I'm studying again cohomology of groups from Brown's book and other books and notes. I want to write some proof of well known statements but I have trouble with this one:
Let be $G$ a group and $M$ a left $G-$module. In order to define the cohomology groups $H^{n}(G,M)$: for $n\geq 0$ let $C^{n}(G,M)$ be the set of all functions from $G^{n}$ to $M$. The coboundary homomorphisms are defined by
$$\delta^{n+1}: C^{n}(G,M)\longrightarrow C^{n+1}(G,M)$$
$$\delta^{n+1}(\phi)(g_{1},...,g_{n+1})=g_{1}\phi(g_{2},...,g_{n})+\sum_{i=1}^{n}(-1)^{i}\phi(g_{1},...,g_{i}g_{i+1},...,g_{n+1})+(-1)^{n+1}\phi(g_{1},...,g_{n})$$
I want to proof that :
$$\delta^{n+1}\circ \delta^{n}=0$$
My Aproach: Let be $C_{n}(G)=G^{n+1}$ and then define $d_{n}:C_{n}\longrightarrow C_{n-1}$ by:
$$d_{n}(g_{1},...,g_{n+1})=\sum_{i=0}^{n}(-1)^{i}(g_{1},...,g_{i-1},g_{i+1},...,g_{n})$$
It is well known that :
$$d_{n}\circ d_{n+1}=0$$
Now apply the functor $F=Hom_{G}(\cdot, M)$ to the complex $C_{n}(G)$ and get the result. Is this a proof?
Can be the equation $\delta^{n+1}\circ\delta^{n}=0$ be proved in another way?