Assume is an abelian category.

Recall that to put an object in into (and ), we just put it in degree . Let us consider the map

We have:

So is automatically the map. A bit strange, but is this a big problem?

What is the cone of ? It is a complex

So for to be a triangle, we need . Is it? Is there a map between the complexes below?

There is always a map down, but not necessarily back up. The existence of the backwards map is related to the existence of splittings of the short exact sequence

For example,

has no map back. But it turns out the canonical downwards map is always a quasi-isomorphism:

If we invert it in , then the answer to whether is yes.

Let be a category, and a “subset”, to be the things we can invert.

An element of is a diagram:

with , thought of as “”, up to an equivalence relation: two of the above diagrams are said to be equivalent if there is a commutative diagram

where . The purple maps in practice often also lie in , but are not forced to by the axioms (sometimes morphisms satisfy a -out-of- property, so when we have a commutative triangle and two of the maps are in then the third is as well).

If and , then we indeed have this -out-of- property. If we have a commutative diagram

and , then since taking cohomology is a functor, we have . Now, we note that if is an isomorphism and is an isomorphism then is an isomorphism. Therefore and being quasi-isomorphisms implies that is a quasi-isomorphisms.

Note that if we force and to both be isomorphisms in our category, then we are forcing to be an isomorphism too.

We have two morphisms

with . By the axiomaxiom of the localising system, we can complete the middle maps into a square

where . Now, the composite is just the composition of the blue maps above

There are two ways to describe :

If we have a quasi-isomorphism and we take its cone

we get a long exact sequence in cohomologylong exact sequence in cohomology:

Having the blue column be is the same as having the green maps be isomorphisms. This gives us a way to talk about localisation in terms of objects instead of in terms of families of morphisms.

Let be a triangulated category. A null system is such that

Acyclic complexes in for abelian. (We need abelian to take cohomology.)

If is a null system, then we can define

is a localising system of morphisms.

Sometimes people use the notation for . Alternative point of view: we are making some objects (the ones in ) zero instead of trying to invert some maps.

By (TR1)(TR1),

is a triangle.

This follows from the octahedral axiomoctahedral axiom:

Since then . The condition that is null implies that . This implies that .

Given , we need to complete the square

so that also. We have the solid diagram

where is a triangle. Now, by (TR2)(TR2), there exists a such that is also a triangle:

Since , then . By (TR4)(TR4),

there exist the red horizontal maps which commute with the rest of the diagram.

The dual version is similar.

We are required to prove: if then the following are equivalent:

Consider the diagram

where we can factorise through by assumption. This lets us fill in the map at the top (by (TR4)(TR4)) and create a triangle (by (TR3)(TR3))

We have because . Last timeLast time, we checked that the composition of two consecutive maps in a triangle is .

So the top composition through is , and therefore the bottom composition .

Let be an abelian category. The derived category is either:

These are the same thing. There are also bounded above/below versions: , , . One might also want to impose additional restrictions on what cohomology groups are allowed.

Let be an abelian category with enough injectives (i.e., every object can be embedded into an injective one). If is a bounded below complex of injectives, then

There is of course a dual statement for projectives.

Let be an abelian category with enough injectives (i.e., every object can be embedded into an injective one). Then

(equivalence of triangulated categories) where is the additive category of injectives.

There is an obvious way to define a functor . To check that the functor is essentially surjective, one constructs injective resolutions.

A bounded below complex of injectives which is acyclic is then in the homotopy category (you can split everything).