I have to apologize for this post: while I have, as of late, been making efforts to make this blog more interesting and useful to outsiders (especially now that I have somewhat more readers than before), the present post will be a somewhat detailed walk-through of one of the first important results in BBD, and, as homological algebra, it is slightly on the technical side. Readers unfamiliar with the material may wish to skim the main result and skip the proof (or just read BBD for it).

**1. A cohomological functor**

We saw that a -structure on a triangulated category always implies that contains an *abelian* category , called the **heart** of . More is true, however:

Theorem 6The functor given by is a cohomological functor.

This is, of course, familiar from the case when is the derived category and then these are just the ordinary cohomology functors. In other words, it is a generalization of the long exact sequence in cohomology from a short exact sequence (triangle) of complexes.

So suppose we have a triangle

where, for starters, are in negative degrees: that is, they belong to . Then the intuition is that, if we pretend we’re working with complexes and that we have a short exact sequence of complexes (which is what a triangle is supposed to be a version of, after all), then will be zero. So we should have an exact sequence

To see this, we have to show that whenever , then the sequence

is exact. This will follow because as is in . Similarly for the others. But we have an exact sequence

because , and that would be the next term.

Thus we have:

Lemma 7If , then is exact in .

Now we want to make the same right-exactness claim when but the others are arbitrary. The intuition is, again, for complexes, that has no cohomology in degree one, which would make this claim true by the long exact sequence in cohomology.

Lemma 8If and is a triangle, then is exact in .

To see this, we shall modify the triangle to get something as in the previous lemma. We start by fixing . We note that . As a result, we have

Thus, represent the same functor on the category . This means that their truncations are isomorphic.

Now we consider the composite

Note that factors through .

We apply the octahedral axiom to this. There are triangles , and , and , where we don’t yet know what is. The octahedral axiom states that there is a triangle

but . As a result, is *canonically* isomorphic to . (We haven’t proved the canonical part, but let’s ignore this; it follows from some more homological algebra.)

But we had a triangle , and . It follows that there is a triangle

which is precisely the triangle we need to apply what we have already done: now we have a triangle all of whose terms live in nonpositive degrees, so is right-exact, and we win.

Similar reasoning (or just plain duality) gives:

Lemma 9 (Dual)Let be a triangle, and suppose . Then is exact in the heart.

Finally, we should handle the general case. Let be a triangle. To show that is exact, the strategy will be to split it into two pieces using the octahedral axiom for the composite . Namely, we can consider and imbed it in a triangle

where we don’t really know what looks like. However, we do get an exact sequence

We have three triangles out of . These give a triangle

We can rotate this to get a triangle

Here the last term is in , which implies that the map

is injective. Since we have an exact sequence and since factors through , we can use the fact that is injective to get the result.

Note that there is a reverse connection between triangles and exact sequences.

Proposition 10If is an exact sequence in the heart, then is also a triangle.

What else could fit in a triangle after ? Suppose some object fit in; then the triangle

shows first that . Moreover by the triangle , and it suffices to show that .

That is, it suffices to show that no object can map in a nonzero way into . Since , it suffices to show by the long exact sequence

that the map is injective. But this follows because these maps are just and is a monomorphism.

**2. -exact functors**

Let be triangulated categories. We have the notion of a **triangulated functor** ; this is an additive functor that takes triangles to triangles. However, in general we *always* want functors between triangulated categories to be of this form: derived functors, for instance.

Yet we want some sort of further distinction. If are abelian categories (say, with enough injectives) and is an additive functor, then the derived functor sends nonnegatively graded chain complexes to nonnegatively concentrated chain complexes.

Definition 11A triangulated functor between triangulated categories with -structures is said to beleft-exactif and only if . Similarly,right-exactnessis defined.

In the next result, we use the symbol to denote the heart of a triangulated category with a -structure.

Proposition 12If is left-exact, then the functor is left-exact.

This follows from the fact that if is exact in , then is a triangle. So is also a triangle. The long exact sequence in cohomology and the fact that now give the result.

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