Homomorphism in each coordinate implies cocycle for trivial group action

Statement

Suppose $G$ is a group and $A$ is an abelian group. Suppose $f : G^{n} \to A$ is a function with the property that, for all $i \in {1, 2, \dots, n}$ , if we fix the entries in all coordinates but the $i^{t h}$ coordinate, the induced function from $G$ to $A$ is a homomorphism of groups.

Then, $f$ is a $n$ -cocycle for trivial group action.

Particular cases

$n$	What being a homomorphism in each coordinate means	What being a $n$ -cocycle means	Link to $n$ -cocycle page
1	being a homomorphism of groups from $G$ to $A$	being a homomorphism of groups from $G$ to $A$	--
2	$f : G \times G \to A$ satisfies, for all $g_{1}, g_{2}, g_{3} \in G$ , both of these: $f (g_{1} g_{2}, g_{3}) = f (g_{1}, g_{3}) + f (g_{2}, g_{3})$ and $f (g_{1}, g_{2} g_{3}) = f (g_{1}, g_{2}) + f (g_{1}, g_{3})$	$f : G \times G \to A$ satisfies, for all $g_{1}, g_{2}, g_{3} \in G$ , the following: $f (g_{2}, g_{3}) + f (g_{1}, g_{2} g_{3}) = f (g_{1} g_{2}, g_{3}) + f (g_{1}, g_{2})$	2-cocycle for trivial group action
3	$f : G \times G \to A$ satisfies, for all $g_{1}, g_{2}, g_{3}, g_{4} \in G$ , all of these: $f (g_{1} g_{2}, g_{3}, g_{4}) = f (g_{1}, g_{3}, g_{4}) + f (g_{2}, g_{3}, g_{4}),$ $f (g_{1}, g_{2} g_{3}, g_{4}) = f (g_{1}, g_{2}, g_{4}) + f (g_{1}, g_{3}, g_{4}),$ $f (g_{1}, g_{2}, g_{3} g_{4}) = f (g_{1}, g_{2}, g_{3}) + f (g_{1}, g_{2}, g_{4})$	$f : G \times G \to A$ satisfies, for all $g_{1}, g_{2}, g_{3}, g_{4} \in G$ , the following: $f (g_{2}, g_{3}, g_{4}) + f (g_{1}, g_{2} g_{3}, g_{4}) + f (g_{1}, g_{2}, g_{3}) = f (g_{1} g_{2}, g_{3}, g_{4}) + f (g_{1}, g_{2}, g_{3} g_{4})$	3-cocycle for trivial group action