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)



Coboundary map on the cochain complex of abelian cosimplicial groups?
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. 


