )$, where the map $f:C'\times C'\to N$ is symmetric and bilinear satisfying the cocycle condition:
$af(b,c)-f(ab,c)+f(a,bc)-f(a,b)c=0$
Then, it's said that this extension is trivial(the exact sequence is trivial) iff there exists a function $g:C'\to N$, s.t. $f(a,b)=ag(b)-g(ab)+g(a)b$.
I don't know why this follows? Hope someone could help. Thanks!
The sequence splits precisely if such $g$ exists. The splitting is given by $a \mapsto (a, -g(a))$. To check it is a homomorphism do: $$ (a, -g(a))(b, -g(b)) = (ab, -ag(a) - g(b)b + f(a,b)) = (ab, -g(ab))$$