Relations between functors in a recollement

Question

Consider a recollement situation like the following

by the very definition of the various functors it follows that $i^* j_*=0$, and $j^! i_* = 0 = j^* i_!$. Also, $j^! i_! = 0 = j^* i_*$ by inspection.

Are these "kernel properties" true in a general recollement situation? More precisely, let $$ \mathbf{D}^0 \underset{\underset{i_R}\leftarrow}{\overset{\overset{i_L}\leftarrow}\to} \mathbf{D} \underset{\underset{q_R}\leftarrow}{\overset{\overset{q_L}\leftarrow}\to} \mathbf{D}^1 $$ be a recollement where $i_L\dashv i\dashv i_R$ and $q_L\dashv q\dashv q_R$. From the axioms of recollement it follows that $qi=0$ implies $i_L q_L = 0 =i_R q_R$.

Is it true that also $i_R q_L = 0 = i_L q_R$?

Answer 1

I'll write my comment again here, so it appears as an answer.

If you have a recollement as in the question with the property that $i_Rq_L=0=i_Lq_R$, then the category $D$ splits as an orthogonal sum of $D^0$ and $D^1$; i.e. every object $d$ of $D$ can be written as a direct sum $d_0 \oplus d_1$ with $d_i \in D^i$, and there are no nonzero morphisms between $D^0$ and $D^1$ either direction. You can see this by looking at one of the distinguished triangles associated to the recollement and observing that the connecting morphism must be zero, and thus the triangle is split.

In particular, it is not true that $j^\ast i_\ast = 0$ in the derived category of sheaves on a space (where $i: U \hookrightarrow X$ is an open embedding and $j: F \hookrightarrow X$ the closed complement). In fact, the (hyper)cohomology of the complex $j^\ast i_\ast \mathbb Z_U$ computes the cohomology of the link of $F$ inside $X$ - an important invariant of the stratification. This is never zero, unless $U$ and $F$ are disjoint components.