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# Cohomologies.

Yoshifumi Tsuchimoto

Let be a group. Let us consider a functor

The functor is left-exact. The derived functor of this functor

is called the -th cohomology of with coefficients in . Let us consider as a -module with trivial -action. Then we may easily verify that

Thus we have

To compute cohomologies of , it is useful to use -resolution of . For any tuples of , we introduce a symbol

and we consider the following sequence

 ( )

where are determined by the following rules.

To see that the sequence is acyclic, we consider a homotopy

EXERCISE 09.1   Show that

LEMMA 09.1
1. Each of the modules that appears in the sequence admits an action of determined by

2. is -free

There are several choices for the -basis of . One such is clearly

It is traditional (and probably useful) to use another basis

where

Conversely we have

DEFINITION 09.2   For any group , the derived functor of a functor

defined by

span

is called the homology of with coefficients in . We denote the homology group by .

LEMMA 09.3

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2010-06-23