We showed that

This allows us to define derived functors. For example, if is left exact, , and has enough injectives, then we get the composition

Similarly, if is right exact and has enough projectives, we can define .

For , we can take an injective resolution

and we can replace in by the chain complex of injectives

Applying , we find that

Finally, the cohomology

Similarly with .

sends distinguished triangles to distinguished triangles.

Let

be a triangle in . WLOG, we can assume that the triangle is a standard triangle

a complex of injectives. Now applying , we have

We compute this termwise (since these are all complexes of injectives).
Claim: (by the explicit construction).
Therefore we have a triangle

To compute , , we can compute with acyclic resolutions instead, i.e., the objects have for all or for all .

In general, and don’t preserve . However, they do if the functors , are homologically bounded, i.e., there exists such that (the “old” derived functors) for all and .

If

is a complex, then the truncation is

Correspondingly there is also the other truncation

There are canonical maps:

Looking at these more closely, we have a short exact sequence of chain complexes

The bottom row is another possible truncation. This breaks up a complex into two complexes which are bounded. But now, a short exact sequence of complexes gives a distinguished triangle in .

Using truncations, we can “build up” objects from iterations of complexes concentrated in degree . If , then there exists a finite number of objects with , plus triangles:

Building using extensions:

and eventually .

Compare this with Jordan-Hölder series: if is a finite-dimensional -module, then there exists

such that is simple. is built from .

If is left exact and homologically bounded in the sense aboveabove, then for , is “built from” for . Since sends triangles to triangles, then is “built from” . Since is homologically bounded, we will have

if or . Since a triangle leads to a long exact sequence in cohomology, one chases these long exact sequences to find that if or . So (because we can truncate , and find that ).

So rather than defining as complexes which are bounded, we can define it as cohomology groups are bounded as well.

Consider

Assume that we have enough injectives and , are left exact, i.e., is left exact. Is

Let . We replace by , a complex of injectives. To compute :

A sufficient condition for this is if sends injectives to -acyclics. This is the same condition as in the Grothendieck spectral sequence.

Let and be abelian categories and … (sufficiently nice so we can derive functors). Let be an adjoint pair, i.e., . Suppose that is exact and is left exact.

Because has an exact left adjoint, then (by PropositionProposition). Let and (need this to derive ).

Suppose that for a complex of injectives which is bounded below. Then (using this theorem from Tuesdaythis theorem from Tuesday)

To see , consider the diagrams

We can certainly convert the top vertical maps to the bottom vertical maps by the adjunction. The question is whether we get chain complexes back. One can check this: this is due to the adjunction being functorial.

One uses the diagonal maps from adjunction to check this. One can also check that homotopy is preserved as well.