I am working through problem 10.16 of Morandi's Field and Galois Theory, which is a guided computation of the second cohomology of a cyclic group of order $n < \infty$.
Let $G =\langle\sigma\rangle$ be cyclic of order $n$, let $G$ act on an Abelian group $M$, and let $f \in Z^2(G,M)$ be a $2$-cocylce. Let $m \in M^G = \{m \in M : \sigma m = m\}$ and define $$ f_m(\sigma^i,\sigma^j) = \begin{cases} 0 &\text{if} \ \ i + j < 0 \\ m &\text{if} \ \ i+j \geq n \end{cases} $$ for $i,j \in \{0,\dots, n-1\}$. I want to show that $f$ is cohomologous to $f_m$, where $m = \sum_{i=0}^{n-1} f(\sigma^i,\sigma)$. It isn't hard to show that $m \in M^G$, but I have no idea how to find a cochain $h : G \to M$ such that $$ \delta_1(h)(\sigma^i, \sigma^j) = \sigma^ih(\sigma^j) - h(\sigma^{i+j}) + h(\sigma^i) = (f - f_m)(\sigma^i,\sigma^j). $$ I have seen some computations of the cohomology of cyclic groups that use a lot of homological algebra, but I am very new to the subject so I haven't managed to use them to solve this problem. I understand that $h$ must have a piece-wise definition since $f_m$ does, but otherwise I feel like I am just taking shots in the dark looking for a good candidate for $h$.
So, I've toyed around with the problem (and checked up on $H^2$ groups), and I've come up with the following argument -- I can't help you see why, but it works up to a sign (are you sure you don't want $f$ cohomologous to $f_{-m}$?).
Let $h(\sigma^i)=\sum_{k=0}^{i-1}{f(\sigma^i,\sigma)}$ for $0 \leq i < n$.
For $0 \leq i,k \leq n$, $\sigma^i(f(\sigma^k,\sigma))=f(\sigma^{i+k},\sigma)+f(\sigma^i,\sigma^{k+1})-f(\sigma^i,\sigma^k)$.
It follows that if $0 \leq i,j < n$, $h(\sigma^i)+\sigma^ih(\sigma^j)=f(\sigma^i,\sigma^j)-f(\sigma^i,e_G)+\sum_{k=0}^{i+j-1}{f(\sigma^k,\sigma)}$.
But it is easy to see from the cocycle equation that $f(e_G,e_G)=0_M$ and then that $f(g,e_G)=0_M$. Therefore $\delta_1(h) = f+f_m$.