Coboundary map on the cochain complex of abelian cosimplicial groups? Maybe I'm looking at the wrong places, but I can't find a definition of
the coboundary map on the cochain complex of abelian cosimplicial groups.
What I have in mind is something similar to the "Moore construction in the
alternating face map complex "
for abelian simplicial groups, for example given in 
http://ncatlab.org/nlab/show/Moore+complex but this time obviously as a 
(co)boundary operator on a cochain complex.
Is there an obstruction to a definition? (I guess not)
If not how is it done? (Definition of the coboundary operator)
 A: If the coface maps are $d^i: C^{n-1} \to C^n$ $(i=0,...,n)$ then the coboundary map is 
$$\delta^n = \sum_{i=0}^n(-1)^i d^i: C^{n-1} \to C^n.$$ 
It might be helpful to keep in mind that the "co" refers to a dual concept, i.e. arrows are reversed. For example a face map $d_i: C_n \to C_{n-1}$  corresponds to a coface map $d^i: C^{n-1} \to C^n$. 
When thinking about what formula should hold or should be used for a definition you can proceed as follows: From linear algebra you know the concept of a dual space and a dual map (I denote them by an upper asterisk). As an example the face maps satisfy  $d_id_j=d_{j-1}d_i$ (if $i < j$). Now just think you had a vector space and apply $\ast$: 
$$(d_i d_j)^\ast = (d_{j-1} d_i)^\ast \;\;\text{ i.e. }\;\; d_j^\ast d_i^\ast = d_i^\ast d_{j-1}^\ast $$
Then set $d^i=d_i^\ast$ and you get the correct formula for the coface maps: $d^j d^i = d^i d^{j-1}$. 
In case of the coboundary map: The boundary map is $\sum_{i=0}^n (-1)^id_i:C_n \to C_{n-1}$. Dualizing yields the formula above. 
