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 6 The 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 7 If
, 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 8 If
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 10 If
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 11 A triangulated functor
between triangulated categories with
-structures is said to be left-exact if and only if
. Similarly, right-exactness is defined.
In the next result, we use the symbol to denote the heart of a triangulated category with a
-structure.
Proposition 12 If
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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