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 $i^{th}$ 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