What is the differential $d^0(f) : C^0(G, A) \to C^1(G, A)$ knowing the general formula?

Question

Romyar Sharifi's Lecture Notes

It's from the first differential formula appearing on page 7 and that formula is:

$$ d^i : C^i(G, A) \to C^{i+1}(G, A), \\ d^i(f)(g_0, \dots, g_i) = g_0 f(g_1, \dots, g_i) + \sum\limits_{j=1}^i (-1)^j f(g_0, \dots, g_{j-2}, g_{j-1}g_j, g_{j+1}, \dots, g_i) + \\ (-1)^{i+1} f(g_1, \dots, g_i) $$

where $A$ is a $\Bbb{Z}[G]$-module, $G$ is a group, and $C^i(G, A) = \{ f: G^i \to A \vert f \text{ is a } \textbf{function} \}$.

Two questions come to mind in proving that $d^{i+1} \circ d^i = 0$ (i.e. $d$ is indeed a differential of a cochain complex).

1. Does the author really mean "function", i.e. any set map even if it's not a group homomorphism?
2. Given the formula, what is $d^0(f)(g_0)$?

I am not sure about the first, but for the second my guess is:

$$ d^0(f)(g_0) = g_0 $$

Is that correct?

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Note $G^i = G \times \dots \times G \ (i \text{ times})$.