# Homomorphism in each coordinate implies cocycle for trivial group action

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.
$n$ 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: $\! f(g_1g_2,g_3) = f(g_1,g_3) + f(g_2,g_3)$ and $\! f(g_1,g_2g_3) = f(g_1,g_2) + f(g_1,g_3)$ $f:G \times G \to A$ satisfies, for all $g_1,g_2,g_3 \in G$, the following: $\! f(g_2,g_3) + f(g_1,g_2g_3) = f(g_1g_2,g_3) + f(g_1,g_2)$ 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_1g_2,g_3,g_4) = f(g_1,g_3,g_4) + f(g_2,g_3,g_4),$$\! f(g_1,g_2g_3,g_4) = f(g_1,g_2,g_4) + f(g_1,g_3,g_4),$ $\! f(g_1,g_2,g_3g_4) = f(g_1,g_2,g_3) + f(g_1,g_2,g_4)$ $f:G \times G \to A$ satisfies, for all $g_1,g_2,g_3,g_4 \in G$, the following: $\! f(g_2,g_3,g_4) + f(g_1,g_2g_3,g_4) + f(g_1,g_2,g_3) = f(g_1g_2,g_3,g_4) + f(g_1,g_2,g_3g_4)$ 3-cocycle for trivial group action