If is an additive category, then is triangulated. So are , , .

Suppose we have . By definition, every distinguished triangle is isomorphic to a “standard” triangle, i.e., one given by

So if we want to rotate this triangle, we must consider . We get the solid commutative diagram (with dashed maps yet to be filled in)

The bottom row is a distinguished triangle by definition. We want the upper row to also be a distinguished triangle. So the question is: are and isomorphic? By construction,

(Clearly these are not isomorphic as chain complexes.) We construct and as follows:

We have . We must show that . Written out, we have

We need some (depicted in blue above) such that

Taking works. (Note that the diagram above commutes.)

For the full proof that is triangulated, read Kashiwara-Shapiro Section 1.4 (and many other places).

A distinguished triangle is denoted

Let be a triangulated category. Let be an abelian category. A functor is cohomological if

Since the rotation of a distinguished triangle is a distinguished triangle, we get a doubly infinite long exact sequence

If , then is cohomological.

Suppose

is a distinguished triangle. We need to prove that

is exact. But as . Suppose now that , so we have the solid diagram:

By (TR3)(TR3) and (TR4)(TR4), there exists such that : So .

If

is a distinguished triangle, then .

By definition,

is a distinguished triangle. So we have the solid diagram:

and (TR4)(TR4) yields the dashed map. By commutativity of , .

is also cohomological once contravariantness is done.

This is well-defined by the fact that homotopy equivalences induce isomorphisms on cohomology.

is cohomological.

It suffices to prove this conditionthis condition for

because triangles can be rotated (TR3)(TR3). This is because given any triangle, we can rotate it, make sure it is isomorphic to , and then rotate it back, at which point it will be isomorphic to . So any triangle is isomorphic to for some .

Now, the result follows from the factfact that there exists a short exact sequence of complexes

A short exact sequence of complexes induces a long exact sequence on .

An issue: In , for , the map of chain complexes

is a quasi-isomorphism, but is not an isomorphism in . Perhaps we would like for these chain complexes to be the same, possibly motivated by the fact that the top complex is a projective resolution of the bottom complex. The derived category forces these to be isomorphic.

For a ring and with and multiplicatively closed. Then can be constructed as the elements ,

Let be an algebra (non-commutative):

Pain.

Let be a category. Let be a family of morphisms in . Then is a (two-sided) multiplicative system if

We think of this diagram as a morphism , and the condition allows us to think of this also as .

Quasi-isomorphisms satisfy the above conditionsabove conditions.

From and , there exists a category with the same objects as . There exists a canonical functor

If is a category and is a functor such that is an isomorphism for all , then