Let $G=〈\sigma〉$ be a cyclic group of order $2$ generated by $\sigma$. Let $A$ be $G$-module. Let $H^1(G,A)$ be a first group cohomology.
I want to prove $H^1(G,A)\cong \text{ker}(\sigma +1)/(\sigma -1)A$.
Let define a map $\phi:H^1(G,A)\to \text{ker}(\sigma+1)/(\sigma-1)A $ by $[f]\to f(\sigma)+(\sigma-1)A$. This is well-defined because $(\sigma+1)f(\sigma)=\sigma f(\sigma)+f(\sigma)=f({\sigma}^2)=f(0)=0$.
Let prove $\phi$ is injective. When $f(\sigma) \in (\sigma-1)A$, there exists $m\in A$ such that $f(\sigma)=\sigma m-m$. This means $f$ is coboundary, thus $[f]=0$.
The last thing I should do is to prove $\phi$ is surjective. What element of $H^1(G,A)$ is presage of arbitrary element in $\text{ker}(\sigma +1)/(\sigma -1)A$ ?
Thank you in advance.
Ok, we can do it this way if you wish. I infer you're using the "derivations / principal derivations" model for first cohomology. By the way, you didn't properly check well-definedness; you need to also check that if $g$ is a principal derivation then $g$ is mapped to zero. Ok, so $g(\sigma)=\sigma a-a$ for some $a$, thus $g$ maps to the class of $(\sigma-1)a$ which is truly zero in the quotient. All is well.
Say $a\in A$ has $\sigma a+a=0$. Simply define $f:G\to A$ by $\sigma\mapsto a,1\mapsto0$. Is this a genuine derivation? Yes; as $G$ has only two elements, the only nontrivial thing to check is whether or not $0=f(1)=f(\sigma^2)=\sigma f(\sigma)+f(\sigma)=\sigma a+a$... but this is true!
Therefore if $[a]\in\ker(\sigma+1)/(\sigma-1)A$ is any class, represented by some $a$, such an $f$ will be mapped to $[a]$ by your map $H^1(G;A)\to\ker(\sigma+1)/(\sigma-1)A$.
The general perspective is that: $$\cdots\to\Bbb ZG\overset{\sigma+1}{\to}\Bbb ZG\overset{\sigma-1}{\to}\Bbb ZG\overset{\sigma+1}{\to}\Bbb ZG\overset{\sigma-1}{\to}\Bbb ZG\overset{\epsilon}{\to}\Bbb Z$$Is a free resolution of the trivial module $\Bbb Z$ (check it) thus $H^1(G;A)$ is the first cohomology of: $$0\to\hom_G(\Bbb ZG,A)\overset{(\sigma-1)^*}{\to}\hom_G(\Bbb ZG,A)\overset{(\sigma+1)*}{\to}\hom_G(\Bbb ZG,A)\to\cdots$$Which is isomorphic to the complex: $$0\to A\overset{\sigma-1}{\to}A\overset{\sigma+1}{\to}A\to\cdots$$So you can compute in general that: $$H^n(G;A)\cong\begin{cases}\ker(\sigma+1)/(\sigma-1)A&n\text{ odd},\,n>0\\\ker(\sigma-1)/(\sigma+1)A&n\text{ even},\,n>0\\\ker(\sigma-1)&n=0\end{cases}$$You can do a similar thing for any finite cyclic group.