I know that the image of the preceding map, when equaling the kernel of the next map, means the complex is exact. Furthermore, this operator, listed below, is linear, but I am having a hard time explicitly calculating it. $E_n(A)$ is the set of all maps $f$ and has an abelian structure where map addition is as $f+g: G^n \to A :: x \mapsto f(x) + g(x) .$ The differntial map used is:
$d_n: E_n(A) \to E_{n+1}(A):: d_n(f)(g_0,\ldots,g_n) = g_0\cdot f(g_1,\ldots,g_n) + \sum_{i=1}^{n}\,(-1)^i f(g_0,\ldots,g_{i-2},(g_{i-1}\,g_i),g_{i+1},\ldots,g_n) + (-1)^{n+1}\,f(g_0,\ldots,g_{n-1})$ where $f: G^n \to A \in E_n(A)$ and the elements $g_0,\ldots,g_n\in G$. If $n=0$, this is meant to reduce to the following: if $a \in E_0(A) = A$ then $d_0(a) : G \to A$ is the map $d_0(f)(g) = g\,a - a.$
So in the explicit calculation, I start by showing, $d_{n+1}(f)(g_0,\ldots,g_{n+1}) = g_0\,f(g_1,\ldots,g_{n+1}) + \sum_{i=1}^{n+1}(-1)^i\,f(g_0,\ldots,g_{i-1}g_i,\ldots,g_{n+1}) + (-1)^{n+2}\,f(g_0,\ldots,g_n)$ and then apply on each of these three terms, $d_n(f)$.
For instance, the first term,
$(d_n \circ d_{n+1})(f)(g_0,\ldots,g_{n+1}) = d_n \circ \big( g_0\,f(g_1,\ldots,g_{n+1}) \big) = g_0\,d_n(f)(g_1,\ldots,g_{n+1}) = g_0\,g_1\,f(g_2,\ldots,g_{n+1}) + g_0\,\sum_{i=1}^n\,(-1)^i\,f(g_1,\ldots,(g_{i-1}g_i),\ldots,g_{n+1}) + g_0\,(-1)^{n+1}f(g_1,\ldots,g_n)$
(with corrections)
On the last term, I get, $(d_n \circ d_{n+1})(f)(g_0,\ldots,g_{n+1}) = d_n \circ \big( (-1)^{n+2}\,f(g_0,\ldots,g_n)\big) = (-1)^{n+2}\,d_n(f)(g_0,\ldots,g_n) = (-1)^{n+2}\,(-1)^{n+1}\,f(g_0,\ldots,g_{n-1}) = (-1)^{n+2}\,g_0\,f(g_1,\ldots,g_{n}) + (-1)^{n+2}\,\sum_{i=1}^n\,(-1)^i f(g_0,\ldots,(g_{i-1}g_i),\ldots,g_n) + (-1)^{n+2}\,(-1)^{n+1}\,f(g_0,\ldots,g_{n-1}) = (-1)\,f(g_0,\ldots,g_{n-1}) .$
(with corrections)
For the middle term I got,
$ d_n(f) \circ \left( \sum_{i=1}^{n+1}\,(-1)^i\,f(g_0,\ldots,(g_{i-1}g_i),\ldots,g_{n+1}) \right) = \sum_{i=1}^{n+1}\,(-1)^i\,d_n(f)(g_0,\ldots,(g_{i-1}g_i),\ldots,g_{n+1}) = nightmare! .$
This is 3 terms for each $i$ - how does one do this in general?
I'm pretty sure the first and last terms will cancel out in the composition of these neighboring differentials and that the summations will cancel out, but I cannot even get the terms not involing the summation to cancel. If one could show this explicitly I would greatly appreciate.
P.S. I forgot to add... For $f: G^2 = G \times G \to A$, how do I show:
Let $Z^2(A)$ be the set of factor systems, denote the set of maps $f$ s.t. $g\cdot f(g',g'') - f(g\,g',g'') + f(g,g'g'') - f(g,g') = 0$ for the $g$'s in $G$. Show that every element in $ker\;d_2$ gives rise to a factor system.
"gives rise"? Thank you for any clarification on that part too!