# 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)

-
If you know some Hochschild cohomology - not that one should be expected to, but I personally think it helps to motivate the Moore complex and Dold-Kan - then I think that what this construction would yield is the usual Hochschild coboundary operator (i.e. an alternating sum of coface maps), restricted in each degree to the subspace of normalized cochains (those cochains which vanish on all degeneracies). –  Yemon Choi Feb 12 '12 at 4:01
I find Chapter 8 of Weibel's Introduction to Homological Algebra a useful crutch for the basics of (co)simplicial stuff in homological algebra, so perhaps that might be worth a look. –  Yemon Choi Feb 12 '12 at 4:04

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.

-