Suppose that are projective and the bottom row is exact.

Then there exists a lift of to a chain map . Any two such lifts and are homotopic.

Given chain maps and , we want to find a chain homotopy such that

Recall: Let be a right exact functor . Let be an -module. We find a projective resolution

We drop , apply and take homology. In this case,

Now if and are two projective resolutions of , we can compare them using the Theorem (Comparison Theorem)Theorem (Comparison Theorem):

Because is a lift, and is also a lift, they are homotopic. Swapping around the roles of and , then we find that and are also homotopic.

Aside: If we have

then applying , we also have

Recall: Homotopic maps induce the same map on homology.

and induce the same map on homology. Similarly, and induce the same map on homology. Therefore . This tells us that is well-defined on objects.

On morphisms: Let and take projective resolutions and . Then we can lift to a chain map :

We apply

A chain map induces a map on homology:

This map is well-defined: two different choices of chain maps are homotopic, and so give the same map on homology.

The same arguments work for right derived functors by changing the directions of all arrow, and we swap:

Let be a group. Let be its group algebra. The category of -modules is the same as the category of -modules, it is the category of abelian groups with a -action that preserve the abelian group structure:

We defined an invariants functor

where . Alternatively,

where is with the trivial -action, i.e. must be sent to an element which is fixed by . Hence, we can also write

Group cohomology is the th right derived functor of invariants. Alternatively, from the point of view,

Instead of using an injective resolution of , we can use a projective resolution of .

Let be any ring and a -algebra (i.e., ). We will write . Consider the following complex:

The differentials are given by:

So the first differential is the multiplication map: it is surjective because we assume that . The above is an exact (to be proved) complex of -bimodules, where the -bimodule structure is the obvious one:

Given an -module , we can apply to the above exact complex to get a projective resolution of (to be proved: why is it exact? why is it projective?).

For example, take , , and . Then

So the bar resolution is

The maps are: