Firstly, the book doesn't really say what $C^{-1}(G, A)$ is so how can I take it's image under $d^{-1}$ in order to compute $H^0(G,A) = \ker d^0 / \text{im } d^{-1}$?
https://www.math.ucla.edu/~sharifi/groupcoh.pdf (remark on page 7)
Can someone please explain this trivial but important remark?
Wikipedia does the same thing as the book:
https://en.wikipedia.org/wiki/Group_cohomology#Definitions
It just states that $H^0(G,A)$ is taken to be $A^G$.
(Edit: They give a proof in the next paragraph ! :)
The notes say explicitly (Definition 1.2.4. ii) that $B^0(G,A)$ should be taken to be $0$ (i.e. the set consisting of the zero map), and so $H^0(G,A) = Z^0(G,A)$. $C^{-1}(G,A)$ and $d^{-1}$ are both undefined. The notes also tell you to consider $G^0$ to be the singleton set, and so $C^0(G,A)$ is identified with $A$ (i.e. $f\in C^0(G,A)$ is being identified with $f(*)$, where $*$ is the unique element of $G^0$). Then the elements of $C^0(G,A)$ which vanish under application of $d^0$ is just the elements $f$ satisfying $(d^0 f)(g_0) = g_0\cdot f(*)- f(*) = 0$ for all $g_0\in G$, i.e. the elements $a\in A$ such that $g\cdot a=a$. This is precisely $A^G$.