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 ${t}$-structure ${(\mathcal{D}_{\geq 0}, \mathcal{D}_{\leq 0})}$ on a triangulated category ${\mathcal{D}}$ always implies that ${\mathcal{D}}$ contains an abelian category ${\mathcal{C} = \mathcal{D}_{\geq 0} \cap \mathcal{D}_{\leq 0}}$, called the heart of ${\mathcal{D}}$. More is true, however:

Theorem 6 The functor ${H^0: \mathcal{D} \rightarrow \mathcal{C}}$ given by ${X \mapsto H^0(X) = \tau_{\geq 0} \tau_{\leq 0} X}$ is a cohomological functor.

This is, of course, familiar from the case when ${\mathcal{D}}$ 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

$\displaystyle X \rightarrow Y \rightarrow Z \rightarrow X[1],$

where, for starters, ${X, Y, Z}$ are in negative degrees: that is, they belong to ${\mathcal{D}_{\leq 0}}$. 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 ${H^1}$ will be zero. So we should have an exact sequence

$\displaystyle H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Z) \rightarrow 0.$

To see this, we have to show that whenever ${W \in \mathcal{C}}$, then the sequence

$\displaystyle 0 \rightarrow \hom(H^0(Z), W) \rightarrow \hom(H^0(Y), W) \rightarrow \hom(H^0(X), W)$

is exact. This will follow because ${\hom(H^0(Z), W) = \hom(Z, W)}$ as ${Z}$ is in ${\mathcal{D}_{\leq 0}}$. Similarly for the others. But we have an exact sequence

$\displaystyle 0 \rightarrow \hom(Z, W) \rightarrow \hom(Y, W) \rightarrow \hom(X, W)$

because ${\hom(X[1], W) = 0}$, and that would be the next term.

Thus we have:

Lemma 7 If ${X, Y, Z \in \mathcal{D}_{\leq 0}}$, then ${H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Z) \rightarrow 0}$ is exact in ${\mathcal{C}}$.

Now we want to make the same right-exactness claim when ${X \in \mathcal{D}_{\leq 0}}$ but the others are arbitrary. The intuition is, again, for complexes, that ${X}$ has no cohomology in degree one, which would make this claim true by the long exact sequence in cohomology.

Lemma 8 If ${X \in \mathcal{D}_{\leq 0}}$ and ${X \rightarrow Y \rightarrow Z \rightarrow X[1]}$ is a triangle, then ${H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Z) \rightarrow 0}$ is exact in ${\mathcal{C}}$.

To see this, we shall modify the triangle to get something as in the previous lemma. We start by fixing ${W \in \mathcal{D}_{\geq 1}}$. We note that ${\hom(X, W) = \hom(X[1], W) = 0}$. As a result, we have

$\displaystyle \hom(Z, W) \simeq \hom(Y, W).$

Thus, ${Z, Y}$ represent the same functor on the category ${\mathcal{D}_{\geq 1}}$. This means that their truncations ${\tau_{\geq 1} Z, \tau_{\geq 1} Y}$ are isomorphic.

Now we consider the composite

$\displaystyle X \rightarrow \tau_{\leq 0}Y \rightarrow Y.$

Note that ${X \in \mathcal{D}_{\leq 0}}$ factors through ${\tau_{\leq 0} Y}$.

We apply the octahedral axiom to this. There are triangles ${ X \rightarrow \tau_{\leq 0}Y \rightarrow Q \rightarrow}$, and ${X \rightarrow Y \rightarrow Z \rightarrow }$, and ${\tau_{\leq 0} Y \rightarrow Y \rightarrow \tau_{\geq 1} Y\rightarrow}$, where we don’t yet know what ${Q}$ is. The octahedral axiom states that there is a triangle

$\displaystyle Q \rightarrow Z \rightarrow \tau_{\geq 1} Y \rightarrow ,.$

but ${\tau_{\geq 1} Y = \tau_{\geq 1} Z}$. As a result, ${Q}$ is canonically isomorphic to ${\tau_{\leq 0} Z}$. (We haven’t proved the canonical part, but let’s ignore this; it follows from some more homological algebra.)

But we had a triangle ${X \rightarrow \tau_{\leq 0} Y \rightarrow Q}$, and ${Q = \tau_{\leq 0} Z}$. It follows that there is a triangle

$\displaystyle X \rightarrow \tau_{\leq 0} Y \rightarrow \tau_{\leq 0} Z,$

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 ${H^0}$ is right-exact, and we win.

Similar reasoning (or just plain duality) gives:

Lemma 9 (Dual) Let ${X \rightarrow Y \rightarrow Z \rightarrow }$ be a triangle, and suppose ${Z \in \mathcal{D}_{\geq 0}}$. Then ${0 \rightarrow H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Z)}$ is exact in the heart.

Finally, we should handle the general case. Let ${X \rightarrow Y \rightarrow Z \rightarrow }$ be a triangle. To show that ${H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Z)}$ is exact, the strategy will be to split it into two pieces using the octahedral axiom for the composite ${\tau_{\leq 0} X \rightarrow X \rightarrow Y}$. Namely, we can consider ${\tau_{\leq 0} X \rightarrow Y}$ and imbed it in a triangle

$\displaystyle \tau_{\leq 0} X \rightarrow Y \rightarrow Q \rightarrow,$

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

$\displaystyle H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Q) \rightarrow 0.$

We have three triangles out of ${\tau_{\leq 0} X \rightarrow X, X \rightarrow Y, \tau_{\leq 0} X \rightarrow Y}$. These give a triangle

$\displaystyle \tau_{\geq 1} X \rightarrow Z \rightarrow Q \rightarrow.$

We can rotate this to get a triangle

$\displaystyle Z \rightarrow Q \rightarrow (\tau_{\geq 1} X)[1] \rightarrow.$

Here the last term is in ${\mathcal{D}_{\geq 0}}$, which implies that the map

$\displaystyle H^0(Z) \rightarrow H^0(Q)$

is injective. Since we have an exact sequence ${H^0(X) \rightarrow H^0(Y) \rightarrow H^0(Q) \rightarrow 0}$ and since ${Y \rightarrow Z}$ factors through ${Q}$, we can use the fact that ${H^0(Q) \rightarrow H^0(Z)}$ is injective to get the result.

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

Proposition 10 If ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$ is an exact sequence in the heart, then ${A \rightarrow B \rightarrow C}$ is also a triangle.

What else could fit in a triangle after ${A \rightarrow B}$? Suppose some object ${X \in \mathcal{D}}$ fit in; then the triangle

$\displaystyle A \rightarrow B \rightarrow X \rightarrow A[1],$

shows first that ${H^0(X) = C}$. Moreover ${X \in \mathcal{D}_{\geq -1} \cap \mathcal{D}_{\leq 0}}$ by the triangle ${B \rightarrow X \rightarrow A[1] \rightarrow }$, and it suffices to show that ${\tau_{\leq -1} X = 0}$.

That is, it suffices to show that no object ${W \in \mathcal{D}_{\leq -1}}$ can map in a nonzero way into ${X}$. Since ${\hom(W, B) = 0}$, it suffices to show by the long exact sequence

$\displaystyle \hom(W, B ) \rightarrow \hom(W, X) \rightarrow \hom(W, A[1]) \rightarrow \hom(W, B[1])$

that the map ${\hom(W, A[1]) \rightarrow \hom(W, B[1])}$ is injective. But this follows because these maps are just ${\hom(H^{-1}W, A) \rightarrow \hom(H^{-1}W, B)}$ and ${A \rightarrow B}$ is a monomorphism.

2. ${t}$-exact functors

Let ${\mathcal{D}, \mathcal{E}}$ be triangulated categories. We have the notion of a triangulated functor ${F: \mathcal{D} \rightarrow \mathcal{E}}$; 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 ${\mathcal{A}, \mathcal{B}}$ are abelian categories (say, with enough injectives) and ${F: \mathcal{A} \rightarrow \mathcal{B}}$ is an additive functor, then the derived functor ${\mathbf{R} F: \mathbf{D}^+(\mathcal{A}) \rightarrow \mathbf{D}^+(\mathcal{B})}$ sends nonnegatively graded chain complexes to nonnegatively concentrated chain complexes.

Definition 11 A triangulated functor ${F: \mathcal{D} \rightarrow \mathcal{E}}$ between triangulated categories with ${t}$-structures is said to be left-exact if and only if ${F(\mathcal{D}_{\geq 0}) \subset \mathcal{E}_{\geq 0}}$. Similarly, right-exactness is defined.

In the next result, we use the symbol ${\flat }$ to denote the heart of a triangulated category with a ${t}$-structure.

Proposition 12 If ${F: \mathcal{D} \rightarrow \mathcal{E}}$ is left-exact, then the functor ${H^0 \circ F: \mathcal{D}^{\flat } \rightarrow \mathcal{E}^{\flat }}$ is left-exact.

This follows from the fact that if ${0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0}$ is exact in ${\mathcal{D}^{\flat }}$, then ${A \rightarrow B \rightarrow C \rightarrow }$ is a triangle. So ${F(A) \rightarrow F(B) \rightarrow F(C) \rightarrow F(A)[1]}$ is also a triangle. The long exact sequence in cohomology and the fact that ${F(C) \in \mathcal{D}_{\geq 0}}$ now give the result.